Thermal Properties of Matter

Given,

• Triple point of neon, T = 24.57 K
• Triple point of carbon dioxide, T = 216.55 K

The Kelvin and the Celsius scales are related as

$\text t_\text C = \text T - 273.15$

and the Celsius and the Fahrenheit scales are related as

$\text t_\text F = \frac{9}{5}\text t_\text C + 32$

For neon :

$\text t_\text C = 24.57 - 273.15 = -248.58^\circ \text C$

$$\begin{aligned}\text t_\text F = \frac{9}{5} \times (-248.58) + 32 \\[1em] = -447.44 + 32 = -415.44^\circ \text F\end{aligned}$$

For carbon dioxide :

$\text t_\text C = 216.55 - 273.15 = -56.60^\circ \text C$

$$\begin{aligned}\text t_\text F = \frac{9}{5} \times (-56.60) + 32 \\[1em] = -101.88 + 32 = -69.88^\circ \text F\end{aligned}$$

Hence, the triple point of neon is − 248.58°C or − 415.44°F, and that of carbon dioxide is − 56.60°C or − 69.88°F.

Given,

• Triple point of water on scale A, T_A = 200 A
• Triple point of water on scale B, T_B = 350 B

Both A and B are absolute scales, so on each of them the zero of the scale is absolute zero. The triple point of water is a single physical state, and therefore the readings 200 A and 350 B represent the same temperature.

If the triple point of water is assigned the value T_tr on the Kelvin scale, then for any absolute scale the temperature of a body is proportional to its reading. Hence, for the triple point,

$\frac{\text T_\text A}{200} = \frac{\text T_\text B}{350}$

Therefore, for any temperature measured on the two scales,

$\text T_\text A = \frac{200}{350}\text T_\text B = \frac{4}{7}\text T_\text B$

Hence, the required relation is $\text T_\text A = \frac{4}{7}\text T_\text B$, that is, 7 T_A = 4 T_B.

Given,

• Resistance at the triple point of water, R₁ = 101.6 Ω at T₁ = 273.16 K
• Resistance at the melting point of lead, R₂ = 165.5 Ω at T₂ = 600.5 K
• Resistance at the unknown temperature, R = 123.4 Ω

The resistance varies linearly with temperature as

$\text R_\text T = \text R_0[1 + \alpha(\text T - \text T_0)]$

Since the relation is linear, the change in resistance is directly proportional to the change in temperature. Therefore,

$\frac{\text R - \text R_1}{\text R_2 - \text R_1} = \frac{\text T - \text T_1}{\text T_2 - \text T_1}$

Substituting the given values,

$\frac{123.4 - 101.6}{165.5 - 101.6} = \frac{\text T - 273.16}{600.5 - 273.16}$

$\frac{21.8}{63.9} = \frac{\text T - 273.16}{327.34}$

$\text T - 273.16 = 0.3412 \times 327.34 = 111.68$

$\text T = 273.16 + 111.68 = 384.84\ \text K$

Hence, the temperature at which the resistance is 123.4 Ω is 384.84 K.

(a) The triple point of water is that unique combination of temperature and pressure at which the solid, the liquid and the vapour phases of water co-exist in stable equilibrium. Since this state occurs at only one value of temperature and one value of pressure, it is perfectly reproducible everywhere.

The melting point of ice and the boiling point of water, on the other hand, depend upon the external pressure. They change if the pressure changes, and so they are not unique. That is why they are unsuitable as standard fixed points, while the triple point of water is taken as the standard fixed point in modern thermometry.

(b) The other fixed point on the Kelvin scale is absolute zero, which is assigned the value 0 K. It is the lowest possible temperature, at which molecular motion would stop completely.

(c) On the Celsius scale the ice point (the melting point of ice at standard atmospheric pressure) is taken as 0°C, and this corresponds to 273.15 K on the Kelvin scale. The triple point of water is not the ice point; it lies 0.01°C above it, that is, at 273.16 K.

Since the relation t_c = T − 273.15 converts a Kelvin reading into a Celsius reading, it must use the Kelvin value of the ice point. Hence 273.15 appears in the relation and not 273.16.

(d) On the Kelvin absolute scale, the triple point of water is 273.16 K and the interval between the ice point and the steam point contains 100 divisions. On the Fahrenheit scale the same interval contains 180 divisions. Hence, one Kelvin division corresponds to $\frac{180}{100} = \frac{9}{5}$ Fahrenheit divisions.

For an absolute scale whose unit interval size equals that of the Fahrenheit scale,

$\text T_\text F = \frac{9}{5} \times 273.16 = 491.69$

Hence, the triple point of water on this absolute scale is 491.69 units.

Given,

• For thermometer A : P_tr = 1.250 × 10⁵ Pa, P = 1.797 × 10⁵ Pa
• For thermometer B : P_tr = 0.200 × 10⁵ Pa, P = 0.287 × 10⁵ Pa
• Triple point of water, T_tr = 273.16 K

(a) In a constant-volume gas thermometer the volume is kept fixed, so the pressure of the gas is directly proportional to its absolute temperature, that is, P ∝ T. Hence

$\text T = \text T_{tr} \times \frac{\text P}{\text P_{tr}}$

For thermometer A :

$$\begin{aligned}\text T = 273.16 \times \frac{1.797 \times 10^5}{1.250 \times 10^5} \\[1em] = 273.16 \times 1.4376 = 392.69\ \text K\end{aligned}$$

For thermometer B :

$$\begin{aligned}\text T = 273.16 \times \frac{0.287 \times 10^5}{0.200 \times 10^5} \\[1em] = 273.16 \times 1.435 = 391.98\ \text K\end{aligned}$$

Hence, the melting point of sulphur is 392.69 K as read by thermometer A and 391.98 K as read by thermometer B.

(b) The two readings differ slightly because oxygen and hydrogen are not perfectly ideal gases. The relation P ∝ T holds exactly only for an ideal gas, and a real gas departs from ideal behaviour, the departure being different for different gases.

The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for different gases agree. Hence, to reduce the discrepancy the experiment should be repeated with smaller and smaller quantities of the gases, that is, at lower and lower pressures, and the readings should be extrapolated to zero pressure. At zero pressure both the gases behave as ideal gases and the two thermometers give the same reading.

Given,

• Reading of the tape at 45°C, l = 63.0 cm
• Temperature of calibration = 27°C
• Temperature on the hot day = 45°C, so Δt = 45 − 27 = 18°C
• Coefficient of linear expansion of steel, α = 1.2 × 10^-5 °C^-1

The tape is correctly calibrated at 27°C. On the hot day the tape itself expands, so each centimetre division of the tape becomes longer than a true centimetre. The true length of one division of the tape at 45°C is

$$\begin{aligned}1 \times (1 + \alpha\Delta \text t) = 1 + (1.2 \times 10^{-5} \times 18) \\[1em] = 1 + 2.16 \times 10^{-4}\ \text{cm}\end{aligned}$$

Since the tape shows a reading of 63.0 cm, the actual length of the rod at 45°C is

$$\begin{aligned}\text l_{45} = 63.0 \times (1 + 2.16 \times 10^{-4}) \\[1em] = 63.0 + 0.0136 = 63.0136\ \text{cm}\end{aligned}$$

The rod is also made of steel, so on cooling from 45°C to 27°C its length becomes

$$\begin{aligned}\text l_{27} = \frac{\text l_{45}}{1 + \alpha\Delta \text t} = \frac{63.0136}{1 + 2.16 \times 10^{-4}} \\[1em] = 63.0\ \text{cm}\end{aligned}$$

Hence, the actual length of the steel rod on the hot day is 63.0136 cm, and at 27°C it is 63.0 cm.

At 27°C the tape is correctly calibrated, so the reading of the tape is itself the true length of the rod.

Given,

• Diameter of the shaft at 27°C, d = 8.70 cm
• Diameter of the hole in the wheel, d' = 8.69 cm
• Coefficient of linear expansion of steel, α = 1.20 × 10^-5 K^-1
• Initial temperature, T₁ = 27°C

The wheel will just slip on the shaft when, on cooling, the diameter of the shaft becomes equal to the diameter of the hole in the wheel. Hence the required change in diameter is

$\Delta \text d = 8.69 - 8.70 = -0.01\ \text{cm}$

For linear expansion,

$\Delta \text d = \text d\alpha\Delta \text T$

$$\begin{aligned}\Delta \text T = \frac{\Delta \text d}{\text d\alpha} = \frac{-0.01}{8.70 \times (1.20 \times 10^{-5})} \\[1em] = \frac{-0.01}{1.044 \times 10^{-4}} = -95.8^\circ \text C\end{aligned}$$

The final temperature of the shaft is

$\text T_2 = \text T_1 + \Delta \text T = 27 - 95.8 = -68.8^\circ \text C$

Hence, the wheel will slip on the shaft when the shaft is cooled to about − 68.8°C.

Given,

• Diameter of the hole at 27°C, d = 4.24 cm
• Rise in temperature, Δt = 227 − 27 = 200°C
• Coefficient of linear expansion of copper, α = 1.70 × 10^-5 °C^-1

On heating, the hole in the sheet expands in exactly the same way as a disc of copper filling the hole would expand. Hence the change in the diameter of the hole is

$\Delta \text d = \text d\alpha\Delta \text t$

Substituting the given values,

$$\begin{aligned}\Delta \text d = 4.24 \times (1.70 \times 10^{-5}) \times 200 \\[1em] = 4.24 \times 3.4 \times 10^{-3} \\[1em] = 1.44 \times 10^{-2}\ \text{cm}\end{aligned}$$

Hence, the diameter of the hole increases by 1.44 × 10^-2 cm.

Given,

• Length of the brass wire, l = 1.8 m
• Fall in temperature, Δt = 27 − (− 39) = 66°C
• Diameter of the wire = 2.0 mm, so radius r = 1.0 mm = 1.0 × 10^-3 m
• Coefficient of linear expansion of brass, α = 2.0 × 10^-5 °C^-1
• Young's modulus, Y = 0.91 × 10¹¹ N m^-2

The area of cross-section of the wire is

$$\begin{aligned}\text A = \pi \text r^2 = 3.14 \times (1.0 \times 10^{-3})^2 \\[1em] = 3.14 \times 10^{-6}\ \text m^2\end{aligned}$$

On cooling, the wire tends to contract by Δl = l α Δt, but the rigid supports do not allow it to contract. The wire therefore behaves as though it has been stretched by this amount, and a tension is developed in it. From the definition of Young's modulus,

$\text Y = \frac{\text F/\text A}{\Delta \text l/\text l} = \frac{\text F\text l}{\text A\Delta \text l}$

Putting Δl = l α Δt,

$\text F = \text{YA}\alpha\Delta \text t$

Substituting the values,

$$\begin{aligned}\text F = (0.91 \times 10^{11}) \times (3.14 \times 10^{-6}) \times (2.0 \times 10^{-5}) \times 66 \\[1em] = 2.857 \times 10^5 \times 2.0 \times 10^{-5} \times 66 \\[1em] = 5.714 \times 66 = 377.1\ \text N\end{aligned}$$

Hence, the tension developed in the wire is about 3.8 × 10² N.

Given,

• Length of each rod, l = 50 cm = 0.50 m
• Rise in temperature, Δt = 250 − 40 = 210°C
• Coefficient of linear expansion of brass, α_b = 2.0 × 10^-5 K^-1
• Coefficient of linear expansion of steel, α_s = 1.2 × 10^-5 K^-1

Since the ends are free to expand, each rod expands independently.

The increase in the length of the brass rod is

$$\begin{aligned}\Delta \text l_\text b = \text l\alpha_\text b\Delta \text t = 0.50 \times (2.0 \times 10^{-5}) \times 210 \\[1em] = 2.1 \times 10^{-3}\ \text m\end{aligned}$$

The increase in the length of the steel rod is

$$\begin{aligned}\Delta \text l_\text s = \text l\alpha_\text s\Delta \text t = 0.50 \times (1.2 \times 10^{-5}) \times 210 \\[1em] = 1.26 \times 10^{-3}\ \text m\end{aligned}$$

The change in the length of the combined rod is

$$\begin{aligned}\Delta \text l = \Delta \text l_\text b + \Delta \text l_\text s \\[1em] = (2.1 \times 10^{-3}) + (1.26 \times 10^{-3}) \\[1em] = 3.36 \times 10^{-3}\ \text m = 0.336\ \text{cm}\end{aligned}$$

Hence, the length of the combined rod increases by 0.336 cm.

No thermal stress is developed at the junction. Thermal stress arises only when the expansion of a rod is prevented. Here the ends of the combined rod are free to expand, so each rod expands freely by its own amount and no restoring force is set up at the junction.

Given,

• Coefficient of volume expansion of glycerine, γ = 49 × 10^-5 °C^-1
• Rise in temperature, Δt = 30°C

When a liquid is heated its volume increases and hence its density decreases. If ρ and ρ′ are the densities at temperatures T and (T + ΔT) respectively, then

$\rho' = \rho(1 - \gamma\Delta \text t)$

Therefore the change in density is

$\rho - \rho' = \rho\gamma\Delta \text t$

and the fractional change in density is

$\frac{\rho - \rho'}{\rho} = \gamma\Delta \text t$

Substituting the values,

$$\begin{aligned}\frac{\Delta \rho}{\rho} = (49 \times 10^{-5}) \times 30 \\[1em] = 1.47 \times 10^{-2}\end{aligned}$$

Hence, the fractional change in the density of glycerine is 1.47 × 10^-2, the density being decreased.

Given,

• Power of the drilling machine, P = 10 kW = 10⁴ W
• Mass of the aluminium block, m = 8.0 kg
• Time, t = 2.5 min = 150 s
• Specific heat of aluminium, c = 0.91 J g^-1 °C^-1 = 910 J kg^-1 °C^-1
• Only 50% of the power is used in heating the block

The useful power supplied to the block is

$\text P' = \frac{50}{100} \times 10^4 = 5 \times 10^3\ \text W$

The heat given to the block in 150 s is

$$\begin{aligned}\text Q = \text P' \times \text t = (5 \times 10^3) \times 150 \\[1em] = 7.5 \times 10^5\ \text J\end{aligned}$$

Using Q = m c ΔT, the rise in temperature is

$$\begin{aligned}\Delta \text T = \frac{\text Q}{\text{mc}} = \frac{7.5 \times 10^5}{8.0 \times 910} \\[1em] = \frac{7.5 \times 10^5}{7280} = 103.0^\circ \text C\end{aligned}$$

Hence, the rise in the temperature of the aluminium block is about 103°C.

Given,

• Mass of the copper block, m = 2.5 kg
• Fall in temperature of the block, ΔT = 500 − 0 = 500°C
• Specific heat of copper, c = 0.39 J g^-1 °C^-1 = 390 J kg^-1 °C^-1
• Latent heat of fusion of ice, L = 335 J g^-1 = 3.35 × 10⁵ J kg^-1

The maximum amount of ice melts when the copper block cools right down to 0°C. The heat given out by the copper block is

$$\begin{aligned}\text Q = \text{mc}\Delta \text T = 2.5 \times 390 \times 500 \\[1em] = 4.875 \times 10^5\ \text J\end{aligned}$$

If m′ be the mass of ice melted, then by the principle of calorimetry the heat lost by the copper block is used up as the latent heat of fusion of the ice,

$\text Q = \text m'\text L$

$$\begin{aligned}\text m' = \frac{\text Q}{\text L} = \frac{4.875 \times 10^5}{3.35 \times 10^5} \\[1em] = 1.455\ \text{kg}\end{aligned}$$

Hence, the maximum amount of ice that can melt is about 1.5 kg.

Given,

• Mass of the metal block, m = 0.20 kg at 150°C
• Water equivalent of the calorimeter, W = 0.025 kg
• Volume of water = 150 cm³, density of water = 10³ kg m^-3, so mass of water m₁ = 0.15 kg at 27°C
• Final temperature of the mixture = 40°C
• Specific heat of water, c₁ = 4.2 × 10³ J kg^-1 °C^-1

Let c be the specific heat of the metal.

The heat lost by the metal block in cooling from 150°C to 40°C is

$\text Q_1 = \text{mc}(150 - 40) = 0.20 \times \text c \times 110 = 22\text c$

The heat gained by the water and the calorimeter in warming from 27°C to 40°C is

$$\begin{aligned}\text Q_2 = (\text m_1 + \text W)\text c_1(40 - 27) \\[1em] = (0.15 + 0.025) \times (4.2 \times 10^3) \times 13 \\[1em] = 0.175 \times 4200 \times 13 = 9555\ \text J\end{aligned}$$

By the principle of calorimetry, heat lost = heat gained,

$22\text c = 9555$

$\text c = \frac{9555}{22} = 434.3$ J kg^-1°C^-1

Hence, the specific heat of the metal is about 0.43 × 10³ J kg^-1 °C^-1.

If the heat losses to the surroundings are not negligible, a part of the heat given out by the metal block escapes to the surroundings and does not reach the water and the calorimeter. The heat gained by the water and the calorimeter is then less than the heat actually lost by the block, and so the calculated value of c comes out to be less than the true value. Hence, the value obtained above is smaller than the actual specific heat of the metal.

All the gases listed in the table are diatomic gases, whereas the value 2.92 cal mol^-1 K^-1 quoted for comparison belongs to a monatomic gas.

A monatomic gas molecule consists of a single atom, so it can possess only translational kinetic energy. The whole of the heat given to it at constant volume is used in increasing this translational energy, and hence its molar specific heat is small.

A diatomic gas molecule, on the other hand, possesses rotational energy in addition to translational energy. When heat is given to a diatomic gas at constant volume, a part of it is used in increasing the rotational energy of the molecules as well. Therefore more heat is required to raise the temperature of one mole of a diatomic gas through 1 K, and its molar specific heat is nearly 5 cal mol^-1 K^-1, which is markedly larger than that of a monatomic gas.

The value for chlorine (6.17 cal mol^-1 K^-1) is larger than that of the remaining diatomic gases. This indicates that in a chlorine molecule, besides the translational and the rotational motions, the vibrational motion of the two atoms about their mean positions is also appreciably excited at room temperature. The additional energy stored in this vibrational mode raises the molar specific heat of chlorine above the value for the other diatomic gases.

Given,

• Mass of the child, m = 30 kg
• Fall in temperature = 101°F − 98°F = 3°F
• Time, t = 20 minutes
• Specific heat of the human body, c = 4.18 × 10³ J kg^-1 °C^-1
• Latent heat of evaporation of water, L = 580 cal g^-1

A temperature difference of 1°F is equal to $\frac{5}{9}$°C. Hence the fall in temperature on the Celsius scale is

$\Delta \text T = 3 \times \frac{5}{9} = 1.667^\circ \text C$

The heat lost by the body of the child is

$$\begin{aligned}\text Q = \text{mc}\Delta \text T = 30 \times (4.18 \times 10^3) \times 1.667 \\[1em] = 2.09 \times 10^5\ \text J\end{aligned}$$

Converting the latent heat of evaporation into joule per kilogram, using 1 cal = 4.2 J,

$$\begin{aligned}\text L = 580 \times 4.2 = 2436\ \text{J g}^{-1} \\[1em] = 2.436 \times 10^6\ \text{J kg}^{-1}\end{aligned}$$

If m′ be the mass of sweat evaporated, then Q = m′L, so

$$\begin{aligned}\text m' = \frac{\text Q}{\text L} = \frac{2.09 \times 10^5}{2.436 \times 10^6} \\[1em] = 0.0858\ \text{kg} = 85.8\ \text g\end{aligned}$$

The average rate of extra evaporation is

$\frac{\text m'}{\text t} = \frac{85.8}{20} = 4.29\ \text{g min}^{-1}$

Hence, the average rate of extra evaporation caused by the drug is about 4.3 g min^-1.

Given,

• Side of the cubical icebox, a = 30 cm = 0.30 m
• Thickness of the wall, l = 5.0 cm = 0.05 m
• Mass of ice put in the box, m = 4.0 kg
• Time, t = 6 h = 6 × 3600 = 21600 s
• Temperature difference, θ₁ − θ₂ = 45 − 0 = 45°C
• Coefficient of thermal conductivity of thermacole, K = 0.01 J s^-1 m^-1 K^-1
• Latent heat of fusion of ice, L = 335 × 10³ J kg^-1

The box has six faces, so the total surface area through which heat enters is

$\text A = 6\text a^2 = 6 \times (0.30)^2 = 0.54\ \text m^2$

The quantity of heat entering the box in time t is

$\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

Substituting the values,

$$\begin{aligned}\text Q = \frac{0.01 \times 0.54 \times 45 \times 21600}{0.05} \\[1em] = \frac{5248.8}{0.05} = 1.05 \times 10^5\ \text J\end{aligned}$$

The mass of ice melted by this heat is

$\text m' = \frac{\text Q}{\text L} = \frac{1.05 \times 10^5}{335 \times 10^3} = 0.313\ \text{kg}$

Therefore the mass of ice remaining after 6 hours is

$4.0 - 0.313 = 3.687\ \text{kg}$

Hence, about 3.7 kg of ice remains in the box after 6 hours.

Given,

• Base area of the boiler, A = 0.15 m²
• Thickness of the base, l = 1.0 cm = 1.0 × 10^-2 m
• Rate of boiling of water = 6.0 kg min^-1
• Thermal conductivity of brass, K = 109 J s^-1 m^-1 °C^-1
• Latent heat of vaporisation of water, L = 2256 × 10³ J kg^-1
• Temperature of the water = 100°C

The heat required per second to convert the water into steam is

$$\begin{aligned}\frac{\text Q}{\text t} = \frac{6.0 \times (2256 \times 10^3)}{60} \\[1em] = 2.256 \times 10^5\ \text{J s}^{-1}\end{aligned}$$

Let θ be the temperature of the flame in contact with the boiler. In the steady state, this heat is conducted through the base of the boiler,

$\frac{\text Q}{\text t} = \frac{\text{KA}(\theta - 100)}{\text l}$

Substituting the values,

$2.256 \times 10^5 = \frac{109 \times 0.15 \times (\theta - 100)}{1.0 \times 10^{-2}}$

$2.256 \times 10^5 = 1635 \times (\theta - 100)$

$\theta - 100 = \frac{2.256 \times 10^5}{1635} = 138$

$\theta = 100 + 138 = 238^\circ \text C$

Hence, the temperature of the part of the flame in contact with the boiler is about 238°C.

(a) A body with large reflectivity reflects back most of the radiant energy falling on it, so only a small part of the incident energy is absorbed by it, that is, its absorptive power is small. According to Kirchhoff's law, at a given temperature the ratio of the emissive power to the absorptive power is the same for all bodies. Hence a body of small absorptive power must also have a small emissive power, that is, a body with large reflectivity is a poor emitter.

(b) Brass is a good conductor of heat while wood is a bad conductor. On a chilly day both the brass tumbler and the wooden tray are at the same low temperature. When we touch the brass tumbler, it conducts heat away from our hand very rapidly, and so it feels much colder. The wooden tray conducts heat away only very slowly, and so it does not feel so cold.

(c) An optical pyrometer is calibrated for the radiation of an ideal black body, for which the radiant energy emitted per second per unit area is σT⁴. A red-hot iron piece in the open is not a perfectly black body, so at the same temperature it emits less energy than a black body, that is, e σT⁴ where e is less than 1. The pyrometer therefore records too low a value of the temperature.

When the same piece is placed inside a furnace, it is surrounded by the walls of the furnace which are at the same temperature. The radiation coming out of the furnace is the radiation of a uniformly heated enclosure, which is very nearly black body radiation. Hence the pyrometer then gives the correct value of the temperature.

(d) The atmosphere is largely transparent to the incoming short-wavelength solar radiation, but it is partly opaque to the long-wavelength infra-red radiation emitted by the warm earth. The gases of the atmosphere absorb this infra-red radiation and re-emit a part of it back towards the surface, which keeps the surface warm. This is the greenhouse effect. In the absence of the atmosphere the whole of the infra-red radiation emitted by the earth would escape into space, and the earth would become inhospitably cold.

(e) Steam at 100°C carries with it the latent heat of vaporisation, which is 22.6 × 10⁵ J kg^-1, over and above the heat carried by water at 100°C. When steam condenses in the radiators of the heating system, it gives out this large amount of latent heat in addition to the heat given out on cooling. Hence, for the same mass of the circulating fluid, a steam heating system supplies much more heat than a hot water system, and is therefore more efficient in warming a building.

Given,

• First case : the body cools from θ₁ = 80°C to θ₂ = 50°C in t = 5 minutes
• Second case : the body cools from θ₁ = 60°C to θ₂ = 30°C in time t′
• Temperature of the surroundings, θ₀ = 20°C

By Newton's law of cooling, the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings. Hence

$\frac{\theta_1 - \theta_2}{\text t} = \text K\left[\frac{\theta_1 + \theta_2}{2} - \theta_0\right]$

First case :

$$\begin{aligned}\frac{80 - 50}{5} = \text K\left[\frac{80 + 50}{2} - 20\right] \\[1em] 6 = \text K(65 - 20) = 45\text K\end{aligned}$$

$\text K = \frac{6}{45} = \frac{2}{15}$

Second case :

$$\begin{aligned}\frac{60 - 30}{\text t'} = \text K\left[\frac{60 + 30}{2} - 20\right] \\[1em] \frac{30}{\text t'} = \text K(45 - 20) = 25\text K\end{aligned}$$

Substituting the value of K,

$\frac{30}{\text t'} = 25 \times \frac{2}{15} = \frac{10}{3}$

$\text t' = \frac{30 \times 3}{10} = 9\ \text{minutes}$

Hence, the body takes 9 minutes to cool from 60°C to 30°C.

Given,

• Specific heat of ice, c = 2100 J kg^-1 °C^-1 = 2.1 J g^-1 °C^-1
• Latent heat of fusion of ice, L = 3.36 × 10⁵ J kg^-1 = 336 J g^-1
• Mass of ice = m gram at − 5°C
• Heat supplied, Q = 420 J
• Mass of ice melted = 1 g

The heat supplied is used up in two stages. First the whole of the ice is warmed from − 5°C to its melting point 0°C, and then 1 g of this ice is melted at 0°C.

Heat used in warming the ice from − 5°C to 0°C is

$\text Q_1 = \text{mc}\Delta \text T = \text m \times 2.1 \times 5 = 10.5\text m$

Heat used in melting 1 g of ice at 0°C is

$\text Q_2 = 1 \times \text L = 336\ \text J$

Since the total heat supplied is 420 J,

$$\begin{aligned}\text Q_1 + \text Q_2 = 420 \\[1em] 10.5\text m + 336 = 420\end{aligned}$$

$10.5\text m = 84$

$\text m = \frac{84}{10.5} = 8\ \text g$

Hence, the value of m is 8 grams.

α₁l₁ = α₂l₂

Reason — Given,

• Coefficient of linear expansion of brass = α₁, length of brass rod = l₁
• Coefficient of linear expansion of steel = α₂, length of steel rod = l₂

For a rise in temperature Δt, the increase in the lengths of the two rods are

$\Delta \text l_1 = \alpha_1 \text l_1 \Delta \text t \quad \text{and} \quad \Delta \text l_2 = \alpha_2 \text l_2 \Delta \text t$

The difference (l₂ − l₁) will remain the same at all temperatures only if both the rods increase in length by exactly the same amount, that is,

$\Delta \text l_1 = \Delta \text l_2$

$\alpha_1 \text l_1 \Delta \text t = \alpha_2 \text l_2 \Delta \text t$

$\alpha_1 \text l_1 = \alpha_2 \text l_2$

25°C; α = 1.85 × 10^-5/°C

Reason — Given,

• The clock loses 12 s per day at 40°C
• The clock gains 4 s per day at 20°C
• Number of seconds in a day = 86400 s

The time period of a pendulum is $\text T = 2\pi\sqrt{\frac{\text l}{\text g}}$, so T ∝ $\sqrt{\text l}$. If the length changes by Δl = l α Δt, the fractional change in the time period is

$\frac{\Delta \text T}{\text T} = \frac{1}{2}\frac{\Delta \text l}{\text l} = \frac{1}{2}\alpha\Delta \text t$

Hence the time lost or gained in one day is

$\Delta \text t' = \frac{1}{2}\alpha\Delta \text t \times 86400$

Let θ be the temperature at which the clock shows correct time.

At 40°C the clock loses 12 s :

$\frac{1}{2}\alpha(40 - \theta) \times 86400 = 12 \qquad \ldots(\text i)$

At 20°C the clock gains 4 s :

$\frac{1}{2}\alpha(\theta - 20) \times 86400 = 4 \qquad \ldots(\text{ii})$

Dividing equation (i) by equation (ii),

$\frac{40 - \theta}{\theta - 20} = \frac{12}{4} = 3$

$40 - \theta = 3\theta - 60 \quad \Rightarrow \quad 4\theta = 100$

$\theta = 25^\circ \text C$

Substituting θ = 25°C in equation (ii),

$$\begin{aligned}\frac{1}{2}\alpha(25 - 20) \times 86400 = 4 \\[1em] \alpha \times 5 \times 43200 = 4\end{aligned}$$

$\alpha = \frac{4}{216000} = 1.85$ × 10^-5°C^-1

8 min 20 s

Reason — Given,

• Volume of water = 2 litre, density = 1 kg litre^-1, so mass m = 2 kg
• Power of the heater, P = 1 kW = 1000 J s^-1
• Rate of heat loss to the surroundings = 160 J s^-1
• Specific heat of water, c = 4.2 kJ kg^-1 K^-1 = 4200 J kg^-1 K^-1
• Rise in temperature, ΔT = 77 − 27 = 50°C

The net rate at which heat is supplied to the water is

$\text P_{net} = 1000 - 160 = 840\ \text{J s}^{-1}$

The heat required to raise the temperature of the water is

$$\begin{aligned}\text Q = \text{mc}\Delta \text T = 2 \times 4200 \times 50 \\[1em] = 4.2 \times 10^5\ \text J\end{aligned}$$

Therefore the time taken is

$\text t = \frac{\text Q}{\text P_{net}} = \frac{4.2 \times 10^5}{840} = 500\ \text s$

$= 8\ \text{min}\ 20\ \text s$

2067

Reason — Given,

• Mass of stored water, m = 120 litre = 120 kg
• Rate of heat generated by the device = 3 kW = 3000 J s^-1
• Initial temperature of water = 10°C, maximum permissible temperature = 30°C
• Time of operation, t = 3 h = 10800 s
• Specific heat of water, c = 4.2 kJ kg^-1 K^-1 = 4200 J kg^-1 K^-1

The heat generated by the device in 3 hours is

$\text Q_{gen} = 3000 \times 10800 = 3.24 \times 10^7\ \text J$

Out of this, a part is removed by the cooler and the rest is absorbed by the stored water, whose temperature is allowed to rise from 10°C to 30°C, that is, through 20°C. The heat absorbed by the stored water is

$$\begin{aligned}\text Q_{water} = \text{mc}\Delta \text T = 120 \times 4200 \times 20 \\[1em] = 1.008 \times 10^7\ \text J\end{aligned}$$

The heat removed by the cooler in 3 hours is P × 10800. Since the system is thermally insulated,

$\text P \times 10800 + 1.008 \times 10^7 = 3.24 \times 10^7$

$\text P \times 10800 = 2.232 \times 10^7$

$\text P = \frac{2.232 \times 10^7}{10800} = 2066.7 \approx 2067\ \text W$

Given,

• Length of the rod AB = 10x, with A at 0°C (ice) and B at 100°C (water)
• Temperature of the point P = 400°C, at a distance λx from A
• Latent heat of evaporation of water, L_v = 540 cal g^-1
• Latent heat of melting of ice, L_f = 80 cal g^-1

Since P is at the highest temperature, heat flows from P towards A as well as from P towards B.

The distance PA = λx and the distance PB = (10 − λ)x. If K be the thermal conductivity and A the area of cross-section of the rod, then in the steady state

Rate of heat flow from P to A,

$\text H_1 = \frac{\text{KA}(400 - 0)}{\lambda \text x} = \frac{400\text{KA}}{\lambda \text x}$

Rate of heat flow from P to B,

$\text H_2 = \frac{\text{KA}(400 - 100)}{(10 - \lambda)\text x} = \frac{300\text{KA}}{(10 - \lambda)\text x}$

Let m gram of ice melt and m gram of water evaporate per unit time. Then

$\text H_1 = \text m \times 80 \quad \text{and} \quad \text H_2 = \text m \times 540$

Therefore,

$\frac{\text H_2}{\text H_1} = \frac{540}{80} = \frac{27}{4}$

Substituting the expressions for H₁ and H₂,

$\frac{300\text{KA}}{(10 - \lambda)\text x} \times \frac{\lambda \text x}{400\text{KA}} = \frac{27}{4}$

$\frac{3\lambda}{4(10 - \lambda)} = \frac{27}{4}$

$$\begin{aligned}3\lambda = 27(10 - \lambda) \\[1em] 3\lambda = 270 - 27\lambda\end{aligned}$$

$30\lambda = 270 \quad \Rightarrow \quad \lambda = 9$

Hence, the value of λ is 9.

Given,

• Radius of A, r_A = 6 cm; radius of B, r_B = 18 cm
• Wavelength of maximum intensity for A, λ_A = 500 nm
• Wavelength of maximum intensity for B, λ_B = 1500 nm

By Wien's displacement law, λ_mT = b (a constant), so

$\frac{\text T_1}{\text T_2} = \frac{\lambda_\text B}{\lambda_\text A} = \frac{1500}{500} = 3$

By Stefan's law, the total energy radiated per second by a black body of surface area A at absolute temperature T is

$\text P = \sigma \text{A}\text T^4 = \sigma(4\pi \text r^2)\text T^4$

Therefore,

$\frac{\text P_\text A}{\text P_\text B} = \left(\frac{\text r_\text A}{\text r_\text B}\right)^2 \left(\frac{\text T_1}{\text T_2}\right)^4$

Substituting the values,

$$\begin{aligned}\frac{\text P_\text A}{\text P_\text B} = \left(\frac{6}{18}\right)^2 \times (3)^4 \\[1em] = \frac{1}{9} \times 81 = 9\end{aligned}$$

Hence, the required ratio is 9.

Given,

• Radius of A, r_A = 400 r_B
• Power emitted by A, P_A = 10⁴ P_B

By Stefan's law, the power radiated by a spherical black body is

$\text P = \sigma(4\pi \text r^2)\text T^4$

Therefore,

$\frac{\text P_\text A}{\text P_\text B} = \left(\frac{\text r_\text A}{\text r_\text B}\right)^2 \left(\frac{\text T_\text A}{\text T_\text B}\right)^4$

Substituting the given values,

$10^4 = (400)^2 \left(\frac{\text T_\text A}{\text T_\text B}\right)^4$

$\left(\frac{\text T_\text A}{\text T_\text B}\right)^4 = \frac{10^4}{1.6 \times 10^5} = \frac{1}{16}$

$\frac{\text T_\text A}{\text T_\text B} = \frac{1}{2}$

By Wien's displacement law, λ_mT = constant, so λ_m is inversely proportional to T,

$\frac{\lambda_\text A}{\lambda_\text B} = \frac{\text T_\text B}{\text T_\text A} = 2$

Hence, the ratio λ_A/λ_B is 2.

Given,

• Initial temperature, T₁ = 487°C = 487 + 273 = 760 K
• Final temperature, T₂ = 2767°C = 2767 + 273 = 3040 K
• Sensor reading at T₁ is log₂(P₁/P₀) = 1

From the given reading,

$\log_2\left(\frac{\text P_1}{\text P_0}\right) = 1 \quad \Rightarrow \quad \frac{\text P_1}{\text P_0} = 2$

By Stefan's law, the power radiated is proportional to the fourth power of the absolute temperature, so

$\frac{\text P_2}{\text P_1} = \left(\frac{\text T_2}{\text T_1}\right)^4 = \left(\frac{3040}{760}\right)^4 = (4)^4 = 256$

Therefore,

$\frac{\text P_2}{\text P_0} = \frac{\text P_2}{\text P_1} \times \frac{\text P_1}{\text P_0} = 256 \times 2 = 512$

The value displayed by the sensor is

$\log_2\left(\frac{\text P_2}{\text P_0}\right) = \log_2 512 = \log_2 2^9 = 9$

Hence, the sensor displays the value 9.

The graph in which the temperature θ falls linearly with the distance x from the hot end.

Reason — In the steady state the temperature of each cross-section of the bar becomes constant with time, so no heat is absorbed by any section. The heat that reaches any cross-section is entirely passed on to the next, and hence the rate of flow of heat is the same all along the bar.

By the law of heat conduction,

$\frac{\text{dQ}}{\text{dt}} = -\text{KA}\frac{\text d\theta}{\text{dx}}$

In the steady state $\frac{\text{dQ}}{\text{dt}}$ is constant, say r. Therefore

$\frac{\text d\theta}{\text{dx}} = -\frac{\text r}{\text{KA}}$

Since K and A are constant for a uniform bar, the temperature gradient is the same at every point. On integrating,

$\theta = -\frac{\text r}{\text{KA}}\text x + \text c$

Comparing this with y = − kx + c, the graph between θ and x is a straight line with a negative slope. Hence the temperature decreases uniformly with distance from the hot end.

4.8 cal/s

Reason — Given,

• Area of cross-section of each rod, A = 4 cm²
• Lengths : l_Cu = 46 cm, l_Br = 13 cm, l_St = 12 cm
• Thermal conductivities : K_Cu = 0.92, K_Br = 0.26, K_St = 0.12 cal s^-1 cm^-1 °C^-1
• Free end of copper at 100°C, free ends of brass and steel at 0°C

Let θ be the steady temperature of the junction. The copper rod carries heat to the junction, and this heat is then shared by the brass and the steel rods. Hence

$\text H_{Cu} = \text H_{Br} + \text H_{St}$

$\frac{\text K_{Cu}\text A(100 - \theta)}{\text l_{Cu}} = \frac{\text K_{Br}\text A\theta}{\text l_{Br}} + \frac{\text K_{St}\text A\theta}{\text l_{St}}$

Cancelling A and substituting the values,

$\frac{0.92(100 - \theta)}{46} = \frac{0.26\theta}{13} + \frac{0.12\theta}{12}$

$0.02(100 - \theta) = 0.02\theta + 0.01\theta$

$2 - 0.02\theta = 0.03\theta \quad \Rightarrow \quad 0.05\theta = 2$

$\theta = 40^\circ \text C$

The rate of heat flow through the copper rod is

$$\begin{aligned}\text H_{Cu} = \frac{0.92 \times 4 \times (100 - 40)}{46} \\[1em] = \frac{0.92 \times 4 \times 60}{46} = 4.8\ \text{cal s}^{-1}\end{aligned}$$

0.78 mm

Reason — Given,

• Length of each wire, l = 1 m at 10°C
• End P at 10°C, end S at 400°C
• Thermal conductivity of PQ = 2 × thermal conductivity of RS
• Coefficient of linear expansion of PQ, α = 1.2 × 10^-5 K^-1

Let the thermal conductivity of RS be K, so that the thermal conductivity of PQ is 2K. Let θ be the steady temperature of the junction QR.

The two wires are joined in series, so the rate of flow of heat through both is the same,

$\frac{2\text{KA}(\theta - 10)}{1} = \frac{\text{KA}(400 - \theta)}{1}$

$2\theta - 20 = 400 - \theta \quad \Rightarrow \quad 3\theta = 420$

$\theta = 140^\circ \text C$

Along the wire PQ the temperature increases uniformly from 10°C at P to 140°C at Q. Hence the average temperature of the wire PQ is

$\theta_{av} = \frac{10 + 140}{2} = 75^\circ \text C$

The whole wire was initially at 10°C, so the effective rise in temperature is

$\Delta \text T = 75 - 10 = 65^\circ \text C$

The change in the length of PQ is

$$\begin{aligned}\Delta \text l = \text l\alpha\Delta \text T = 1 \times (1.2 \times 10^{-5}) \times 65 \\[1em] = 7.8 \times 10^{-4}\ \text m = 0.78\ \text{mm}\end{aligned}$$

2.0 s

Reason — Given,

• Thermal conductivities of the two blocks are K and 2K
• The blocks have identical dimensions, say length l and area of cross-section A
• Same temperature difference Δθ in both configurations
• Time taken in configuration (i), t₁ = 9 s

In configuration (i) the two blocks are joined in series along the X-axis. The equivalent thermal resistance is

$\text R_1 = \frac{\text l}{\text{KA}} + \frac{\text l}{2\text{KA}} = \frac{3\text l}{2\text{KA}}$

In configuration (ii) the two blocks are joined in parallel. The rate of flow of heat is the sum of the rates through the two blocks,

$\text H_2 = \frac{\text{KA}\Delta \theta}{\text l} + \frac{2\text{KA}\Delta \theta}{\text l} = \frac{3\text{KA}\Delta \theta}{\text l}$

The rate of flow of heat in configuration (i) is

$\text H_1 = \frac{\Delta \theta}{\text R_1} = \frac{2\text{KA}\Delta \theta}{3\text l}$

Therefore,

$\frac{\text H_2}{\text H_1} = \frac{3\text{KA}\Delta \theta}{\text l} \times \frac{3\text l}{2\text{KA}\Delta \theta} = \frac{9}{2} = 4.5$

For the same quantity of heat Q, the time taken is inversely proportional to the rate of flow of heat, so

$\text t_2 = \frac{\text t_1}{4.5} = \frac{9}{4.5} = 2.0\ \text s$

The graph which is a straight line with a negative slope, having a positive intercept on the log_e(θ − θ₀) axis.

Reason — By Newton's law of cooling, the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings,

$-\frac{\text d\theta}{\text{dt}} = \text K(\theta - \theta_0)$

Separating the variables,

$\frac{\text d\theta}{\theta - \theta_0} = -\text K\text{dt}$

On integrating,

$\log_e(\theta - \theta_0) = -\text{Kt} + \text C$

where C is the constant of integration. This is of the form y = mx + c, with slope − K and intercept C. Hence the graph between log_e(θ − θ₀) and t is a straight line having a negative gradient.

The graph in which T falls exponentially from θ_i and approaches the value θ₀ asymptotically, without cutting the time axis.

Reason — Given,

• Initial temperature of the metal piece = θ_i
• Temperature of the surroundings (room) = θ₀
• Temperature of the metal piece at any instant t = T
• Mass of the metal piece = m and specific heat capacity of its material = c

By Newton's law of cooling, the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings,

$-\frac{\text{dQ}}{\text{dt}} = \text k(\text T - \theta_0) \qquad \ldots(\text i)$

where k is a positive constant depending upon the area and the nature of the surface of the body.

If in the small time interval dt the temperature of the body changes by dT, then the heat lost is

$\text{dQ} = \text m\text c\text{dT} \qquad \ldots(\text{ii})$

From equations (i) and (ii),

$-\frac{\text m\text c\text{dT}}{\text{dt}} = \text k(\text T - \theta_0)$

Separating the variables,

$\frac{\text{dT}}{\text T - \theta_0} = -\frac{\text k}{\text{mc}}\text{dt}$

The body cools from its initial temperature θ_i at t = 0 to the temperature T at time t. Integrating between these limits,

$\int_{\theta_i}^{\text T} \frac{\text{dT}}{\text T - \theta_0} = -\frac{\text k}{\text{mc}}\int_{0}^{\text t} \text{dt}$

Since θ₀ is a constant, the integral on the left is a standard logarithmic integral,

$\left[\log_e(\text T - \theta_0)\right]_{\theta_i}^{\text T} = -\frac{\text k}{\text{mc}}\left[\text t\right]_{0}^{\text t}$

Evaluating the limits on both sides,

$\log_e(\text T - \theta_0) - \log_e(\theta_i - \theta_0) = -\frac{\text k}{\text{mc}}(\text t - 0)$

Combining the two logarithms,

$\log_e\left(\frac{\text T - \theta_0}{\theta_i - \theta_0}\right) = -\frac{\text k}{\text{mc}}\text t$

Taking antilogarithms,

$\frac{\text T - \theta_0}{\theta_i - \theta_0} = \text e^{-\frac{\text k}{\text{mc}}\text t}$

$\text T - \theta_0 = (\theta_i - \theta_0)\text e^{-\frac{\text k}{\text{mc}}\text t}$

$\text T = \theta_0 + (\theta_i - \theta_0)\text e^{-\frac{\text k}{\text{mc}}\text t} \qquad \ldots(\text{iii})$

Equation (iii) shows that the temperature of the body decreases exponentially with time. The graph of T versus t is obtained by adding the graph of the constant line T = θ₀ and the decaying exponential $\text T = (\theta_i - \theta_0)\text e^{-\frac{\text k}{\text{mc}}\text t}$.

At t = 0 : Since e⁰ = 1,

$\text T = \theta_0 + (\theta_i - \theta_0) = \theta_i$

so the curve starts from the initial temperature θ_i on the temperature axis.

As t → ∞ : Since $\text e^{-\frac{\text k}{\text{mc}}\text t} \rightarrow 0$,

$\text T \rightarrow \theta_0$

so the curve approaches the line T = θ₀.

Slope of the curve : Differentiating equation (iii) with respect to t,

$\frac{\text{dT}}{\text{dt}} = -\frac{\text k}{\text{mc}}(\theta_i - \theta_0)\text e^{-\frac{\text k}{\text{mc}}\text t}$

At t = 0 the slope is $-\frac{\text k}{\text{mc}}(\theta_i - \theta_0)$, which is the largest in magnitude, so the curve falls steeply in the beginning. As t increases, the exponential factor decreases and the slope becomes smaller and smaller, so the curve flattens out.

The body can never become colder than its surroundings, so the curve approaches the line T = θ₀ asymptotically and does not meet the time axis.

$\left(\frac{97}{2}\right)^{\frac{1}{4}}\text T$

Reason — Given,

• Temperature of the first plate = 2T
• Temperature of the third plate = 3T
• Let the steady temperature of the middle (second) plate be T′
• All three plates are ideal black surfaces of the same area A

Since the third plate is the hottest and the first plate the coldest, heat flows from the third plate to the middle plate, and from the middle plate to the first plate.

By Stefan's law, the energy radiated per second by a black body of surface area A at absolute temperature T is

$\frac{\text Q}{\text t} = \sigma \text A\text T^4$

Heat gained by the middle plate from the third plate : The third plate at 3T radiates towards the middle plate, and the middle plate at T′ radiates back towards it. Hence the net rate of heat gained by the middle plate on this side is

$\text H_1 = \sigma \text A(3\text T)^4 - \sigma \text A(\text T')^4$

Heat lost by the middle plate to the first plate : Similarly, the middle plate at T′ radiates towards the first plate, and the first plate at 2T radiates back. Hence the net rate of heat lost by the middle plate on this side is

$\text H_2 = \sigma \text A(\text T')^4 - \sigma \text A(2\text T)^4$

Steady state condition : In the steady state the temperature of the middle plate remains constant, so the heat it gains per second must be exactly equal to the heat it loses per second,

$\text H_1 = \text H_2$

$\sigma \text A(3\text T)^4 - \sigma \text A(\text T')^4 = \sigma \text A(\text T')^4 - \sigma \text A(2\text T)^4$

Cancelling σA throughout,

$(3\text T)^4 - (\text T')^4 = (\text T')^4 - (2\text T)^4$

Taking the terms containing T′ to one side,

$(3\text T)^4 + (2\text T)^4 = 2(\text T')^4$

Expanding the fourth powers,

$81\text T^4 + 16\text T^4 = 2(\text T')^4$

$97\text T^4 = 2(\text T')^4$

$(\text T')^4 = \frac{97}{2}\text T^4$

Taking the fourth root of both sides,

$\text T' = \left(\frac{97}{2}\right)^{\frac{1}{4}}\text T$

The middle plate radiates from both of its faces, which is why the factor 2 appears on the right-hand side. Numerically $\left(\frac{97}{2}\right)^{\frac{1}{4}} = 2.64$, so T′ = 2.64 T, which lies between 2T and 3T as expected.

4°C

Reason — Water exhibits an anomalous property; it contracts on heating between 0°C and 4°C and expands on further heating. Hence, as water is heated from 0°C its volume first decreases, becomes minimum at 4°C, and then increases. Since the mass remains the same, the density of water is maximum at 4°C, being equal to 1 g cm^-3.

Calorimeter

Reason — A device in which heat measurement can be made is called a calorimeter. It consists of a cylindrical copper vessel provided with a copper stirrer and a lid, kept inside a wooden jacket filled with heat-insulating material. By applying the principle of calorimetry to the mixture prepared in it, the specific heat of a given solid or liquid is determined. A thermometer measures temperature, a barometer measures atmospheric pressure and a pyrometer measures very high temperatures.

radiation

Reason — Radiation is the process by which heat is transferred directly from one body to another without affecting the intervening medium (if any). The heat travels in the form of electromagnetic waves with the speed of light, so no medium is needed at all, and if a medium is present it does not itself become hot.

In convection the material particles of the fluid themselves move from one place to another, so the medium clearly moves. Evaporation is a change of state and not a mode of heat transfer.

Kelvin

Reason — The Kelvin scale is the absolute temperature scale commonly used in science. Its zero is taken at absolute zero (0 K = − 273.15°C), the lowest possible temperature at which molecular motion would theoretically stop. On this scale temperature is always positive and is proportional to the average kinetic energy of the molecules. The zero points of the Celsius and Fahrenheit scales are arbitrary.

W/m K

Reason — From the relation $\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$, the coefficient of thermal conductivity is

$\text K = \frac{\text Q\text l}{\text A(\theta_1 - \theta_2)\text t}$

Its SI unit is therefore J s^-1 m^-1 K^-1, that is, W m^-1 K^-1. The unit J/kg K belongs to specific heat capacity and N/m² to pressure or stress.

Water

Reason — Water has a very high specific heat capacity, 1 kcal kg^-1 K^-1 or 4.18 × 10³ J kg^-1 K^-1. It is nearly ten times the specific heat of copper and five times that of sand or earth. It is because of this high specific heat that water is used as an effective coolant and in hot water bottles for fomentation.

Water at 4°C

Reason — When water is heated from 0°C its density first increases up to 4°C and then decreases above 4°C. Thus the density of water is maximum at 4°C and is equal to 1 g cm^-3. When water at 0°C freezes into ice, the volume of the ice is greater than that of the water, so the density of ice is less than the density of water and ice floats on water.

Solids

Reason — In a solid the molecules are held together by strong intermolecular forces and are able only to vibrate about their fixed mean positions. Hence, for the same rise in temperature, a solid expands the least. Liquids expand more than solids, and gases at ordinary temperature expand much more than both solids and liquids, the coefficient of volume expansion of an ideal gas being $\gamma = \frac{1}{\text T}$, that is, 3.3 × 10^-3 K^-1 at room temperature.

Convection

Reason — When the temperature of the lower region of a fluid becomes higher than that of the upper region, the density of the fluid in the lower region becomes less than that in the upper region. Hence the warmer and lighter particles rise up and their places are taken by the colder and heavier particles from above. This process of heat-transmission in which the material particles of the fluid themselves move is called convection.

Both 1 and 3

Reason — The specific heat capacity of a substance is the amount of heat required to raise the temperature of the unit mass of the substance through a unit degree (1°C or 1 K). Since a temperature difference of 1°C is exactly equal to a temperature difference of 1 K, both the statements 1 and 3 describe the same quantity, whose SI unit is J kg^-1 K^-1.

In a vacuum

Reason — In radiation, heat energy is transferred in the form of electromagnetic waves which travel with the speed of light. No medium is required for the propagation of these waves, and if a medium is present it does not become hot. This is how heat from the sun reaches the earth across the empty space.

Kelvin

Reason — The SI temperature unit is the kelvin, abbreviated K, which is written without a degree sign. It is defined by assigning the value 273.16 K to the triple point of water. The degree Celsius and the degree Fahrenheit are units of other thermometric scales, and the joule is the unit of heat and energy.

It requires a medium

Reason — In conduction the particles of a body pass on heat to the neighbouring particles by mutual contact without leaving their own positions. A material medium is therefore essential for conduction. It takes place chiefly in solids and in mercury, and it cannot occur in a vacuum. Transfer of heat by the bulk movement of particles is convection.

Convection

Reason — In convection the heated portion of a fluid expands, becomes lighter and rises up, and its place is taken by the colder and heavier fluid from above. Thus the material particles carrying heat themselves move from one place to another. Convection takes place only in liquids and gases and not in solids.

Glass wool

Reason — A material for which the coefficient of thermal conductivity K is small is a poor conductor or a good insulator. Copper, aluminium and steel are metals which possess a large number of free electrons and are therefore very good conductors of heat. Glass wool has hardly any free electrons and also traps air in it, so its thermal conductivity is very small and it makes the best insulator.

Mercury thermometer

Reason — In a mercury thermometer, mercury is used as the thermometric substance because it expands uniformly with a rise in temperature. The expansion of the mercury column in a fine capillary is used as a measure of the temperature. A thermocouple works on the thermo-emf produced at a junction, a thermistor on the change in electrical resistance, and an infra-red thermometer on the radiation emitted by a body.

4°C

Reason — Water contracts on heating from 0°C to 4°C and expands on further heating. Since a given mass of water occupies the minimum volume at 4°C, its density is maximum at 4°C and is equal to 1 g cm^-3.

Fourier's law

Reason — The law of heat conduction states that the rate of flow of heat through a material is directly proportional to the area of cross-section and to the temperature gradient,

$\frac{\text{dQ}}{\text{dt}} = -\text{KA}\frac{\text d\theta}{\text{dx}}$

This is Fourier's law of heat conduction, the negative sign showing that the temperature always decreases in the direction of the flow of heat.

heat lost by the hot object

Reason — When bodies at different temperatures are brought in contact, then the heat lost by the hot body must be equal to the heat gained by the cold body, provided no heat escapes to the surroundings. This is known as the principle of calorimetry, and it is a consequence of the principle of conservation of energy. Thus, heat lost = heat gained.

is absorbed without changing the temperature

Reason — During the change of state, the heat supplied to the substance does not produce any rise in its temperature. The heat supplied is used up in overcoming the intermolecular force of attraction between the molecules and increasing the potential energy of the substance. Hence the latent heat of fusion is absorbed at the constant temperature of 0°C, and the average kinetic energy of the molecules, which determines the temperature, remains unchanged.

Vacuum

Reason — Thermal expansion occurs because on heating, the molecules of a substance start vibrating more vigorously and the average distance between the molecules increases. A vacuum contains no molecules at all, so there is nothing that can expand. Solids, liquids and gases all expand on heating, gases expanding the most and solids the least.

Metals

Reason — The heat-conduction in metals mostly takes place by the 'free electrons' present within the metals. These electrons are not bound to any one molecule of the metal but are free to move within it. When one end of a metal is heated, the average kinetic energy of the free electrons at that end increases, and these electrons quickly give energy to their neighbouring electrons by colliding with them. Hence the transmission of heat in metals is very rapid. Bad conductors have hardly any free electrons.

°C^-1

Reason — The coefficient of linear expansion is

$\alpha = \frac{\Delta \text L}{\text L \times \Delta \text t} = \frac{\text{Increase in length}}{\text{Original length} \times \text{Rise in temperature}}$

Since the increase in length and the original length have the same unit, they cancel out, leaving only the reciprocal of temperature. Hence the unit of α is per °C, that is, °C^-1 (or K^-1).

the fourth power of the temperature

Reason — According to Stefan's law, the total radiant energy emitted per second per unit surface-area of a black body is proportional to the fourth power of the absolute temperature of the body,

$\frac{\text Q}{\text t} \propto \text T^4 \quad \text{or} \quad \frac{\text Q}{\text t} = \sigma \text T^4$

where σ is Stefan's constant.

W/m K

Reason — The SI unit of the coefficient of thermal conductivity is J s^-1 m^-1 K^-1, which is the same as W m^-1 K^-1. In the CGS system it is expressed in cal s^-1 cm^-1 °C^-1. The unit J kg^-1 K^-1 belongs to specific heat capacity.

convection

Reason — Convection is that process of heat-transmission in which the material particles of the fluid themselves move from one place to another carrying heat with them. The water in a vessel placed on fire becomes hot by convection. Convection takes place only in liquids and gases and not in solids.

Both 2 and 3

Reason — From $\text c = \frac{\text Q}{\text m \times \Delta \text T}$, the unit of specific heat capacity is the unit of heat divided by the units of mass and temperature difference, that is, J kg^-1 K^-1. Since a temperature difference of 1°C is exactly equal to a temperature difference of 1 K, J kg^-1 °C^-1 and J kg^-1 K^-1 are numerically and dimensionally the same unit.

It increases

Reason — From the ideal gas equation PV = nRT, at constant volume

$\frac{\Delta \text P}{\text P} = \frac{\Delta \text T}{\text T}$

so that P ∝ T. Hence the pressure of a gas increases in direct proportion to its absolute temperature when the volume is kept constant. This is the principle on which the constant-volume gas thermometer works.

Constant-volume gas thermometer

Reason — A gas thermometer uses the thermometric property of a gas, whose volume or pressure changes with temperature according to Charles' and Boyle's laws. In the constant-volume gas thermometer, a vessel of constant volume filled with gas is subjected to temperature changes, and the measured temperature is proportional to the change in pressure. Such thermometers are used mostly as standards to calibrate other thermometers.

Note: The option refers to a gas thermometer in general. In the constant-volume type named here, the volume of the gas is held fixed and it is the pressure of the gas, and not its expansion, that is actually measured.

specific heat

Reason — The specific heat capacity of a substance is the amount of heat required to raise the temperature of the unit mass of the substance through a unit degree (1°C or 1 K). Latent heat is the heat needed for a change of state without any change of temperature, and thermal capacity is the heat required to raise the temperature of the whole body through 1 K.

Newton's law of cooling

Reason — Newton's law of cooling states that the rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small,

$-\frac{\text{dQ}}{\text{dt}} = \text k(\text T - \text T_0)$

Stefan's law holds for all temperatures of the hot body and relates the emitted energy to T⁴, while Wien's displacement law relates the wavelength of maximum emission to the temperature.

Temperature

Reason — All bodies continuously emit energy by virtue of their temperature. The rate of radiant energy emitted does not depend upon the shape, size or material of the body; it depends upon the temperature, being proportional to the fourth power of the absolute temperature. The nature of the surface affects only the emissivity of the body.

0°C to 4°C

Reason — Almost all substances expand on heating, but water is an exception in the range 0°C to 4°C, in which it contracts on heating instead of expanding. Above 4°C it expands normally. This behaviour is called the anomalous expansion of water, and it is because of this that the density of water is maximum at 4°C.

Pyrometer

Reason — A pyrometer measures temperature from the radiation emitted by a hot body, so it need not be in contact with the body. It is therefore suitable for measuring the very high temperatures of furnaces and molten metals. A mercury thermometer can be used only up to 357°C, the boiling point of mercury, and an alcohol thermometer is useful only at low temperatures.

electromagnetic waves

Reason — In radiation, heat energy is emitted from a hot body in the form of electromagnetic waves which travel with the speed of light. Electromagnetic waves need no material medium for their propagation, and if a medium is present it does not itself become hot. This is why heat from the sun reaches the earth through the empty space.

conduction

Reason — According to kinetic theory, the molecules in a solid vibrate about their equilibrium positions and the amplitude of vibration depends upon the temperature. When the temperature of one part of the solid increases, the amplitude of vibration of the molecules of this part increases, and these molecules give some energy to the adjacent molecules through mutual interaction. Thus energy is transmitted from the hotter part to the colder part without the molecules leaving their positions, which is conduction.

It can occur in a vacuum

Reason — Radiation is the process by which heat is transferred directly from one body to another without affecting the intervening medium (if any). It travels in the form of electromagnetic waves with the speed of light, and since these waves need no material medium, radiation can take place in a vacuum.

different thermal expansions

Reason — A bimetallic strip is made by firmly joining two thin strips of different metals, such as brass and iron, which are chosen because they have different coefficients of thermal expansion. Before heating the strip is straight, but on heating it bends with brass on the outside of the curve, since brass expands more than iron for the same rise in temperature. This bending is used as the sensitive element in thermostats and in bimetallic thermometers.

equal to the heat lost by the hot body

Reason — By the principle of calorimetry, when bodies at different temperatures are brought in contact in an isolated system, heat lost = heat gained, provided no heat escapes to the surroundings. The heat given out by the body at the higher temperature is entirely received by the body at the lower temperature. This principle follows from the conservation of energy.

Alcohol

Reason — The expansion of a liquid for a given rise in temperature is decided by its coefficient of volume expansion γ. Among the liquids listed, alcohol has by far the largest value of γ, of the order of 11 × 10^-4 °C^-1, as against about 4.9 × 10^-4 °C^-1 for glycerine, 2.1 × 10^-4 °C^-1 for water and 1.8 × 10^-4 °C^-1 for mercury. Hence, for the same rise in temperature and the same original volume, alcohol expands the most.

bulk movement of molecules

Reason — In convection the heated fluid expands, becomes less dense and rises, while the colder and denser fluid sinks to take its place. Heat is thus carried from one place to another by the actual bodily movement of the material particles of the fluid. Transfer by molecular collisions without change of position is conduction, and transfer by electromagnetic waves is radiation.

Hydrogen

Reason — For an ideal gas the coefficient of volume expansion at constant pressure is $\gamma_\text p = \frac{1}{\text T}$, which is the same for all gases and does not depend upon the nature of the gas. Real gases, however, depart slightly from ideal behaviour. Hydrogen, being the lightest gas with the weakest intermolecular forces, behaves most nearly like an ideal gas and shows the largest expansion of the gases listed.

Note: Strictly, all the four gases expand very nearly equally at constant pressure, since γ_p depends only on the temperature. The option is meaningful only if the small departures of real gases from ideal behaviour are taken into account.

survival of aquatic life in cold regions

Reason — In cold weather, when the temperature falls below 4°C, the water at the surface of a pond becomes less dense and remains at the surface where it freezes. Since ice is a poor conductor of heat, the ice layer checks the flow of heat from the water below to the atmosphere, and the water well below the ice layer remains at 4°C. As a result, fish and other aquatic creatures remain alive in the water of the pond even though the surface has frozen into ice.

Heat is energy and temperature is the measure of that energy

Reason — Heat is energy in transit, which flows from a hotter body to a colder one on account of a temperature difference. Temperature, on the other hand, is a measure of the average kinetic energy of the particles of a body, and it tells how hot or cold the body is. Temperature is an intensive property and does not measure the total energy of a system.

independent of its temperature

Reason — The coefficient of linear expansion $\alpha = \frac{\Delta \text L}{\text L \times \Delta \text t}$ describes how much the length of a material changes per unit length per degree rise in temperature. It is a characteristic property of the material of the rod and does not depend on the original length of the rod. Over ordinary ranges it is taken to be constant, that is, independent of the temperature.

double

Reason — According to Charles' law, at constant pressure the volume of a gas is directly proportional to its absolute temperature,

$\frac{\text V}{\text T} = \text{constant}$

Hence if the absolute temperature is doubled at constant pressure, the volume of the gas also becomes double.

Colour of the material

Reason — From $\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$, the quantity of heat conducted depends upon the area of cross-section, the temperature difference, the thickness of the slab, the time and the nature of the material through K. The colour of the material has no effect on conduction; colour affects only the emission and absorption of radiation.

less than the real expansion of the liquid

Reason — A liquid is always taken in a vessel, and when it is heated the vessel also expands along with the liquid. Due to the expansion of the vessel, the observed expansion of the liquid is somewhat less than its real expansion. The two are related as

$\text{Real expansion} = \text{Apparent expansion} + \text{Expansion in the volume of the vessel}$

that is, γ_r = γ_a + γ_g. Hence the apparent expansion is less than the real expansion.

become four times

Reason — The rate of flow of heat in the steady state is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

Thus H is directly proportional to the area of cross-section A and inversely proportional to the length l. On doubling A the rate becomes twice, and on halving l it becomes twice again. Hence the rate of heat conduction becomes 2 × 2 = 4 times its original value.

Melting of ice and cooling of water

Reason — The heat given out by the water is used up first in warming the ice from − 10°C to 0°C and then in melting a part of the ice at 0°C. Since the final temperature is 0°C and some ice is still left, the water has cooled from 20°C to 0°C and a part of the ice has melted at 0°C. The temperature of the ice-water mixture cannot rise above 0°C so long as any ice remains.

its speed of rotation decreases

Reason — On heating, the length of the rod increases due to linear expansion. The moment of inertia of a rod about its perpendicular bisector is $\text I = \frac{\text{Ml}^2}{12}$, so the moment of inertia increases with the length.

No external torque acts on the rod, so its angular momentum L = Iω remains conserved. Since I increases, the angular speed ω must decrease.

Both 1 and 3

Reason — In the heating curve, the horizontal portion AB is the melting stage, during which the temperature stays constant at the melting point while the solid changes into liquid. Thus, in the region AB both the solid and the liquid states of the substance co-exist in thermal equilibrium.

The horizontal portion CD is the vaporisation stage, during which the temperature remains constant at the boiling point while the liquid changes into vapour. Hence CD represents water and steam co-existing in equilibrium at the boiling point.

At B the whole of the ice has just melted and the temperature of the water begins to rise; boiling starts only at C. Hence option 2 is incorrect and both statements 1 and 3 are correct.

both 2 and 3

Reason — From the equation of heat conduction in the steady state,

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

the rate of flow of heat is directly proportional to the area of cross-section A and directly proportional to the temperature difference (θ₁ − θ₂). It is inversely, and not directly, proportional to the thickness l.

It decreases

Reason — The temperature of a body is a measure of the average kinetic energy of its particles. As water is cooled from 10°C to 4°C its temperature falls, so the average kinetic energy of its molecules decreases. The fact that the volume also decreases over this range is a consequence of the anomalous behaviour of water and does not alter the fall in kinetic energy.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Heat is energy in transit. When two bodies at different temperatures interact, both exchange energy with each other, but the net flow of energy is always from the hotter body to the colder one, and it continues until both reach the same temperature, a condition called thermal equilibrium.

Reason (R) is also correct: The second law of thermodynamics fixes the direction in which heat can flow of its own accord, namely from a body at a higher temperature to a body at a lower temperature.

Since it is the second law that decides the direction of the net heat transfer, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The density of water increases as it is heated from 0°C to 4°C and decreases above 4°C. Hence the density of water is maximum at 4°C, being equal to 1 g cm^-3.

Reason (R) is also correct: Water exhibits an anomalous property; it contracts on heating between 0°C and 4°C and expands on further heating.

Since a given mass of water occupies the least volume at 4°C on account of this contraction, its density there is the greatest. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: Calorimetry deals with the measurement of the quantity of heat absorbed or given out by a body during a physical or chemical process.

Reason (R) is also correct: In a calorimeter the vessel is kept inside a wooden jacket filled with heat-insulating material, and its surfaces are polished, so that the loss of heat to the surroundings by conduction, convection and radiation is minimised.

The Reason describes the construction of the instrument used, but it does not explain what calorimetry as a science means. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Mercury does not have a high coefficient of expansion; its coefficient of volume expansion, about 1.8 × 10^-4 °C^-1, is in fact smaller than that of most other liquids such as alcohol and glycerine. Mercury is preferred for other reasons — it has a high thermal conductivity, a wide temperature range from − 39°C to 357°C, it does not wet glass, it is opaque and shiny, and it does not evaporate easily.

Reason (R) is correct: Mercury expands uniformly over a wide range of temperatures. This linear relationship between temperature and expansion simplifies the calibration and the interpretation of thermometers.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Specific heat capacity is a characteristic property of the material of a body and is entirely independent of its mass. Two pieces of copper of different masses have exactly the same specific heat capacity. It is the thermal capacity (m c) which depends upon the mass.

Reason (R) is correct: From $\text c = \frac{\text Q}{\text m \times \Delta \text T}$, if m = 1 and ΔT = 1 then c = Q. Hence the specific heat capacity is the amount of heat required to raise the temperature of unit mass of the substance through a unit degree.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: During the change of state the heat supplied to the substance does not produce any rise in its temperature so long as the change of state takes place. Both the solid and the liquid states of the substance co-exist in thermal equilibrium at the melting point.

Reason (R) is also correct: The heat supplied during the change of state is used up in overcoming the intermolecular force of attraction between the molecules and in increasing the potential energy of the substance.

Since the heat goes into the potential energy of the molecules and not into their kinetic energy, the average kinetic energy and hence the temperature remains unchanged. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Almost all substances, solids as well as liquids, expand on heating. In fact liquids expand more than solids for the same rise in temperature. The only well-known exception is water, which contracts on heating between 0°C and 4°C.

Reason (R) is correct: The intermolecular forces in solids are indeed stronger than those in liquids, which is why the molecules of a solid can only vibrate about fixed mean positions while those of a liquid can move about within the liquid.

Therefore, assertion is false but reason is true.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: By experiment it is observed that on heating a rod, the increase in its length is directly proportional to its original length and to the increase in its temperature,

$\Delta \text L \propto \text L \times \Delta \text t \quad \text{or} \quad \Delta \text L = \alpha\text L\Delta \text t$

Reason (R) is also correct: The constant of proportionality α is called the coefficient of linear expansion, and it depends only upon the material of the rod.

The Reason tells us what the constant α stands for, but the proportionality itself is an experimental result and is not a consequence of α being a material property. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Gases at ordinary temperature expand much more than solids and liquids. For an ideal gas the coefficient of volume expansion at constant pressure is $\gamma_\text p = \frac{1}{\text T}$, which is about 3.3 × 10^-3 K^-1 at room temperature, far larger than the value for any liquid.

Reason (R) is also correct: The intermolecular forces in a gas are extremely weak compared with those in a liquid, so the molecules of a gas are almost free.

Because the molecules are so weakly bound, they can move much farther apart when heated, and this is precisely why a gas expands more than a liquid. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The temperature at which a liquid changes into vapour is called its boiling point, and this temperature depends upon the pressure. By increasing the pressure over the water, it can be made to boil at a temperature higher than 100°C.

Reason (R) is also correct: Like the melting point, the boiling point of a liquid also depends upon pressure, and it rises as the external pressure is increased.

Since the boiling point itself rises with pressure, the temperature of the boiling water rises with it. The Reason therefore explains the Assertion, and this is the principle of the pressure cooker.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Metals are very good conductors of heat, and in solids and in mercury the transmission of heat takes place mainly by conduction. In non-metallic solids the conduction of heat is very small, so they are bad conductors.

Reason (R) is also correct: The heat-conduction in metals mostly takes place by the free electrons present within the metals. These electrons are not bound to any one molecule but are free to move within the metal, and they behave just like the molecules of a gas.

When one end of a metal is heated, the average kinetic energy of the free electrons at that end increases, and these electrons rapidly pass on energy to their neighbouring electrons by colliding with them. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Radiation is the only mode of heat transfer that can take place through empty space. Heat from the sun reaches the earth through the vacuum of space by radiation.

Reason (R) is also correct: Radiant energy travels in the form of electromagnetic waves with the speed of light, and no medium is required for the propagation of these waves. If a medium is present, it does not itself become hot.

Since no material medium is needed, radiation can proceed in a vacuum. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The coefficient of thermal conductivity of water is much greater than that of air, so water conducts heat better than air. It is because of the very low conductivity of air that trapped air is used as an insulator in woollen clothing and in thermos flasks.

Reason (R) is also correct: The density of water is about eight hundred times that of air.

However, the thermal conductivity of a substance is decided by the way in which its molecules and free electrons transfer energy, and not merely by its density. Mercury, for example, is far denser than many metals and yet conducts heat less well than copper. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: A thermos flask is designed to cut down the loss of heat by all the three modes of heat transfer, so that a hot liquid kept in it remains hot for a long time.

Reason (R) is also correct: The vacuum between the double walls of the flask contains no material particles, so it stops the transfer of heat by conduction and by convection, both of which require a medium.

But the vacuum by itself cannot stop radiation, which needs no medium. The loss by radiation is checked separately by silvering the walls of the flask, which makes them good reflectors and poor emitters. Hence the Reason explains only a part of the Assertion and is not its complete explanation.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When common salt is sprinkled over ice, the ice begins to melt even though the surroundings are at the same temperature as before.

Reason (R) is also correct: The dissolved salt lowers the freezing point of water below 0°C.

Since the freezing point has been lowered, the ice at 0°C now finds itself above its new melting point and therefore melts, drawing the necessary latent heat from itself and from the surroundings. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In the steady state the rate of flow of heat through a slab is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

so it does depend upon the thickness l of the slab.

Reason (R) is also correct: Since H is inversely proportional to l, a thicker slab offers a greater thermal resistance $\left(\text R = \frac{\text l}{\text{KA}}\right)$ and hence a smaller rate of heat transfer.

The Reason states the exact manner of this dependence and therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: A solid has a definite shape of its own and expands in all three dimensions, so the change in its length can be measured directly and its expansion is easily noticed.

Reason (R) is also correct: Solids have fixed shapes, which is why their expansion is easier to observe than that of a liquid, which has no shape of its own and takes the shape of the vessel containing it.

However, the ease of observing the expansion is not the reason why the expansion of a solid is noticeable; the two statements describe different things. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A surface which is a good absorber of a particular wavelength of radiation is also a good emitter of that wavelength. A perfectly black body, being a perfect absorber, is also a perfect radiator.

Reason (R) is also correct: By Kirchhoff's law, at a definite temperature and for a given wavelength, the ratio of the emissive power to the absorptive power is the same for all bodies and is equal to the emissive power of a perfectly black body at that temperature,

$\frac{\text e_\lambda}{\text a_\lambda} = \text E_\lambda$

Since this ratio is a constant at a given temperature, a large absorptive power must be accompanied by a large emissive power. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A metal rod conducts heat far more rapidly than a wooden stick, because the heat-conduction in metals takes place through the free electrons present within them, whereas wood has hardly any free electrons.

Reason (R) is false: Metals have a much higher coefficient of thermal conductivity than wood, not a lower one. It is precisely this high conductivity that makes the transfer of heat faster in a metal rod.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: During the change of state, the heat supplied to the substance does not produce any rise in its temperature so long as the change of state continues. The heat so supplied is called the latent heat.

Reason (R) is also correct: The heat supplied is used up in overcoming the intermolecular force of attraction between the molecules and in increasing the potential energy of the substance.

Since the temperature depends on the average kinetic energy of the molecules, and the supplied heat goes entirely into their potential energy, the temperature stays constant. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a body absorbs heat its temperature rises. From Q = m c ΔT, for a given body the rise in temperature increases as more heat is absorbed.

Reason (R) is also correct: The specific heat capacity does decide the amount of heat required to produce a given rise in temperature, since ΔT = $\frac{\text Q}{\text{mc}}$.

But the specific heat capacity only fixes how much heat is needed; it is not itself the cause of the rise in temperature, and the rise is inversely, not directly, related to it. Hence the Reason is not the correct explanation of the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The rate of flow of heat in the steady state is

$\text H = \text{KA}\frac{(\theta_1 - \theta_2)}{\text l}$

which is directly proportional to the temperature gradient $\frac{\theta_1 - \theta_2}{\text l}$. Hence a larger temperature gradient means a larger rate of conduction.

Reason (R) is also correct: For a slab of given thickness, the quantity of heat conducted is directly proportional to the temperature-difference between the two faces.

Since the temperature gradient is just the temperature difference per unit length, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On the Celsius scale the freezing point of water is taken as 0°C and the boiling point as 100°C, the interval between them being divided into 100 equal divisions.

Reason (R) is also correct: A thermometric scale is created by selecting certain fixed reference points which are reproducible. For the Celsius scale these fixed points are the freezing point and the boiling point of pure water at standard atmospheric pressure.

Since the zero of the scale is fixed by the freezing point of water at standard atmospheric pressure, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The coefficient of thermal conductivity of air is very much smaller than that of water. Air is one of the poorest conductors of heat.

Reason (R) is also correct: A material for which K is small is a poor conductor, that is, a good insulator. Since air has a very small K, it is a better insulator than water.

Being a good insulator is the same thing as having a low thermal conductivity, so the Reason supports the Assertion directly. It is because of this that trapped air is used for insulation in woollen clothing and in double-glazed windows.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At thermal equilibrium a body continues to emit and to absorb radiant energy, but the rate of emission becomes equal to the rate of absorption, so there is no net exchange of heat.

Reason (R) is also correct: Thermal equilibrium is that condition in which the body and its surroundings have attained the same temperature.

Heat flows only on account of a temperature difference. When the temperatures become equal, the driving cause of the net flow disappears. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The air inside a greenhouse becomes warmer than the air outside, which is why greenhouses are used for growing plants in cold weather.

Reason (R) is also correct: Glass is largely transparent to the incoming short-wavelength radiation of sunlight, but it is opaque to the long-wavelength infra-red radiation emitted by the warm objects inside.

The short-wavelength radiation passes in and warms the interior, while the long-wavelength radiation emitted by the interior cannot pass out through the glass. The energy is thus trapped inside and the temperature rises. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On a cold day a metal object feels much colder than a wooden or a plastic object lying beside it, although both are at the same temperature.

Reason (R) is also correct: Being a good conductor, a metal conducts heat away from the hand very rapidly, whereas wood and plastic, being bad conductors, conduct it away only very slowly.

The sensation of coldness depends on the rate at which heat leaves the hand, and this rate is governed by the conductivity of the material touched. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Steam at 100°C carries much more heat than water at the same temperature of 100°C. This is why burns from steam are far more serious than those from boiling water.

Reason (R) is also correct: To convert 1 kg of water at 100°C into steam at 100°C, 539 kcal or 22.6 × 10⁵ J of latent heat has to be given. Steam at 100°C therefore carries this latent heat over and above the heat contained in water at 100°C.

Since the extra energy possessed by the steam is exactly this latent heat, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A solid on heating expands in all directions, so that all the three of its length, area and volume increase.

Reason (R) is also correct: In a solid the expansion in length is called linear expansion, the expansion in area is called superficial expansion and the expansion in volume is called cubical or volume expansion, the corresponding coefficients being related as α : β : γ = 1 : 2 : 3.

Since the solid possesses all the three kinds of expansion, it must expand in all three dimensions. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Food is cooked much more quickly in a pressure cooker than in an open vessel.

Reason (R) is also correct: The steam produced inside the closed cooker cannot escape freely, so the pressure inside rises well above the atmospheric pressure. Since the boiling point of a liquid rises with an increase in pressure, the water inside the cooker boils at a temperature well above 100°C.

The food is therefore cooked at a higher temperature, and the rate of cooking is correspondingly faster. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Yes, the size of the hole increases.

On heating, the hole in the disc expands in exactly the same way as a disc of the same metal filling the hole would expand. Every linear dimension of the disc, including the diameter of the hole, increases in the ratio (1 + α Δt). Hence the hole becomes larger and does not shrink.

For a rise in temperature Δt, the fractional increases in the radius, the surface area and the volume of the ball are

$\frac{\Delta \text r}{\text r} = \alpha\Delta \text t, \quad \frac{\Delta \text A}{\text A} = \beta\Delta \text t = 2\alpha\Delta \text t, \quad \frac{\Delta \text V}{\text V} = \gamma\Delta \text t = 3\alpha\Delta \text t$

since α : β : γ = 1 : 2 : 3.

The radius undergoes the least percentage increase and the volume undergoes the largest percentage increase. This is because the coefficient of superficial expansion is twice, and the coefficient of volume expansion is three times, the coefficient of linear expansion of the material.

(i) Just below the lower surface of ice, the temperature of the water is 0°C. At this surface the ice and the water are in contact and are in thermal equilibrium, so the water there must be at the freezing point.

(ii) At the bottom of the lake, the temperature of the water is 4°C. Water has its maximum density at 4°C, so the densest water sinks to the bottom and stays there. Since ice is a poor conductor of heat, the ice layer checks the flow of heat from this water to the atmosphere, and the water at the bottom remains at 4°C. This is how aquatic life survives in cold regions.

Water has a very high specific heat capacity, 1 kcal kg^-1 K^-1, which is nearly ten times that of copper and five times that of sand.

Therefore, when the temperature of the water in the bottle falls by 1°C, the patient receives 1 calorie of heat per gram of water. This is much greater than the heat that would be obtained from any solid or liquid of the same mass. Moreover, on account of its high specific heat, the water cools slowly and supplies this heat steadily over a long time. Hence hot water bottles are used for fomentation.

Water is allowed to flow in pipes around the heated parts of machines as an effective coolant, as in the radiators of cars and generating sets.

Due to its high specific heat, water absorbs a large amount of heat for only a small rise in its own temperature. Consequently it carries away a great deal of heat from the machine, and the temperature of the machine does not rise beyond its safe limit.

Yes, gravity plays an essential part in convection.

In convection the heated portion of the fluid expands and becomes lighter, and it therefore rises up under the action of the buoyant force, while the colder and heavier fluid sinks to take its place. Both the rising of the lighter fluid and the sinking of the heavier fluid are due to gravity, and so convection currents cannot be set up in the absence of gravity.

In conduction the particles do not leave their positions, and in radiation the energy travels as electromagnetic waves, so gravity plays no part in conduction and radiation.

In the beginning, each cross-section of the rod absorbs a part of the heat reaching it, so its temperature goes on rising with time. This is the variable state, in which Q = Q₁ + Q₂ + Q₃.

After some time a state is reached in which no heat is absorbed by any cross-section of the rod, that is, Q₁ = 0. The heat that reaches any section is entirely transferred to the next section, except for the part which escapes from the sides by convection and radiation. This state of the rod, in which the temperature of each section becomes constant, is called the steady state.

Hence in the steady state $\frac{\text d\theta}{\text{dt}} = 0$ for every section, and the temperature at each point of the rod remains constant even though heat continues to flow in from the hot end.

The thermal conductivity of metals is very large because the heat-conduction in metals mostly takes place by the 'free electrons' present within the metals.

These electrons are not bound to any one molecule of the metal, but are free to move within it, and they behave just like the molecules of a gas. When one end of a metal is heated, the average kinetic energy of the free electrons at that end increases. These electrons give some energy to their neighbouring free electrons by colliding with them, and these in turn pass on energy to their nearby electrons.

Since the free electrons move very rapidly and can travel over relatively large distances between collisions, the transmission of energy through a metal is extremely fast. Bad conductors have hardly any free electrons, and so heat-conduction in them takes place very slowly.

No. By steady state it does not mean that the temperature of the whole rod is the same.

In the steady state the temperatures of different parts of the rod are different; as we go away from the hot end, the temperature falls. What the steady state means is that the temperature of each particular section, whatever it may be, remains constant with time, that is, $\frac{\text d\theta}{\text{dt}} = 0$ for every section.

Coefficient of thermal conductivity : The coefficient of thermal conductivity K of a material is defined as the amount of heat that flows in unit time through unit area of the material perpendicular to the flow under unit temperature-gradient, when the steady state has been reached. From

$\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

if A = 1, $\frac{\theta_1 - \theta_2}{\text l} = 1$ and t = 1, then K = Q.

Its SI unit is J s^-1 m^-1 K^-1 or W m^-1 K^-1, and in the CGS system it is expressed in cal s^-1 cm^-1 °C^-1.

Thermal resistance : The ratio of the temperature difference (θ₁ − θ₂) and the rate of heat flow H is called the thermal resistance R of the conductor,

$\text R = \frac{\theta_1 - \theta_2}{\text H} = \frac{\text l}{\text{KA}}$

Its SI unit is °C/W or K W^-1, and it is also expressed as second-°C/kilocalorie. Its dimensional formula is [M^-1 L^-2 T³ θ].

Copper is a good conductor of heat while glass is a bad conductor.

Both the pieces are at the same temperature, but when we touch the copper piece it conducts heat to our hand very rapidly, since heat-conduction in metals takes place through the free electrons. Hence a large quantity of heat is received by the hand in a short time and the copper feels hotter.

The glass piece conducts heat to the hand only very slowly, so much less heat reaches the hand in the same time and it does not feel so hot.

They would appear equally cold or equally hot when both are at the temperature of our body, that is, at about 37°C (98.6°F).

The sensation of hotness or coldness depends upon the direction and the rate at which heat flows between the hand and the object touched. If the block is at the temperature of the body, there is no temperature difference between the hand and the block, so no heat flows in either direction whatever be the conductivity of the material. Hence both the wood and the metal feel neither hot nor cold.

Both the hands receive heat from the bulb by radiation, and since they are at the same distance from the bulb, the heat received by radiation is the same for both.

The right hand, however, is placed above the bulb. The air in contact with the bulb becomes hot, expands, becomes lighter and rises upward, carrying heat with it. Thus the right hand receives heat by convection in addition to radiation, while the left hand, being below the bulb, receives heat by radiation only.

Hence the right hand feels more heat.

A perfectly black body is one which absorbs completely all the radiation falling on its surface, whatever be the wavelength. For a perfectly black body the absorptive power is

$\text a_\lambda = 1 \quad \text{and hence} \quad \text a = 1$

Since a perfectly black body is a perfect absorber, by Kirchhoff's law it is also a perfect radiator, so when heated to a high temperature it emits radiation of all possible wavelengths.

Wien's displacement law states that the wavelength λ_m, corresponding to which the energy emitted by a black body is maximum, is inversely proportional to the absolute temperature T of the body,

$\lambda_\text m \times \text T = \text b\ (\text{a constant})$

where b = 2.9 × 10^-3 m K is called Wien's constant.

It shows that as the temperature of the black body rises, the maximum-energy emitted radiation shifts towards shorter wavelength. On heating a body to an ordinary temperature only long-wavelength radiation is emitted, but as the temperature is raised, radiation of shorter and shorter wavelength is emitted in increasing quantity. That is why, when iron is heated, it first becomes light-red, then dark-red, then yellow, and ultimately it becomes white.

The calorie is the unit of heat in the CGS system and the joule is the SI unit of heat and energy. They are related as

$1\ \text{calorie} = 4.18\ \text{joule}$

and therefore

$1\ \text{kilocalorie} = 4180\ \text{joule} = 4.18 \times 10^3\ \text J$

The specific heat capacity of water at 15°C is

$1\ \text{cal g}^{-1}\space^\circ \text C^{-1} = 1\ \text{kcal kg}^{-1}\space^\circ\text C^{-1} = 4180\ \text{J kg}^{-1}\space\text K^{-1}$

This value is very high, being nearly ten times the specific heat of copper and five times that of sand or earth, and it is responsible for the use of water as a coolant and in hot water bottles.

In the steady state, the quantity of heat Q flowing through a slab of area of cross-section A and thickness l, whose faces are maintained at temperatures θ₁ and θ₂, in time t is

$\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

Therefore, the coefficient of thermal conductivity is

$\text K = \frac{\text Q\text l}{\text A(\theta_1 - \theta_2)\text t}$

From $\text K = \frac{\text Q\text l}{\text A(\theta_1 - \theta_2)\text t}$, substituting the dimensions of heat [M L² T^-2], length [L], area [L²], temperature [θ] and time [T],

$[\text K] = \frac{[\text M \text L^2 \text T^{-2}][\text L]}{[\text L^2][\theta][\text T]}$

Hence, the dimensional formula of thermal conductivity is [M L T^-3 θ^-1].

By Stefan's law, $\frac{\text Q}{\text t} = \sigma \text T^4$ per unit surface area. Hence the unit of Stefan's constant is

$\text{J m}^{-2}\text s^{-1}\text K^{-4} \quad \text{or} \quad \text W\text m^{-2}\text K^{-4}$

Its dimensions are [M T^-3 θ^-4], and its value is 5.67 × 10^-8 W m^-2 K^-4.

When a drop of water falls on a very hot iron, the layer of water in contact with the iron is instantly converted into vapour, and a thin film of vapour is formed between the drop and the hot iron.

Vapour is a very bad conductor of heat. This film therefore separates the drop from the hot surface and checks the flow of heat from the iron to the drop. Hence the heat reaches the drop very slowly and it takes quite a long time to evaporate.

If the iron is only moderately hot, no such vapour film is formed, the drop remains in direct contact with the iron, and it evaporates quickly.

The statement is correct.

It is not essential that a perfectly black body should appear black. A perfectly black body is one which absorbs externally-incident radiation of all wavelengths, and by Kirchhoff's law it therefore also emits radiation of all wavelengths when heated to a high temperature.

A mixture of the radiations of all wavelengths of the visible region appears white to the eye. Hence the radiation coming from a perfectly black body at a high temperature appears white. The sun emits radiation of all wavelengths and may be called a black body, even though it looks white.

In Stefan's law, Q is the radiant heat energy emitted by the black body. Since heat is a form of energy, its unit is the same as that of energy.

Hence the SI unit of Q is the joule (J). In the CGS system it may also be expressed in calorie, where 1 calorie = 4.18 joule.

Copper and steel have different coefficients of linear expansion, that of copper being greater than that of steel.

Before heating the bimetallic strip remains straight. On heating, both the strips tend to expand, but for the same rise in temperature the copper strip becomes longer than the steel strip. Since the two are firmly rivetted together, they cannot slide over each other, and so the strip bends into an arc with the copper strip on the outside (convex side) of the curve and the steel strip on the inside.

On cooling below the original temperature, the reverse bending occurs, with copper on the concave side. This property of unequal expansion is put to practical use in the electric thermostat.

The function of a coolant is to carry away the large quantity of heat produced in the plant so that the temperature of the plant does not rise beyond its safe limit.

The heat absorbed by a coolant of mass m is Q = m c ΔT. For a given rise in temperature ΔT, the greater the specific heat c, the greater is the quantity of heat absorbed.

Hence a coolant of high specific heat absorbs a very large amount of heat for only a small rise in its own temperature, and so a small quantity of it is able to remove a great deal of heat. This is why water, whose specific heat is very high, is used as a coolant in the radiators of cars and in generating sets.

The specific heat of water is nearly five times that of sand or earth.

Therefore, in peak summer, during the day time the temperature of the sea water rises much more slowly than that of the earth remote from the sea. During the night both the earth and the sea water radiate heat energy, but the temperature of the sea water falls more slowly than that of the earth.

Hence, at places remote from the sea, as in a desert, there is a great difference between the temperatures of day and night, while in coastal areas this temperature difference is small. Clearly, in coastal areas the days are not very hot and the nights are not very cold, so the climate is more temperate.

During the change of state, the heat supplied to a substance does not produce any rise in its temperature; it is used up in overcoming the intermolecular force of attraction between the molecules. This heat is called the latent heat, and it is given by Q = m L.

Latent heat of fusion : It is the heat given (or taken out) to convert unit mass (1 kg) of a substance from the solid state to the liquid state (or from the liquid state to the solid state) at its melting point, without change of temperature. For ice it is 3.34 × 10⁵ J kg^-1 or 80 kcal kg^-1.

Latent heat of vaporisation : It is the heat given (or taken out) to convert unit mass (1 kg) of a substance from the liquid state to the vapour state (or from the vapour state to the liquid state) at its boiling point, without change of temperature. For water at 100°C it is 22.6 × 10⁵ J kg^-1 or 539 kcal kg^-1.

The SI unit of latent heat is J kg^-1.

For water the latent heat of fusion is 3.34 × 10⁵ J kg^-1, whereas the latent heat of vaporisation is 22.6 × 10⁵ J kg^-1, which is nearly seven times as large.

In fusion, the substance passes from the solid to the liquid state. Only the rigid arrangement of the molecules has to be broken down; the molecules still remain close to one another, and the volume changes very little.

In vaporisation, the molecules of the liquid have to be completely separated from one another against the whole of the intermolecular force of attraction, so that a very much larger amount of work has to be done against these forces. Besides this, the volume increases enormously, so additional work has to be done by the substance against the external atmospheric pressure in expanding.

Hence the latent heat of vaporisation of a substance is much greater than its latent heat of fusion.

No, water cannot be boiled by convection inside an earth satellite.

In convection, the heated fluid expands, becomes lighter and rises up under the action of the buoyant force, while the colder and heavier fluid sinks down to take its place. Both these motions are produced by gravity.

Inside an earth satellite the condition of weightlessness prevails, so there is no buoyant force acting on the heated water. The heated water therefore does not rise and the colder water does not sink, and no convection currents can be set up. Hence water cannot be boiled by convection in a satellite.

The heat which reaches us from a hot body placed at a distance travels mainly by radiation and not by conduction. Radiant energy travels in the form of electromagnetic waves, and if a medium is present, it does not itself become hot.

Air is transparent to this radiation, so the heat-rays pass freely through the air and fall upon us, and we feel the heat of the hot body. The fact that air is a bad conductor is of no consequence here, because the heat is not being carried by conduction.

Wood, on the other hand, is opaque to this radiation. When a wooden screen is interposed, the heat-rays are absorbed by the screen and cannot pass through it. Since wood is also a bad conductor, the absorbed heat is not conducted through the screen to the other side. Hence the heat does not reach us.

During the night, the earth loses heat by radiating it into the atmosphere.

On a cloudy night, the clouds absorb this radiation emitted by the earth and re-radiate a large part of it back towards the earth's surface. The heat is thus prevented from escaping into space, and the surface of the earth remains warm.

On a clear-sky night, there are no clouds to send the radiation back, so the heat radiated by the earth escapes freely into space and the surface of the earth cools rapidly.

Hence a cloudy night is hotter than a clear-sky night.

Emissive power : The total amount of radiant energy emitted per unit area of a surface per second is called the emissive power e of that surface. Its unit is joule per metre² per second (J m^-2 s^-1).

The amount of radiant energy emitted per unit area of a surface per second per unit wavelength-range at wavelength λ is called the spectral emissive power e_λ of that surface at wavelength λ.

Kirchhoff's law : According to this law, the ratio of the emissive power to the absorptive power for radiation of a given wavelength is the same for all bodies at the same temperature, and is equal to the emissive power of a perfectly black body at that temperature. Thus,

$\frac{\text e_\lambda}{\text a_\lambda} = \text E_\lambda$

where E_λ is the spectral emissive power of a perfectly black body at that temperature. It follows from this law that if a surface is a good absorber of a particular wavelength of radiation, it is also a good emitter of that wavelength.

The radiant energy emitted per second by a body depends upon the following factors :

(i) The nature of the surface of the body. A rough and black surface emits more strongly than a polished and shining surface of the same area at the same temperature. This is expressed through the emissivity e of the surface.

(ii) The area of the surface. The larger the surface area A, the greater is the total energy emitted per second.

(iii) The temperature of the body. By Stefan's law, the energy emitted per second per unit area is proportional to the fourth power of the absolute temperature. At low temperature the rate of emission is small, and as the temperature is raised, the rate of emission increases rapidly.

Combining these, the heat energy emitted per second by the whole body is

$\frac{\text Q}{\text t} = \text e\sigma\text A\text T^4$

Statement : According to Stefan's law, the total radiant energy emitted per second per unit surface-area of a black body is proportional to the fourth power of the absolute temperature of the body.

Thus, if the absolute temperature of a black body is T, then the heat energy emitted per second per unit surface area is

$\frac{\text Q}{\text t} \propto \text T^4 \quad \text{or} \quad \frac{\text Q}{\text t} = \sigma \text T^4$

where σ is a constant, called Stefan's constant. Its value is 5.67 × 10^-8 W m^-2 K^-4 and its dimensions are [M T^-3 θ^-4].

In 1884 Boltzmann gave its theoretical proof and showed that this law applies only to a perfectly black body, and so it is now called the Stefan-Boltzmann law.

If the black body at absolute temperature T is surrounded by a black enclosure at temperature T₀, then it emits σT⁴ and absorbs σT₀⁴ per second per unit area. Hence the net heat energy emitted per second per unit area is

$\frac{\text Q}{\text t} = \sigma(\text T^4 - \text T_0^4)$

The rate of radiant energy emitted does not depend upon the shape, size or material of the body; it depends only upon its temperature.

This happens on account of the greenhouse effect.

The sunlight reaching the car is mostly short-wavelength radiation, and the glass windows are transparent to it. This radiation therefore passes into the car and is absorbed by the seats and the other objects inside, whose temperature rises.

These warm objects also radiate energy, but being at a much lower temperature they radiate long-wavelength infra-red radiation. Glass is opaque to this long-wavelength radiation, so it cannot pass out through the closed windows and remains trapped inside the car.

Since energy keeps entering but cannot leave, the temperature inside the car goes on rising and becomes higher than that of the air outside.

No. It is not essential that a perfectly black body should appear black, nor is every black-coloured body a perfectly black body.

A body appears black to the eye if it absorbs all the wavelengths of the visible region of the spectrum. But a perfectly black body must absorb completely all the radiation falling on it, whatever be the wavelength, including the infra-red and the ultraviolet regions.

Lamp-black and platinum-black, for example, look black because they absorb all the visible and near infra-red parts of the incident radiation, but they reflect the far infra-red parts, and so they are not perfectly black bodies. Conversely, the sun emits radiation of all wavelengths and may be called a black body, even though it looks white.

The Fraunhofer lines are the best example of the illustration of Kirchhoff's law.

The light coming from the very hot photosphere of the sun passes through the comparatively cooler vapours of the chromosphere surrounding it. These vapours absorb exactly those wavelengths which they themselves are capable of emitting. Hence these wavelengths are missing from the light reaching us, and they appear as dark lines in the continuous solar spectrum.

At the time of a total solar eclipse, the bright photosphere is covered by the moon, so the strong continuous background is cut off. The vapours of the chromosphere, however, are still hot and continue to emit radiation of those very wavelengths. Against the dark background these emitted wavelengths are now seen as bright lines exactly in the positions where the dark Fraunhofer lines were formerly seen.

Yes, it is possible.

Let the lengths of the copper and the steel rods at a given temperature be l_C and l_S, and let their coefficients of linear expansion be α_C and α_S. For a rise in temperature Δt, the increases in their lengths are

$\Delta \text l_\text C = \alpha_\text C \text l_\text C \Delta \text t \quad \text{and} \quad \Delta \text l_\text S = \alpha_\text S \text l_\text S \Delta \text t$

The difference (l_S − l_C) will remain the same at all temperatures only if both the rods increase in length by exactly the same amount, that is, Δl_C = Δl_S. Therefore

$\alpha_\text C \text l_\text C = \alpha_\text S \text l_\text S \quad \text{or} \quad \frac{\text l_\text C}{\text l_\text S} = \frac{\alpha_\text S}{\alpha_\text C}$

Hence the condition is that the lengths of the two rods must be in the inverse ratio of their coefficients of linear expansion.

When the two bodies are brought in thermal contact, by the principle of calorimetry the heat lost by the hotter body is equal to the heat gained by the colder body. If T is the final common temperature, then

$\text m_1\text c_1(\text T_1 - \text T) = \text m_2\text c_2(\text T - \text T_2)$

which gives

$\text T = \frac{\text m_1\text c_1\text T_1 + \text m_2\text c_2\text T_2}{\text m_1\text c_1 + \text m_2\text c_2}$

This becomes equal to the mean temperature $\frac{\text T_1 + \text T_2}{2}$ only in the special case when m₁c₁ = m₂c₂, that is, when the two bodies have the same thermal capacity.

Hence, in general the final temperature is not the mean temperature, because the two bodies may differ in mass and in specific heat.

Since the rods are of the same metal and of the same area of cross-section, K and A are the same for both. Hence the rate of flow of heat is proportional to the temperature gradient,

$\text H \propto \frac{\Delta \theta}{\Delta \text x}$

(i) Let the length of B be 2l, so that the length of A is l.

$\text H_\text A \propto \frac{20 - 0}{\text l} = \frac{20}{\text l}, \qquad \text H_\text B \propto \frac{30 - 0}{2\text l} = \frac{15}{\text l}$

Since $\frac{20}{\text l} \gt \frac{15}{\text l}$, the rate of heat-conduction is more in rod A.

(ii) Let the length of B be l, so that the length of A is 2l.

$\text H_\text A \propto \frac{20 - 0}{2\text l} = \frac{10}{\text l}, \qquad \text H_\text B \propto \frac{10 - 0}{\text l} = \frac{10}{\text l}$

Since the two temperature gradients are equal, the rate of heat-conduction is the same in both the rods.

The spectral distribution curves of black body radiation at different temperatures show the following changes when the temperature is increased.

(i) The area enclosed by the curve goes on increasing. The area enclosed by a curve represents the total radiant energy Q of all wavelengths emitted by the body at that temperature, and these areas are found to be proportional to the fourth power of the corresponding absolute temperatures, Q ∝ T⁴. Thus the curves verify Stefan's law.

(ii) The peak of the distribution curve shifts towards shorter wavelength. By Wien's displacement law λ_mT = b, so the wavelength λ_m at which the emitted energy is maximum is inversely proportional to the absolute temperature.

Hence, on heating a body to ordinary temperature only long-wavelength radiation is emitted, but as the temperature is raised, radiation of shorter and shorter wavelength is emitted in increasing quantity. That is why, when iron is heated, it first becomes light-red, then dark-red, then yellow, and ultimately it becomes white.

Let V be the initial volume of the body and ΔV the decrease in its volume under the uniform pressure p. Then the bulk modulus of the material is

$\text B = \frac{\text p}{\Delta \text V/\text V} \quad \Rightarrow \quad \frac{\Delta \text V}{\text V} = \frac{\text p}{\text B}$

To restore the body to its initial volume, the same volume ΔV must be regained by heating. If the temperature is raised through ΔT and γ is the coefficient of volume expansion, then

$\frac{\Delta \text V}{\text V} = \gamma\Delta \text T$

Equating the two expressions,

$\gamma\Delta \text T = \frac{\text p}{\text B}$

$\Delta \text T = \frac{\text p}{\gamma \text B}$

Since γ = 3α, where α is the coefficient of linear expansion,

$\Delta \text T = \frac{\text p}{3\alpha \text B}$

Hence, the temperature of the body should be raised by $\frac{\text p}{3\alpha \text B}$.

The statement means that when 1 kg of ice at 0°C melts into water at 0°C, it absorbs 80 kilocalorie or 3.34 × 10⁵ joule of heat, and during this whole process there is no change of temperature; both the ice and the water remain at 0°C.

Conversely, when 1 kg of water at 0°C freezes to form ice at 0°C, the same amount of heat, 80 kcal or 3.34 × 10⁵ J, is liberated.

This heat is used up in overcoming the intermolecular force of attraction between the molecules and in increasing the potential energy of the substance, and not in raising its temperature.

The statement means that to convert 1 kg of water at 100°C into steam at 100°C, 539 kilocalorie or 22.6 × 10⁵ joule of heat has to be given, and during this whole process the temperature does not change; both the water and the steam remain at 100°C.

Conversely, in liquefying 1 kg of steam at 100°C into water at 100°C, the same 539 kcal of heat is taken out.

It is on account of this large latent heat carried by steam that burns from steam are more serious than those from boiling water, and that heating systems based on the circulation of steam are more efficient than those based on the circulation of hot water.

1 calorie is the amount of heat required to raise the temperature of 1 gram of water through 1°C. It is the unit of heat in the CGS system, and 1 calorie = 4.18 joule.

1 kilocalorie is the amount of heat required to raise the temperature of 1 kilogram of water through 1°C. Since 1 kg = 1000 g,

$1\ \text{kilocalorie} = 1000\ \text{calorie} = 4180\ \text{joule}$

The thermal capacity of a body is defined as the amount of heat required to raise the temperature of the body through 1 K (or 1°C).

If in the formula Q = m c ΔT we put ΔT = 1 K, then Q = m c. Thus the thermal capacity of a body is equal to the product of the mass of the body and the specific heat of the substance of that body.

Its unit is cal/°C or kcal/K, and its SI unit is J/K (or J/°C).

The water equivalent of a calorimeter is the mass of water which would increase in temperature by the same amount as the calorimeter if given the same amount of heat.

If the mass of a calorimeter (with stirrer) is m gram and the specific heat of its material is c cal g^-1 °C^-1, then the heat required to raise the temperature of the calorimeter by 1°C is m c calorie. The same amount of heat will raise the temperature of m c gram of water by 1°C. Hence, by definition, the water equivalent of the calorimeter is

$\text W = \text m\text c\ \text{gram}$

The specific heat capacity of a substance is the amount of heat required to raise the temperature of the unit mass of the substance through a unit degree (1°C or 1 K).

From Q = c m ΔT, we have

$\text c = \frac{\text Q}{\text m \times \Delta \text T}$

so that if m = 1 and ΔT = 1, then c = Q.

Its SI unit is J kg^-1 K^-1, and a commonly used unit is cal g^-1 °C^-1 or kcal kg^-1 °C^-1, where 1 cal g^-1 °C^-1 = 4180 J kg^-1 K^-1.

When radiant energy falls on a body, it is partly reflected from the surface, partly absorbed by the surface and the rest is transmitted through the body.

The ratio of the radiant energy absorbed by a surface in a given time to the total radiant energy incident on the surface in the same time is called the 'absorptive power' or 'absorption coefficient' a of the surface.

Since a is a ratio of two energies, it is a pure number and has no unit.

A perfectly black body is one which absorbs completely all the radiation falling on its surface, whatever be the wavelength. Hence for it the absorbed energy is equal to the incident energy, and so

$\text a_\lambda = 1 \quad \text{and therefore} \quad \text a = 1$

The absorptive power of a perfectly black body is 1, which is the maximum possible value.

According to Kirchhoff's law, the ratio of the emissive power to the absorptive power for radiation of a given wavelength is the same for all bodies at the same temperature, and is equal to the emissive power of a perfectly black body at that temperature.

Mathematically, at a definite temperature and for a wavelength λ,

$\left(\frac{\text e_\lambda}{\text a_\lambda}\right)_1 = \left(\frac{\text e_\lambda}{\text a_\lambda}\right)_2 = \ldots = \text E_\lambda$

where E_λ is the spectral emissive power of a perfectly black body at that temperature.

The important conclusion from this law is that if a surface is a good absorber of a particular wavelength of radiation, it is also a good emitter of that wavelength of radiation.

The rate of flow of heat in the steady state is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

For the two rods the length l and the temperature difference (θ₁ − θ₂) are the same. Hence, for the rates of heat conduction to be equal,

$\frac{\text K_1\text A_1(\theta_1 - \theta_2)}{\text l} = \frac{\text K_2\text A_2(\theta_1 - \theta_2)}{\text l}$

$\text K_1\text A_1 = \text K_2\text A_2$

Hence, the required condition is K₁A₁ = K₂A₂, that is, the areas of cross-section must be in the inverse ratio of the thermal conductivities.

The estimated spectral distribution curves at 2000 K and 4000 K are drawn below.

In drawing them, the following two characteristics of black body radiation are used.

(i) By Wien's displacement law, λ_mT = b, so the wavelength of maximum emission is inversely proportional to the absolute temperature. Since 4000 K is twice 2000 K, the peak of the 4000 K curve occurs at half the wavelength of the peak of the 2000 K curve, that is, it is shifted towards the shorter wavelength side.

(ii) By Stefan's law, the area enclosed by a curve represents the total radiant energy of all wavelengths and is proportional to T⁴. Since the temperature is doubled, the area under the 4000 K curve is 2⁴ = 16 times the area under the 2000 K curve, so the 4000 K curve lies entirely above the 2000 K curve and is much taller.

Both curves have the same general shape — the energy first increases with wavelength, reaches a maximum at λ_m, and then decreases.

The rod of circular cross-section will be hotter.

Both the rods have the same length, the same area of cross-section and are of the same metal, so in the ideal case the rate of conduction along them would be the same. In practice, however, a part of the heat escapes from the curved side surfaces of the rod by convection and radiation, and the heat that finally reaches the far end depends on how much is lost from the sides.

For a given area of cross-section, the circle has the least perimeter. Hence the rod of circular cross-section has a smaller side surface area than the rod of square cross-section, and so it loses less heat from its sides. More heat therefore reaches its far end, and that end is found to be hotter.

A cavity in a piece of coal behaves very nearly as a perfectly black body, on the same principle as Fery's black body.

Any radiation entering the cavity suffers multiple reflections at its inner walls and has little chance of coming out again, so it is almost completely absorbed. Hence the absorptive power of the cavity is very nearly 1.

By Kirchhoff's law, a good absorber is also a good emitter. Therefore, when the coal is heated, the cavity emits the maximum possible radiation for that temperature, that is, black body radiation, while the outer surface of the coal, not being a perfectly black body, emits less than this.

Hence the cavities look brighter than the surface of the coal, although both are at the same temperature.

The hollow sphere will cool faster.

Both spheres are of the same material and have the same external radius, so their surface areas are equal and both are at the same temperature. Hence, by Stefan's law, both lose heat by radiation at exactly the same rate, $\frac{\text Q}{\text t} = \text e\sigma \text A\text T^4$.

The hollow sphere, however, contains less material and so has a smaller mass, and therefore a smaller thermal capacity (m c). The rate of fall of temperature is

$\frac{\text{dT}}{\text{dt}} = \frac{\text Q/\text t}{\text m\text c}$

Since the numerator is the same for both while m is smaller for the hollow sphere, its temperature falls more rapidly. Hence the hollow sphere cools faster.

(i) The cooler body radiates energy but less than it absorbs from the surroundings.

According to Prevost's theory of heat exchange, all bodies, whatever their temperature, radiate heat to their surroundings. A cooler body placed in warmer surroundings therefore does not stop radiating; it continues to emit energy, but its rate of emission is lower than its rate of absorption, and so its temperature rises.

(ii) Net heat transfer is one-way, but actual energy exchange is continuous and bidirectional.

Both bodies radiate energy towards each other all the time, so the actual exchange of energy takes place in both directions. Since the warmer body radiates more than it receives, the net result is a one-way flow of heat from the warmer body to the cooler one.

(iii) Both bodies continue to exchange heat equally, resulting in no net transfer.

At thermal equilibrium the emission and absorption do not stop; the rate of radiation becomes equal to the rate of absorption for each body. Hence the exchange of energy continues, but there is no net transfer of heat in either direction.

(iv) Only at absolute zero.

At absolute zero (0 K), molecular motion theoretically ceases, so no thermal energy can be radiated at all. This is the only condition in which heat exchange stops entirely; at thermal equilibrium it merely becomes equal in both directions.

(v) Because it refers to net heat transfer, but does not consider the fact that energy is radiated in all directions by both bodies.

The description of heat flow as one-way is a simplification that applies only to the net transfer between bodies at different temperatures. In reality, as Prevost's theory shows, energy is radiated in all directions by every body, and this exchange persists even at thermal equilibrium.

(i) It is based on the triple point of water and absolute zero.

The Kelvin scale is an absolute temperature scale whose zero point is absolute zero, and which is defined by assigning the value 273.16 K to the triple point of water. Such a scale is convenient in science because several physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The triple point is preferred as the defining fixed point because, unlike the freezing and boiling temperatures, it occurs at only one temperature.

(ii) 100°C

On the Celsius scale the freezing point of water is taken as 0°C and the boiling point as 100°C, the interval between them being divided into 100 equal divisions.

(iii) 1 degree Celsius is 1.8 times larger

The interval between the freezing and boiling points of water spans 100 divisions on the Celsius scale and 180 divisions on the Fahrenheit scale. Hence

$1^\circ \text C = \frac{180}{100}\ ^\circ \text F = 1.8\ ^\circ \text F$

that is, a temperature difference of one degree on the Celsius scale is 1.8 times as large as a difference of one degree on the Fahrenheit scale.

(iv) Constant-volume gas thermometer

Different thermometers give somewhat different results for temperatures other than the fixed points, so a standard thermometer is required. The constant-volume gas thermometer has been chosen for this purpose, since the measured temperature is proportional to the change in pressure of a fixed volume of gas. Such thermometers are big and come to equilibrium slowly, so they are used mostly as standards to calibrate other thermometers.

(i) It predicted infinite energy at high frequencies.

According to the classical Rayleigh-Jeans law, the energy distribution at high frequencies was predicted to increase indefinitely. This unphysical result is known as the "ultraviolet catastrophe", and it was to remove this difficulty that Planck introduced his theory.

(ii) quanta

Planck proposed that energy is quantized, that is, it can only be emitted or absorbed in discrete packets called quanta or photons, and not continuously.

(iii) Inverse temperature

By Wien's displacement law, λ_mT = b, so

$\lambda_\text m = \frac{\text b}{\text T}$

Thus the peak wavelength is directly proportional to the reciprocal of the absolute temperature, that is, to the inverse temperature. As the temperature increases, the peak shifts towards shorter wavelengths.

(iv) temperature and frequency

In Planck's law the spectral radiance is written as E(v, T), that is, as a function of the frequency v of the radiation and the absolute temperature T of the black body.

(v) Gravitational constant

The formula for Planck's law contains Planck's constant h, the speed of light c and Boltzmann's constant k. The gravitational constant does not appear in it at all.

Note: In the passage the expression $\frac{8\pi \text h\nu^3}{\text c^3}\cdot\frac{1}{\text e^{\text h\nu/\text{kT}}-1}$ is called the spectral radiance. It is in fact the spectral energy density u_ν, whose unit is J m^-3 Hz^-1. The spectral radiance is $\text B_\nu = \frac{2\text h\nu^3}{\text c^2}\cdot\frac{1}{\text e^{\text h\nu/\text{kT}}-1}$.

(i) The size of the hole increases on heating the metal disc.

On heating, the hole in the disc expands in exactly the same way as a disc of the same metal filling the hole would expand. Every linear dimension of the disc, including the diameter of the hole, increases in the ratio (1 + α Δt), so the hole becomes larger.

(ii) (a) Just below the lower surface of ice the temperature of the water is 0°C, since at that surface the ice and the water are in contact and are in thermal equilibrium at the freezing point.

(b) At the bottom of the lake the temperature of the water is 4°C. Water has its maximum density at 4°C, so the densest water sinks to the bottom and remains there. Since ice is a poor conductor of heat, the ice layer checks the flow of heat from this water to the atmosphere, and so the water at the bottom stays at 4°C. This is how fish and other aquatic creatures remain alive in a frozen pond.

(iii) Thermal expansion : Almost all substances expand on heating and contract on cooling. The expansion of a substance on heating is called 'thermal expansion' of that substance.

Every substance is made up of molecules. When a substance is heated, its molecules start vibrating and so the kinetic energy of the molecules increases. Further, the average distance between the molecules increases, and the substance expands. A solid on heating expands in all directions, that is, in length, area and volume, whereas liquids expand in volume only.

(i) 'Black body radiation is white' — the statement is correct.

It is not essential that a perfectly black body should appear black. A perfectly black body absorbs externally-incident radiation of all wavelengths, and by Kirchhoff's law it therefore also emits radiation of all wavelengths when heated to a high temperature. A mixture of the radiations of all the wavelengths of the visible region appears white to the eye. Hence the radiation coming from a perfectly black body at a high temperature appears white. The sun emits radiation of all wavelengths and may be called a black body, though it looks white.

(ii) Given,

• Temperature of the first black body, T₁ = 727°C = 727 + 273 = 1000 K
• Temperature of the second black body, T₂ = 327°C = 327 + 273 = 600 K

By Stefan's law, the energy radiated per second per unit area is proportional to the fourth power of the absolute temperature, E ∝ T⁴. Therefore

$\frac{\text E_1}{\text E_2} = \left(\frac{\text T_1}{\text T_2}\right)^4 = \left(\frac{1000}{600}\right)^4$

$= \left(\frac{5}{3}\right)^4 = \frac{625}{81} = 7.72$

Hence, the ratio of the energy radiated per second by them is 625 : 81, that is, about 7.72 : 1.

(iii) Newton's law of cooling : The rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small.

If T is the temperature of the body and T₀ that of the surroundings, then

$-\frac{\text{dQ}}{\text{dt}} = \text k(\text T - \text T_0)$

where k is a positive constant depending upon the area and the nature of the surface of the body.

In solids all the three of length, area and volume increase on heating. A solid on heating expands in all directions. Thus, in a solid, expansion in length is called linear expansion, expansion in area is called superficial expansion and expansion in volume is called cubical or volume expansion.

Coefficient of linear expansion (α) : By experiment it is observed that on heating a rod, the increase in its length is directly proportional to its original length and to the increase in its temperature. If the length of a rod is L and its temperature rises by Δt, then

$\Delta \text L \propto \text L \times \Delta \text t \quad \text{or} \quad \Delta \text L = \alpha\text L\Delta \text t$

$\alpha = \frac{\Delta \text L}{\text L \times \Delta \text t} = \frac{\text{Increase in length}}{\text{Original length} \times \text{Rise in temperature}}$

If Δt = 1°C and L = 1, then α = ΔL. Thus, the coefficient of linear expansion of the material of a rod is equal to the increase in unit length of the rod when its temperature rises by 1°C. Its unit is per °C.

Coefficient of superficial expansion (β) : Like linear expansion, the superficial expansion of a solid also depends upon its original area, the rise in temperature and the material of the solid. If the original area of a lamina is A and its area becomes A + ΔA on raising the temperature by Δt°C, then

$\beta = \frac{\Delta \text A}{\text A \times \Delta \text t} = \frac{\text{Increase in area}}{\text{Original area} \times \text{Rise in temperature}}$

Thus, the coefficient of superficial expansion of the material of a lamina is equal to the increase in unit area of the lamina when its temperature rises by 1°C.

Coefficient of volume expansion (γ) : If the original volume of a solid is V and it becomes V + ΔV on raising the temperature by Δt°C, then

$\gamma = \frac{\Delta \text V}{\text V \times \Delta \text t} = \frac{\text{Increase in volume}}{\text{Original volume} \times \text{Rise in temperature}}$

Thus, the coefficient of volume expansion of the material of a solid is equal to the increase in unit volume of the solid when its temperature rises by 1°C.

Relation between β and α : Let each side of a square lamina be 1 cm at a given temperature, so that its area is 1 cm². Let the coefficient of linear expansion of its material be α.

Let the temperature of the lamina be increased by 1°C. Then at the new temperature,

$\text{Each side of the lamina} = (1 + \alpha)\ \text{cm}$

$\text{Area of the lamina} = (1 + \alpha)^2\ \text{cm}^2$

$\text{Increase in area} = (1 + \alpha)^2 - 1 = 1 + 2\alpha + \alpha^2 - 1 = 2\alpha + \alpha^2$

Since α is much less than 1, α² may be neglected. Then the increase in area = 2α, and

$\beta = \frac{\text{Increase in area}}{\text{Original area} \times \text{Rise in temperature}} = \frac{2\alpha}{1 \times 1} = 2\alpha$

Thus, the coefficient of superficial expansion is twice the coefficient of linear expansion.

Relation between γ and α : Let each side of a cube be 1 cm at a given temperature, so that its volume is 1 cm³.

Let the temperature of the cube be raised by 1°C. Then at the new temperature,

$\text{Each side of the cube} = (1 + \alpha)\ \text{cm}$

$\text{Volume of the cube} = (1 + \alpha)^3\ \text{cm}^3$

$$\begin{aligned}\text{Increase in volume} = (1 + \alpha)^3 - 1 \\[1em] = 1 + 3\alpha + 3\alpha^2 + \alpha^3 - 1 = 3\alpha + 3\alpha^2 + \alpha^3\end{aligned}$$

Since α is much less than 1, the higher power terms of α may be neglected. Then the increase in volume = 3α, and

$\gamma = \frac{\text{Increase in volume}}{\text{Original volume} \times \text{Rise in temperature}} = \frac{3\alpha}{1 \times 1} = 3\alpha$

Thus, the coefficient of volume expansion is three times the coefficient of linear expansion.

Relation among α, β and γ : Since β = 2α and γ = 3α,

$\alpha : \beta : \gamma = \alpha : 2\alpha : 3\alpha = 1 : 2 : 3$

Steady state : When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end. In the process of conduction each cross-section of the rod receives heat Q from the adjacent cross-section towards the heated end. A part of this heat (Q₁) is absorbed by the cross-section itself, due to which its temperature increases, another part (Q₂) goes into the atmosphere by convection and radiation, and the rest (Q₃) is conducted to the next cross-section. In this variable state, Q = Q₁ + Q₂ + Q₃.

After some time a state is reached when the temperature of each section of the rod becomes constant. In this state no heat is absorbed by the rod, that is, Q₁ = 0. This state of the rod in which the temperature of each section of the rod becomes constant is called the 'steady state'.

By steady state it does not mean that the temperature of the whole rod is the same; the temperatures of different parts are different, but the temperature of each part remains constant with time, so that for every section

$\frac{\text d\theta}{\text{dt}} = 0$

Rate of heat flow in the steady state : Consider a parallel-faced slab of cross-sectional area A and length l, whose faces are maintained at steady temperatures θ₁ and θ₂ (θ₁ > θ₂). Experiment shows that the heat Q flowing through the slab in time t is

(i) directly proportional to the area of cross-section A,

(ii) directly proportional to the temperature difference (θ₁ − θ₂),

(iii) directly proportional to the time t, and

(iv) inversely proportional to the length l of the slab.

Combining these,

$\text Q \propto \frac{\text A(\theta_1 - \theta_2)\text t}{\text l} \quad \text{or} \quad \text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

where K is the coefficient of thermal conductivity of the material. Hence the rate of flow of heat is

$\text H = \frac{\text Q}{\text t} = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

Dimensional equation of K : From the above equation,

$\text K = \frac{\text Q\text l}{\text A(\theta_1 - \theta_2)\text t}$

Substituting the dimensions of heat Q as [M L² T^-2], of length l as [L], of area A as [L²], of temperature difference as [θ] and of time t as [T],

$[\text K] = \frac{[\text M \text L^2 \text T^{-2}][\text L]}{[\text L^2][\theta][\text T]} = [\text M \text L \text T^{-3} \theta^{-1}]$

Hence, the dimensional formula of the coefficient of thermal conductivity is [M L T^-3 θ^-1], and its SI unit is J s^-1 m^-1 K^-1 or W m^-1 K^-1.

Emissive power : When the temperature of a body is higher than that of its surroundings, radiant energy is continuously emitted from the surface of the body. The total amount of radiant energy emitted per unit area of a surface per second is called the 'emissive power' e of that surface. Its unit is joule per metre² per second (J m^-2 s^-1).

The amount of radiant energy emitted per unit area of a surface per second per unit wavelength-range at wavelength λ is called the 'spectral emissive power' e_λ of that surface at wavelength λ.

Absorptive power : When radiant energy falls on a body, it is partly reflected from the surface, partly absorbed by the surface and the rest is transmitted through the body. The ratio of the radiant energy absorbed by a surface in a given time to the total radiant energy incident on the surface in the same time is called the 'absorptive power' or 'absorption coefficient' a of the surface. Since a is a ratio, it has no unit.

The ratio of the radiant energy absorbed by a surface per unit wavelength-range at wavelength λ in a given time to the incident radiant energy in that time is called the 'spectral absorptive power' a_λ of that surface at wavelength λ.

Proof of Kirchhoff's law : Let a body be placed in a uniformly-heated enclosure at a constant temperature T. In the equilibrium state, the temperature of the body will be equal to that of the enclosure.

Suppose at this temperature the absorptive power of the body for the wavelength λ is a_λ. Let an amount of radiant energy ΔQ between the wavelengths λ and (λ + Δλ) be incident per second on the unit surface area of the body. Then the amount of energy absorbed by unit surface area of the body per second is

$= \text a_\lambda\Delta \text Q$

Now, if at temperature T and wavelength λ the emissive power of the surface of the body be e_λ, then the amount of energy emitted by unit surface area of the body per second between the wavelengths λ and (λ + Δλ) is

$= \text e_\lambda\Delta \lambda$

In a uniform-temperature enclosure, the quantity and quality of radiation is not affected by the presence of any body inside it. Hence, whatever radiant energy of any wavelength the body absorbs, it emits exactly an equal amount of radiant energy of the same wavelength. Thus

$\text a_\lambda\Delta \text Q = \text e_\lambda\Delta \lambda \qquad \ldots(\text i)$

Now, a perfectly black body absorbs the whole of the radiant energy incident on it, that is, its absorptive power a_λ = 1. So, if a perfectly black body whose emissive power is E_λ is placed inside the enclosure, then for it the equation corresponding to equation (i) will be

$\Delta \text Q = \text E_\lambda\Delta \lambda \qquad \ldots(\text{ii})$

Dividing equation (i) by equation (ii), we get

$\text a_\lambda = \frac{\text e_\lambda}{\text E_\lambda} \quad \text{or} \quad \frac{\text e_\lambda}{\text a_\lambda} = \text E_\lambda$

Since, for a given temperature, E_λ is a constant, so for all materials at that temperature

$\left(\frac{\text e_\lambda}{\text a_\lambda}\right)_1 = \left(\frac{\text e_\lambda}{\text a_\lambda}\right)_2 = \left(\frac{\text e_\lambda}{\text a_\lambda}\right)_3 = \ldots = \text E_\lambda$

Hence, at a definite temperature and for a given wavelength, the ratio of the emissive power to the absorptive power is the same for all surfaces and is equal to the emissive power of a perfectly black body at that temperature. This is Kirchhoff's law.

Importance : A conclusion from Kirchhoff's law is that if a surface is a good absorber of a particular wavelength of radiation, it is also a good emitter of that wavelength of radiation.

Newton's law of cooling : The rate of loss of heat from a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small.

Stefan's law holds for all temperatures of the hot body, but Newton's law of cooling is applicable only when the temperature difference (T − T₀) is small.

Deduction from Stefan's law : Let T = T₀ + ΔT, where ΔT is the small difference between the temperature T of the body and the temperature T₀ of the surroundings, so that ΔT << T₀. Then, from Stefan's law,

$-\frac{\text Q}{\text t} = \sigma\left[(\text T_0 + \Delta \text T)^4 - \text T_0^4\right]$

the negative sign being taken because there is a loss of heat. Expanding by the binomial theorem and retaining only the first-order term,

$-\frac{\text Q}{\text t} = \sigma\left[\text T_0^4\left(1 + \frac{4\Delta \text T}{\text T_0} + \cdots\right) - \text T_0^4\right] = 4\sigma\Delta \text T\text T_0^3$

Since T is only slightly greater than T₀, we can assume T₀ to be practically constant during cooling. Hence

$-\frac{\text Q}{\text t} = \text k\Delta \text T = \text k(\text T - \text T_0), \quad \text{where } \text k = 4\sigma \text T_0^3$

Proof that the temperature decreases exponentially with time : Let at any instant the temperature of the body be T and that of the surroundings be T₀. Then the loss of heat in time dt is − dQ, and

$\frac{-\text{dQ}}{\text{dt}} = \text k(\text T - \text T_0) \qquad \ldots(\text i)$

If the mass of the body is m and the specific heat capacity of its material is c, and in the infinitesimally small time interval dt the temperature of the body falls by dT, then

$\text{dQ} = \text m\text c\text{dT} \qquad \ldots(\text{ii})$

From equations (i) and (ii),

$\frac{-\text m\text c\text{dT}}{\text{dt}} = \text k(\text T - \text T_0)$

Separating the variables,

$\frac{\text{dT}}{\text T - \text T_0} = -\frac{\text k}{\text m\text c}\text{dt} = -\text K\text{dt}$

where $\text K = \frac{\text k}{\text{mc}}$ is a constant for a given body.

On integrating,

$\log_e(\text T - \text T_0) = -\text{Kt} + \text C \qquad \ldots(\text{iii})$

where C is the constant of integration. If the initial temperature of the body at t = 0 is T′, then substituting in equation (iii),

$\log_e(\text T' - \text T_0) = \text C$

Substituting this value of C in equation (iii),

$\log_e(\text T - \text T_0) = -\text{Kt} + \log_e(\text T' - \text T_0)$

$\log_e\left(\frac{\text T - \text T_0}{\text T' - \text T_0}\right) = -\text{Kt}$

Taking antilogarithms,

$\text T - \text T_0 = (\text T' - \text T_0)\text e^{-\text{Kt}}$

Clearly, the temperature of the body decreases exponentially with time.

Also, from equation (iii), if a graph is drawn between t and log_e(T − T₀), it is a straight line having a negative gradient.

(a) Anomalous expansion of water : Almost all substances expand on heating and contract on cooling. Water, however, exhibits an anomalous property; it contracts on heating between 0°C and 4°C. The volume of a given amount of water decreases when it is cooled from room temperature to 4°C, and below 4°C the volume increases instead of decreasing.

The volume of water first decreases when water is heated from 0°C to 4°C and then increases. Correspondingly, when water is heated from 0°C the density of water first increases from 0°C to 4°C and then decreases above 4°C. Thus, the density of water is maximum at 4°C and is equal to 1 g cm^-3.

When the temperature of water becomes 0°C, water freezes into ice, and then the volume of the ice is greater than the volume of the water. It is for this reason that the density of ice is less than the density of water and ice floats on water.

Consequences in everyday life :

(i) In preserving the aquatic life during very cold weather : In cold weather, when the temperature starts falling below 4°C in winter, the water at the surface of a pond or lake near the surface begins to lose heat energy to the atmosphere, becomes denser and sinks, and the warmer less dense water near the bottom rises. However, once the colder water on the top reaches a temperature below 4°C it becomes less dense and remains at the surface where it freezes. Since ice is a poor conductor of heat, the ice now checks the flow of heat from the water of the pond to the atmosphere, and the water well below the ice layer remains at 4°C. As a result fish and other aquatic creatures remain alive in the water of the pond, though the water of the surface has frozen into ice.

(ii) Water pipe lines, nerves of plants and rocks burst during the very cold nights : When the atmospheric temperature in winter starts falling below 4°C, water expands and it exerts large pressure on the water pipe lines, so they burst. The plants also die, as their nerves (or capillaries) burst when water expands below 4°C. For this reason farmers fill their fields with water to save the crop from this effect.

(b) Variation of the density of liquids with temperature : When a given mass of a liquid is heated, its volume increases; accordingly the density of the liquid decreases.

Let V and V′ be the volumes of a liquid at temperatures T and (T + ΔT) respectively, and γ the coefficient of volume expansion. Then

$\Delta \text V = \text V' - \text V = \gamma\text V\Delta \text T$

$\text V' = \text V(1 + \gamma\Delta \text T) \qquad \ldots(\text i)$

If ρ and ρ′ are the densities of the liquid at temperatures T and (T + ΔT) respectively, then since the mass is the same,

$\text V\rho = \text V'\rho' \quad \text{or} \quad \frac{\rho}{\rho'} = \frac{\text V'}{\text V}$

Making this substitution in equation (i),

$\rho = \rho'(1 + \gamma\Delta \text T) \quad \text{or} \quad \rho' = \rho(1 + \gamma\Delta \text T)^{-1}$

Expanding by the binomial theorem and neglecting the smaller terms,

$\rho' = \rho(1 - \gamma\Delta \text T)$

Thus, the density of liquids, which expand on heating, decreases with rise in temperature. However, the density of water increases with rise in temperature in the range 0°C to 4°C.

When a body is heated, its temperature rises. Experiments show that the amount of heat given to a body is directly proportional to the mass of the body and to the temperature-rise. Thus, if an amount of heat Q given to a body of mass m raises its temperature by ΔT, then

$\text Q \propto \text m \times \Delta \text T \quad \text{or} \quad \text Q = \text c \times \text m \times \Delta \text T$

where c is a constant depending upon the material of the body. It is called the 'specific heat capacity' of the material. From the above equation,

$\text c = \frac{\text Q}{\text m \times \Delta \text T}$

Definition : If m = 1 and ΔT = 1, then c = Q. Hence, the specific heat capacity of a substance is the amount of heat required to raise the temperature of the unit mass of the substance through a unit degree (1°C or 1 K).

SI unit : The SI unit of specific heat capacity is J kg^-1 K^-1. A commonly used unit is cal g^-1 °C^-1 or kcal kg^-1 °C^-1, the numerical value of the specific heat in these two units being the same. Since 1 kcal is equivalent to 4180 J and a temperature difference of 1°C is equal to a temperature difference of 1 K,

$1\ \text{cal g}^{-1}\space^\circ\text C^{-1} = 1\ \text{kcal kg}^{-1}\space^\circ\text C^{-1} = 4180\ \text{J kg}^{-1}\space\text K^{-1}$

The specific heat capacity of water at 15°C is 1 cal g^-1 °C^-1 or 4180 J kg^-1 K^-1.

Principle of calorimetry : A system is said to be isolated if no exchange of heat takes place between the system and its surroundings. When different parts of an isolated system are at different temperatures, then some heat is transferred from the part at higher temperature to the part at lower temperature.

When bodies at different temperatures are brought in contact, then the heat lost by the hot body must be equal to the heat gained by the cold body, provided no heat escapes to the surroundings. This is known as the principle of calorimetry. Mathematically,

$\text{heat lost} = \text{heat gained}$

This principle is a consequence of the principle of conservation of energy.

Determination of the specific heat of a solid : Let us consider a hot solid of mass m and specific heat c at a temperature t. It is dropped into water of mass m₁ at a temperature t₁ (lower than t) contained in a calorimeter of water equivalent W.

The contents of the calorimeter are stirred by the stirrer continuously till the temperature of the mixture becomes steady. Let it be t₂. In this process the solid has been cooled from temperature t to t₂, while the water and the calorimeter have been heated from temperature t₁ to t₂.

Now,

$\text{heat lost by the solid} = \text m\text c(\text t - \text t_2)$

and

$\text{heat gained by the water and the calorimeter} = (\text m_1 + \text W)\text c_1(\text t_2 - \text t_1)$

where c₁ is the specific heat of water.

By the principle of calorimetry, heat lost = heat gained,

$\text m\text c(\text t - \text t_2) = (\text m_1 + \text W)\text c_1(\text t_2 - \text t_1)$

$\text c = \frac{(\text m_1 + \text W)\text c_1(\text t_2 - \text t_1)}{\text m(\text t - \text t_2)}$

Knowing m, m₁, W, c₁, t, t₁ and t₂, the specific heat c of the solid is calculated.

Latent heat : During the change of state of a substance, the heat supplied to the substance does not produce any rise in the temperature of the substance so long as the change of state takes place. The heat supplied is used up in overcoming the intermolecular force of attraction between the molecules and increasing the potential energy of the substance.

If a mass m of a substance undergoes a change from one state to the other, then the amount of heat required is given by

$\text Q = \text m\text L$

where L is known as the latent heat and is a characteristic of the substance. Thus $\text L = \frac{\text Q}{\text m}$. The latent heat for a solid-liquid state change is called the 'latent heat of fusion'.

Definition : The latent heat of fusion of a substance is the heat given (or taken out) to convert unit mass (1 kg) of the substance from the solid state to the liquid state (or from the liquid state to the solid state) at its melting point, without change of temperature.

Explanation : The latent heat of ice is 3.34 × 10⁵ J kg^-1 or 80 kcal kg^-1 or 80 cal g^-1. It means that when 1 kg of ice at 0°C melts, then it absorbs 3.34 × 10⁵ J (or 80 kcal) of latent heat. On the other hand, when 1 kg of water at 0°C freezes to form ice at 0°C, then the same amount of heat is liberated. It may be noted that the melting point of a solid and the freezing point of its liquid are the same; for example, the melting point of ice and the freezing point of water are both 0°C.

SI unit : The SI unit of latent heat is J kg^-1. Other units are kcal kg^-1 or cal g^-1. The value of L also depends upon pressure.

Just as charge flows in an electrical circuit due to a potential difference between two points of the circuit, in the same way heat flows in a conductor due to a temperature-difference between two points of the conductor. Hence, like electrical resistance, there is also a thermal resistance in a material.

Definition : Let l be the length and A the area of cross-section of a rod, and let θ₁ and θ₂ be the temperatures of the hot and the cold ends of the rod in the steady state. Then the rate of flow of heat in the rod is

$\text H = \frac{\text Q}{\text t} = \text{KA}\frac{\theta_1 - \theta_2}{\text l}$

The ratio of the temperature difference (θ₁ − θ₂) and the rate of heat flow (H) is called the 'thermal resistance' R of the conductor. Thus,

$\text R = \frac{\theta_1 - \theta_2}{\text H} = \frac{\text l}{\text{KA}}$

Factors on which it depends : From the above expression, the thermal resistance of a rod depends upon the following factors.

(i) The length of the rod (l) : R is directly proportional to the length, so a longer rod offers a greater thermal resistance.

(ii) The area of cross-section (A) : R is inversely proportional to the area of cross-section, so a thicker rod offers a smaller thermal resistance.

(iii) The thermal conductivity of the material (K) : R is inversely proportional to K. Thus, greater the coefficient of thermal conductivity of a material, smaller is the thermal resistance of a rod of that material.

Its SI unit is °C/W or K W^-1, and it is also expressed as second-°C/kilocalorie. Its dimensional formula is [M^-1 L^-2 T³ θ].

Perfectly black body : A perfectly black body is one which absorbs completely all the radiation falling on its surface, whatever be the wavelength. Thus, for a perfectly black body,

$\text a_\lambda = 1 \quad \text{and hence} \quad \text a = 1$

Since a perfectly black body is a perfect absorber, according to Kirchhoff's law it will also be a perfect radiator. So, when a perfectly black body is heated to a high temperature, it emits radiation of all possible wavelengths.

No known body is a perfectly black body. Lamp-black and platinum-black, which absorb all the visible and near infra-red parts of the incident radiation, reflect the far infra-red parts. For practical purposes, Fery designed a perfectly black body which is a double-walled metallic hollow sphere with a narrow hole, whose inner wall is blackened.

Spectral distribution curves : Lummer and Pringsheim heated a black body to different temperatures and drew distribution curves between the emitted energy and the wavelength for all the temperatures on the same graph paper.

Effect of rise in temperature :

(i) As the temperature of the black body rises, the area enclosed by the distribution curve goes on increasing. The area enclosed by a curve represents the total radiant energy Q (of all wavelengths) emitted by the black body at that temperature. When the areas enclosed by different curves are measured, they are found to be proportional to the fourth power of the corresponding absolute temperatures, that is, Q ∝ T⁴. Thus these curves verify Stefan's law.

(ii) At a given temperature T, with increase in wavelength λ, the energy Q_λ first increases, reaches a maximum and then decreases. Thus for a given temperature the radiant energy emitted by a black body is maximum for a particular wavelength λ_m.

(iii) As the temperature of the black body rises, the peak of the distribution curve shifts towards shorter wavelength (λ). In 1896 Wien established the relation

$\lambda_\text m \times \text T = \text b\ (\text{a constant})$

where b = 2.9 × 10^-3 m K. This is called Wien's displacement law. It shows that λ_m is inversely proportional to the absolute temperature T. That is why, when iron is heated, it first becomes light-red, then dark-red, then yellow and ultimately it becomes white.

Let us consider two slabs of the same area of cross-section A joined in series. Let l₁ and l₂ be their lengths and K₁ and K₂ their thermal conductivities. Let heat be allowed to flow through this combination. After the steady state is reached, let θ₁ be the temperature of the open face of the first slab, θ₂ the temperature of the open face of the second slab, and θ the steady temperature of the interface.

Let R₁ be the thermal resistance of the first slab and R₂ that of the second slab. In the steady state, the rate of flow of heat H in both the slabs is the same, since whatever heat passes through the first slab must pass through the second.

The rate of flow of heat in the first slab is

$\text H = \frac{\text K_1\text A(\theta_1 - \theta)}{\text l_1} \quad \text{or} \quad \frac{\theta_1 - \theta}{\text H} = \frac{\text l_1}{\text K_1\text A} = \text R_1 \qquad \ldots(\text i)$

Similarly, the rate of flow of heat in the second slab is

$\text H = \frac{\text K_2\text A(\theta - \theta_2)}{\text l_2} \quad \text{or} \quad \frac{\theta - \theta_2}{\text H} = \frac{\text l_2}{\text K_2\text A} = \text R_2 \qquad \ldots(\text{ii})$

Adding equations (i) and (ii), we get

$\frac{\theta_1 - \theta + \theta - \theta_2}{\text H} = \text R_1 + \text R_2$

$\frac{\theta_1 - \theta_2}{\text H} = \text R_1 + \text R_2$

But $\frac{\theta_1 - \theta_2}{\text H} = \text R$, the equivalent thermal resistance of the combined rod. Hence

$\text R = \text R_1 + \text R_2$

Thus, the equivalent thermal resistance of rods joined in series is equal to the sum of their individual thermal resistances. This is the series law of thermal resistances. For n such rods joined in series,

$\text R = \text R_1 + \text R_2 + \text R_3 + \ldots + \text R_\text n$

Let two slabs of the same length l be connected in parallel. Let A₁ and A₂ be their areas of cross-section and K₁ and K₂ their thermal conductivities. Let heat be allowed to flow from left to right, and in the steady state let the temperatures of the left and right cross-sectional surfaces of the slabs be θ₁ and θ₂.

Since the two slabs are connected side by side, the temperature difference across each of them is the same, but the rates of flow of heat in the two slabs are different.

The rate of flow of heat in the first slab is

$\text H_1 = \frac{\text K_1\text A_1(\theta_1 - \theta_2)}{\text l} = \frac{\theta_1 - \theta_2}{\text R_1}, \quad \text{where } \text R_1 = \frac{\text l}{\text K_1\text A_1}$

The rate of flow of heat in the second slab is

$\text H_2 = \frac{\text K_2\text A_2(\theta_1 - \theta_2)}{\text l} = \frac{\theta_1 - \theta_2}{\text R_2}, \quad \text{where } \text R_2 = \frac{\text l}{\text K_2\text A_2}$

The total rate of flow of heat in the combined slab is the sum of the two,

$\text H = \text H_1 + \text H_2 = \frac{\theta_1 - \theta_2}{\text R_1} + \frac{\theta_1 - \theta_2}{\text R_2}$

$\text H = \left[\frac{1}{\text R_1} + \frac{1}{\text R_2}\right](\theta_1 - \theta_2) \qquad \ldots(\text i)$

If the combined equivalent thermal resistance of both the slabs is R, then

$\text H = \frac{\theta_1 - \theta_2}{\text R} \qquad \ldots(\text{ii})$

Comparing equations (i) and (ii), we get

$\frac{1}{\text R} = \frac{1}{\text R_1} + \frac{1}{\text R_2}$

Hence, when two conductors are connected in parallel, the reciprocal of the equivalent thermal resistance is equal to the sum of the reciprocals of the individual thermal resistances. This is the parallel law of thermal resistances.

Let A be the area of cross-section of each plate, and let θ₁ and θ₂ be the temperatures of the outer faces of the first and the second plate respectively (θ₁ > θ₂). Let θ be the steady temperature of the interface.

Since the plates are joined in series, in the steady state the rate of flow of heat through both the plates is the same,

$\text H = \frac{\text K_1\text A(\theta_1 - \theta)}{\text l_1} = \frac{\text K_2\text A(\theta - \theta_2)}{\text l_2}$

From the first two members,

$\frac{\text{Hl}_1}{\text K_1\text A} = \theta_1 - \theta \qquad \ldots(\text i)$

and from the first and the third members,

$\frac{\text{Hl}_2}{\text K_2\text A} = \theta - \theta_2 \qquad \ldots(\text{ii})$

Adding equations (i) and (ii), the interface temperature θ cancels out,

$\frac{\text H}{\text A}\left(\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}\right) = \theta_1 - \theta_2$

$\text H = \frac{\text A(\theta_1 - \theta_2)}{\left(\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}\right)} \qquad \ldots(\text{iii})$

Now, if K be the equivalent thermal conductivity of the composite slab, then treating the combination as a single slab of total thickness (l₁ + l₂),

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l_1 + \text l_2} \qquad \ldots(\text{iv})$

Comparing equations (iii) and (iv),

$\frac{\text{KA}(\theta_1 - \theta_2)}{\text l_1 + \text l_2} = \frac{\text A(\theta_1 - \theta_2)}{\left(\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}\right)}$

$\text K = \frac{\text l_1 + \text l_2}{\left(\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}\right)}$

which is the required result.

For two plates of the same area of cross-section joined in series, the equivalent thermal conductivity is

$\text K = \frac{\text l_1 + \text l_2}{\left(\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}\right)}$

Here both the plates have the same thickness, so l₁ = l₂ = l (say). Substituting this,

$\text K = \frac{\text l + \text l}{\left(\frac{\text l}{\text K_1} + \frac{\text l}{\text K_2}\right)} = \frac{2\text l}{\text l\left(\frac{1}{\text K_1} + \frac{1}{\text K_2}\right)}$

Cancelling l,

$\text K = \frac{2}{\frac{1}{\text K_1} + \frac{1}{\text K_2}} = \frac{2}{\frac{\text K_2 + \text K_1}{\text K_1\text K_2}}$

$\text K = \frac{2\text K_1\text K_2}{\text K_1 + \text K_2}$

which is the required result.

Let A be the area of cross-section of each plate. Let θ₁ be the temperature of the open face of the first plate and θ₂ that of the open face of the second plate (θ₁ > θ₂), and let θ be the steady temperature of the plane of contact.

Temperature at the plane of contact : In the steady state, the rate of flow of heat in both the plates is the same. Hence

$\text H = \frac{\text Q}{\text t} = \frac{\text K_1\text A(\theta_1 - \theta)}{\text l_1} = \frac{\text K_2\text A(\theta - \theta_2)}{\text l_2} \qquad \ldots(\text i)$

Cancelling A,

$\frac{\text K_1}{\text l_1}(\theta_1 - \theta) = \frac{\text K_2}{\text l_2}(\theta - \theta_2)$

$\frac{\text K_1\theta_1}{\text l_1} + \frac{\text K_2\theta_2}{\text l_2} = \theta\left(\frac{\text K_1}{\text l_1} + \frac{\text K_2}{\text l_2}\right)$

$\theta = \frac{\frac{\text K_1\theta_1}{\text l_1} + \frac{\text K_2\theta_2}{\text l_2}}{\frac{\text K_1}{\text l_1} + \frac{\text K_2}{\text l_2}} = \frac{\text K_1\theta_1\text l_2 + \text K_2\theta_2\text l_1}{\text K_1\text l_2 + \text K_2\text l_1} \qquad \ldots(\text{ii})$

This is the required expression for the temperature of the interface. If the lengths of both the plates are the same (l₁ = l₂ = l), then

$\theta = \frac{\text K_1\theta_1 + \text K_2\theta_2}{\text K_1 + \text K_2}$

Equivalent thermal conductivity : Substituting the value of θ from equation (ii) in equation (i), the rate of flow of heat in the composite slab is

$\text H = \frac{\text K_1\text A}{\text l_1}\left[\theta_1 - \frac{\text K_1\theta_1\text l_2 + \text K_2\theta_2\text l_1}{\text K_1\text l_2 + \text K_2\text l_1}\right]$

$= \frac{\text K_1\text A}{\text l_1}\left[\frac{\text K_2\text l_1\theta_1 - \text K_2\theta_2\text l_1}{\text K_1\text l_2 + \text K_2\text l_1}\right]$

$= \frac{\text K_1\text K_2\text A(\theta_1 - \theta_2)}{\text K_1\text l_2 + \text K_2\text l_1} = \frac{\text A(\theta_1 - \theta_2)}{\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}} \qquad \ldots(\text{iii})$

If K be the equivalent thermal conductivity of the composite slab of total length (l₁ + l₂), then

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l_1 + \text l_2} \qquad \ldots(\text{iv})$

Comparing equations (iii) and (iv),

$\text K = \frac{\text l_1 + \text l_2}{\frac{\text l_1}{\text K_1} + \frac{\text l_2}{\text K_2}} = \frac{\text K_1\text K_2(\text l_1 + \text l_2)}{\text K_2\text l_1 + \text K_1\text l_2}$

If the lengths of the two plates are equal (l₁ = l₂ = l), this reduces to

$\text K = \frac{2\text K_1\text K_2}{\text K_1 + \text K_2}$

Given,

• Reading of the barometer at 27°C, l = 75 cm
• Rise in temperature, Δt = 27 − 0 = 27°C
• Coefficient of linear expansion of brass, α = 2.0 × 10^-5 °C^-1

The brass scale gives correct reading at 0°C. At 27°C the scale itself expands, so each division of the scale becomes longer than a true centimetre. The true length of one division of the scale at 27°C is

$1 \times (1 + \alpha\Delta \text t) = 1 + (2.0 \times 10^{-5} \times 27)$

Hence the correct height of the mercury column is

$$\begin{aligned}\text l' = 75 \times (1 + \alpha\Delta \text t) \\[1em] = 75 \times \left[1 + (2.0 \times 10^{-5} \times 27)\right] \\[1em] = 75 \times (1 + 5.4 \times 10^{-4}) \\[1em] = 75 + 0.0405 = 75.04\ \text{cm}\end{aligned}$$

Hence, the correct atmospheric pressure at 27°C is 75.04 cm of mercury.

Given,

• Length of the zinc rod at 0°C, l = 100 cm
• Rise in temperature, Δt = 100 − 0 = 100°C
• α_iron = 1.2 × 10^-5 °C^-1, α_zinc = 2.6 × 10^-5 °C^-1

At 100°C the true length of the zinc rod is

$\text l_{zinc} = 100\left[1 + (2.6 \times 10^{-5} \times 100)\right]\ \text{cm}$

The iron scale also expands, so the true length of one division of the scale at 100°C is

$= 1 \times \left[1 + (1.2 \times 10^{-5} \times 100)\right]\ \text{cm}$

The reading of the scale is the true length of the rod divided by the true length of one division,

$\text{Reading} = \frac{100\left[1 + (2.6 \times 10^{-5} \times 100)\right]}{1 + (1.2 \times 10^{-5} \times 100)}$

Since the expansions are very small, this may be written as

$\text{Reading} = 100\left[1 + (\alpha_{zinc} - \alpha_{iron})\Delta \text t\right]$

$$\begin{aligned}= 100\left[1 + (2.6 \times 10^{-5} - 1.2 \times 10^{-5}) \times 100\right] \\[1em] = 100\left[1 + (1.4 \times 10^{-5} \times 100)\right] \\[1em] = 100(1 + 1.4 \times 10^{-3}) = 100.14\ \text{cm}\end{aligned}$$

Hence, the length of the zinc rod as measured by the iron scale at 100°C is 100.14 cm.

Given,

• Diameter of the steel cylinder at 30°C = 1.000 cm
• Diameter of the hole at 30°C, d = 0.99970 cm
• Coefficient of linear expansion of steel, α = 1.1 × 10^-5 °C^-1
• Initial temperature, t₁ = 30°C

On heating the plate, the hole expands in exactly the same way as a circular plate of steel filling the hole would expand. The cylinder will just fit into the hole when the diameter of the hole becomes 1.000 cm.

If Δt be the required rise in temperature, then

$\text d(1 + \alpha\Delta \text t) = 1.000$

$0.99970\left[1 + (1.1 \times 10^{-5})\Delta \text t\right] = 1.000$

$1 + (1.1 \times 10^{-5})\Delta \text t = \frac{1.000}{0.99970} = 1.00030$

$(1.1 \times 10^{-5})\Delta \text t = 3.0 \times 10^{-4}$

$\Delta \text t = \frac{3.0 \times 10^{-4}}{1.1 \times 10^{-5}} = 27.28^\circ \text C$

The final temperature of the plate is

$\text t_2 = \text t_1 + \Delta \text t = 30 + 27.28 = 57.28^\circ \text C$

Hence, the plate should be heated to about 57.3°C.

Given,

• Temperatures of A, B and C are 12°C, 19°C and 28°C respectively
• Masses of the three liquids are equal, say m
• On mixing A and B, the temperature is 16°C
• On mixing B and C, the temperature is 23°C

Let c_A, c_B and c_C be the specific heats of the three liquids.

Mixing A and B : Here A is heated from 12°C to 16°C and B is cooled from 19°C to 16°C. By the principle of calorimetry, heat gained = heat lost,

$\text m\text c_\text A(16 - 12) = \text m\text c_\text B(19 - 16)$

$4\text c_\text A = 3\text c_\text B \quad \Rightarrow \quad \text c_\text A = \frac{3}{4}\text c_\text B \qquad \ldots(\text i)$

Mixing B and C : Here B is heated from 19°C to 23°C and C is cooled from 28°C to 23°C,

$\text m\text c_\text B(23 - 19) = \text m\text c_\text C(28 - 23)$

$4\text c_\text B = 5\text c_\text C \quad \Rightarrow \quad \text c_\text C = \frac{4}{5}\text c_\text B \qquad \ldots(\text{ii})$

Mixing A and C : Let the final temperature be T. Here A is heated from 12°C to T and C is cooled from 28°C to T,

$\text m\text c_\text A(\text T - 12) = \text m\text c_\text C(28 - \text T)$

Substituting the values of c_A and c_C from equations (i) and (ii),

$\frac{3}{4}\text c_\text B(\text T - 12) = \frac{4}{5}\text c_\text B(28 - \text T)$

Multiplying throughout by 20 and cancelling c_B,

$15(\text T - 12) = 16(28 - \text T)$

$15\text T - 180 = 448 - 16\text T$

$31\text T = 628 \quad \Rightarrow \quad \text T = 20.26^\circ \text C$

Hence, the temperature when A and C are mixed is about 20.3°C.

Given,

• Length of the rod, l = 75 cm = 0.75 m
• Area of cross-section, A = 10^-2 m²
• Temperature difference, θ₁ − θ₂ = 90°C
• Thermal conductivity, K = 80 cal m^-1 s^-1 °C^-1

(i) The thermal resistance of the rod is

$$\begin{aligned}\text R = \frac{\text l}{\text{KA}} = \frac{0.75}{80 \times 10^{-2}} \\[1em] = \frac{0.75}{0.8} = 0.94\ ^\circ\text C\text s\text{cal}^{-1}\end{aligned}$$

(ii) The rate of flow of heat is

$$\begin{aligned}\text H = \frac{\theta_1 - \theta_2}{\text R} = \frac{90}{0.9375} \\[1em] = 96\ \text{cal s}^{-1}\end{aligned}$$

Hence, the thermal resistance of the rod is 0.94 °C s cal^-1 and the rate of flow of heat is 96 cal s^-1.

Given,

• Length of the copper rod, l = 0.25 m
• Area of cross-section, A = 10^-4 m²
• Temperatures of the ends, θ₁ = 125°C and θ₂ = 0°C
• Thermal conductivity of copper, K = 92 cal s^-1 m^-1 K^-1

(i) The rate of flow of heat in the steady state is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

Substituting the values,

$$\begin{aligned}\text H = \frac{92 \times 10^{-4} \times (125 - 0)}{0.25} \\[1em] = \frac{1.15}{0.25} = 4.6\ \text{cal s}^{-1}\end{aligned}$$

(ii) Since no heat escapes from the sides, the temperature falls uniformly along the rod. The temperature gradient is

$\frac{\theta_1 - \theta_2}{\text l} = \frac{125 - 0}{0.25} = 500\ ^\circ\text C\text m^{-1}$

At a point 0.1 m from the higher temperature end, the fall in temperature is 500 × 0.1 = 50°C. Hence the temperature at that point is

$\theta = 125 - 50 = 75^\circ \text C$

Hence, the rate of heat flow is 4.6 cal s^-1 and the temperature at the given point is 75°C.

Given,

• Thickness of the sheet, l = 0.4 cm = 4 × 10^-3 m
• Temperature difference, θ₁ − θ₂ = 32°C
• Area, A = 5 cm² = 5 × 10^-4 m²
• Rate of transmission of heat = 200 kcal per hour

The rate of flow of heat in kilocalorie per second is

$\text H = \frac{200}{3600} = 5.556 \times 10^{-2}\ \text{kcal s}^{-1}$

From the equation of heat conduction, $\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$, we get

$\text K = \frac{\text H\text l}{\text A(\theta_1 - \theta_2)}$

Substituting the values,

$$\begin{aligned}\text K = \frac{(5.556 \times 10^{-2}) \times (4 \times 10^{-3})}{(5 \times 10^{-4}) \times 32} \\[1em] = \frac{2.222 \times 10^{-4}}{1.6 \times 10^{-2}} \\[1em] = 1.39 \times 10^{-2}\ \text{kcal m}^{-1}\text s^{-1} ^{\circ} \text C^{-1}\end{aligned}$$

Hence, the coefficient of thermal conductivity of nickel is 1.39 × 10^-2 kcal m^-1 s^-1 °C^-1.

Given,

• Side of the cubical pot, a = 10 cm = 0.1 m
• Thickness of the pot, l = 0.2 cm = 2 × 10^-3 m
• Temperature difference, θ₁ − θ₂ = 100 − 0 = 100°C
• Thermal conductivity, K = 0.02 cal m^-1 s^-1 °C^-1
• Mass of ice, m = 1 kg; latent heat of ice, L = 80 kcal kg^-1

The pot has six faces, so the total surface area through which heat enters is

$\text A = 6\text a^2 = 6 \times (0.1)^2 = 0.06\ \text m^2$

The rate of flow of heat into the pot is

$$\begin{aligned}\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l} = \frac{0.02 \times 0.06 \times 100}{2 \times 10^{-3}} \\[1em] = \frac{0.12}{2 \times 10^{-3}} = 60\ \text{cal s}^{-1}\end{aligned}$$

The total heat required to melt 1 kg of ice is

$\text Q = \text{mL} = 1 \times 80 = 80\ \text{kcal} = 80000\ \text{cal}$

Therefore the time taken is

$\text t = \frac{\text Q}{\text H} = \frac{80000}{60} = 1333.3\ \text s$

$= \frac{1333.3}{60} = 22.2\ \text{minutes}$

Hence, the time taken to melt the total ice is about 1333 s, that is, 22.2 minutes.

Given,

• First rod : l₁ = 0.6 m, temperature difference = 90 − 60 = 30°C
• Second rod : l₂ = 0.8 m, temperature difference = 150 − 110 = 40°C
• Both rods are of the same metal and of uniform cross-section

Since the rods are of the same metal and of the same area of cross-section, K and A are the same for both. Hence the rate of flow of heat is proportional to the temperature gradient,

$\text H = \text{KA}\frac{\Delta \theta}{\text l} \quad \Rightarrow \quad \text H \propto \frac{\Delta \theta}{\text l}$

For the first rod :

$\frac{\Delta \theta_1}{\text l_1} = \frac{30}{0.6} = 50\ ^\circ\text C\text m^{-1}$

For the second rod :

$\frac{\Delta \theta_2}{\text l_2} = \frac{40}{0.8} = 50\ ^\circ\text C\text m^{-1}$

Since the temperature gradients in the two rods are equal,

$\text H_1 = \text H_2$

Hence, neither rod has a higher rate of heat-conduction; the rate of heat-conduction is the same in both the rods.

Given,

• Length of the first rod, l₁ = 0.5 m, temperature difference Δθ₁ = 40°C
• Length of the second rod, l₂ = 0.8 m
• The rate of heat conduction is the same in both the rods

Since the rods are of the same metal and of the same cross-section, K and A are the same for both. For equal rates of heat conduction,

$\frac{\text{KA}\Delta \theta_1}{\text l_1} = \frac{\text{KA}\Delta \theta_2}{\text l_2}$

$\frac{\Delta \theta_1}{\text l_1} = \frac{\Delta \theta_2}{\text l_2}$

Substituting the values,

$\frac{40}{0.5} = \frac{\Delta \theta_2}{0.8}$

$\Delta \theta_2 = \frac{40 \times 0.8}{0.5} = 64^\circ \text C$

Hence, the temperature difference across the second rod is 64°C.

Given,

• Ratio of the areas of cross-section, A₁ : A₂ = 1 : 2
• Ratio of the thermal conductivities, K₁ : K₂ = 4 : 3
• The temperature difference Δθ is the same for both, and the rates of heat conduction are equal

The rate of flow of heat is

$\text H = \frac{\text{KA}\Delta \theta}{\text l}$

Since H₁ = H₂ and Δθ is the same for both,

$\frac{\text K_1\text A_1}{\text l_1} = \frac{\text K_2\text A_2}{\text l_2}$

$\frac{\text l_1}{\text l_2} = \frac{\text K_1\text A_1}{\text K_2\text A_2}$

Substituting the given ratios,

$\frac{\text l_1}{\text l_2} = \frac{4 \times 1}{3 \times 2} = \frac{4}{6} = \frac{2}{3}$

Hence, the ratio of the lengths of the rods is 2 : 3.

Given,

• Thermal conductivities of the two vessels, K₁ = 5 × 10^-2 and K₂ = 3 × 10^-2 kcal m^-1 s^-1 °C^-1
• The vessels are identical in size and shape, and contain equal masses of boiled water

Since both the vessels contain equal masses of boiled water, the quantity of heat Q required to convert the whole of the water into steam is the same for both.

The heat conducted through the base of a vessel in time t is

$\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

The vessels are identical in size and shape, and are placed on the same heater, so A, l and (θ₁ − θ₂) are the same for both. Since Q is also the same,

$\text K_1\text t_1 = \text K_2\text t_2$

$\frac{\text t_1}{\text t_2} = \frac{\text K_2}{\text K_1} = \frac{3 \times 10^{-2}}{5 \times 10^{-2}} = \frac{3}{5}$

Hence, the ratio of the timings taken by the water in the two vessels to convert into steam is 3 : 5.

Given,

• Surface area of each block, A = 100 cm² = 10^-2 m²
• Thickness of wood, l₁ = 25 cm = 0.25 m; thickness of cork, l₂ = 4 cm = 0.04 m
• K_wood = 2.0 × 10^-5 kcal m^-1 s^-1 °C^-1; K_cork = 4.0 × 10^-5 kcal m^-1 s^-1 °C^-1

The thermal resistance of a slab is $\text R = \frac{\text l}{\text{KA}}$.

Thermal resistance of the wood block :

$$\begin{aligned}\text R_1 = \frac{0.25}{(2.0 \times 10^{-5}) \times 10^{-2}} = \frac{0.25}{2.0 \times 10^{-7}} \\[1em] = 1.25 \times 10^6\ ^\circ\text C\text s\text{kcal}^{-1}\end{aligned}$$

Thermal resistance of the cork block :

$$\begin{aligned}\text R_2 = \frac{0.04}{(4.0 \times 10^{-5}) \times 10^{-2}} = \frac{0.04}{4.0 \times 10^{-7}} \\[1em] = 1.0 \times 10^5\ ^\circ\text C\text s\text{kcal}^{-1}\end{aligned}$$

The two blocks are placed in contact, that is, joined in series. Hence the combined thermal resistance is

$$\begin{aligned}\text R = \text R_1 + \text R_2 = (1.25 \times 10^6) + (1.0 \times 10^5) \\[1em] = 1.35 \times 10^6\ ^\circ\text C\text s\text{kcal}^{-1}\end{aligned}$$

Hence, the combined thermal resistance is 1.35 × 10⁶ °C s kcal^-1.

Given,

• Initial temperature, T₁ = 273°C = 273 + 273 = 546 K
• Final temperature, T₂ = 819°C = 819 + 273 = 1092 K

By Stefan's law, the rate of emission of energy per unit area of a black body is proportional to the fourth power of its absolute temperature,

$\frac{\text Q}{\text t} = \sigma \text T^4$

Therefore,

$\frac{(\text Q/\text t)_2}{(\text Q/\text t)_1} = \left(\frac{\text T_2}{\text T_1}\right)^4$

Substituting the values,

$= \left(\frac{1092}{546}\right)^4 = (2)^4 = 16$

Hence, the rate of emission of energy becomes 16 times its original value.

Given,

• Initial temperature, T₁ = 127°C = 127 + 273 = 400 K
• Initial rate of radiation, E₁ = 1.0 × 10⁶ J s^-1 m^-2
• Final rate of radiation, E₂ = 16.0 × 10⁶ J s^-1 m^-2

By Stefan's law, E = σT⁴, so

$\frac{\text E_2}{\text E_1} = \left(\frac{\text T_2}{\text T_1}\right)^4$

Substituting the values,

$\frac{16.0 \times 10^6}{1.0 \times 10^6} = \left(\frac{\text T_2}{400}\right)^4$

$16 = \left(\frac{\text T_2}{400}\right)^4 \quad \Rightarrow \quad \frac{\text T_2}{400} = (16)^{\frac{1}{4}} = 2$

$\text T_2 = 800\ \text K = 800 - 273 = 527^\circ \text C$

Hence, the required temperature is 800 K, that is, 527°C.

Given,

• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Initial rate of radiation, P₁ = 1 kJ s^-1
• Final rate of radiation, P₂ = 16 kJ s^-1

By Stefan's law, the energy radiated per second is proportional to the fourth power of the absolute temperature,

$\frac{\text P_2}{\text P_1} = \left(\frac{\text T_2}{\text T_1}\right)^4$

Substituting the values,

$\frac{16}{1} = \left(\frac{\text T_2}{300}\right)^4$

$\frac{\text T_2}{300} = (16)^{\frac{1}{4}} = 2$

$\text T_2 = 600\ \text K = 600 - 273 = 327^\circ \text C$

Hence, the black body will radiate 16 kJ per second at 600 K, that is, at 327°C.

Given,

• Temperature of the first black body, T₁ = 727°C = 727 + 273 = 1000 K
• Temperature of the second black body, T₂ = 327°C = 327 + 273 = 600 K

By Stefan's law, the energy radiated per second per unit area is E = σT⁴. Hence

$\frac{\text E_1}{\text E_2} = \left(\frac{\text T_1}{\text T_2}\right)^4$

Substituting the values,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \left(\frac{1000}{600}\right)^4 = \left(\frac{5}{3}\right)^4 \\[1em] = \frac{625}{81} = 7.72\end{aligned}$$

Hence, the ratio of the energy radiated per second by them is 625 : 81, that is, about 7.72 : 1.

Given,

• Temperature of the first body, T₁ = 327°C = 600 K
• Temperature of the second body, T₂ = 427°C = 700 K
• Temperature of the surroundings, T₀ = 27°C = 300 K
• Both bodies are assumed to have equal surface areas and equal emissivities

When a body at absolute temperature T is surrounded by an enclosure at temperature T₀, the net heat energy lost per second per unit area is

$\frac{\text Q}{\text t} = \sigma(\text T^4 - \text T_0^4)$

For the first body :

$\left(\frac{\text Q}{\text t}\right)_1 \propto (600)^4 - (300)^4$

Taking out (100)⁴ as a common factor,

$\propto (6^4 - 3^4) = 1296 - 81 = 1215$

For the second body :

$\left(\frac{\text Q}{\text t}\right)_2 \propto (7^4 - 3^4) = 2401 - 81 = 2320$

Therefore,

$\frac{(\text Q/\text t)_1}{(\text Q/\text t)_2} = \frac{1215}{2320} = \frac{243}{464} = 0.524$

Hence, the rates of heat loss from the two bodies are in the ratio 243 : 464.

Given,

• Temperature of the filament, T = 2450 K
• Effective area, A = 0.408 cm² = 0.408 × 10^-4 m²
• Emissive power, e = 0.30
• Stefan's constant, σ = 5.67 × 10^-8 W m^-2 K^-4

The energy radiated per second by a body of emissive power e is

$\text P = \text e\sigma\text A\text T^4$

Substituting the values,

$\text P = 0.30 \times (5.67 \times 10^{-8}) \times (0.408 \times 10^{-4}) \times (2450)^4$

Now, (2450)⁴ = 3.603 × 10¹³. Therefore,

$$\begin{aligned}\text P = 0.30 \times (5.67 \times 10^{-8}) \times (4.08 \times 10^{-5}) \times (3.603 \times 10^{13}) \\[1em] = (6.940 \times 10^{-13}) \times (3.603 \times 10^{13}) \\[1em] = 25.0\ \text W\end{aligned}$$

Hence, the wattage of the lamp is about 25 W.

Given,

• Radius of the sphere, r = 5 cm = 0.05 m
• Temperature, T = 527°C = 527 + 273 = 800 K
• Emissive power, e = 0.04
• Stefan's constant, σ = 5.67 × 10^-8 W m^-2 K^-4

The surface area of the sphere is

$$\begin{aligned}\text A = 4\pi \text r^2 = 4 \times 3.14 \times (0.05)^2 \\[1em] = 3.142 \times 10^{-2}\ \text m^2\end{aligned}$$

The energy radiated per second is

$\text P = \text e\sigma\text A\text T^4$

Now, (800)⁴ = 4.096 × 10¹¹. Substituting the values,

$$\begin{aligned}\text P = 0.04 \times (5.67 \times 10^{-8}) \times (3.142 \times 10^{-2}) \times (4.096 \times 10^{11}) \\[1em] = (7.125 \times 10^{-11}) \times (4.096 \times 10^{11}) \\[1em] = 29.2\ \text{J s}^{-1}\end{aligned}$$

Hence, the energy of the emitted radiation is about 29.2 J per second.

Given,

• Temperature of the surface, T = 227°C = 227 + 273 = 500 K
• Emissive power, e = 0.60
• Stefan's constant, σ = 5.67 × 10^-8 W m^-2 K^-4

The energy emitted per second per unit area is

$\text E = \text e\sigma\text T^4$

Substituting the values, with (500)⁴ = 6.25 × 10¹⁰,

$$\begin{aligned}\text E = 0.60 \times (5.67 \times 10^{-8}) \times (6.25 \times 10^{10}) \\[1em] = (3.402 \times 10^{-8}) \times (6.25 \times 10^{10}) \\[1em] = 2126.25\ \text{W m}^{-2}\end{aligned}$$

Since 1 m² = 10⁴ cm²,

$\text E = \frac{2126.25}{10^4} = 0.213\ \text{W cm}^{-2}$

Hence, the rate of emission of energy is 0.213 W cm^-2.

Given,

• Volume of the bulb, V = 0.300 cm³
• Diameter of the stem = 0.0100 cm, so radius r = 0.00500 cm
• Rise in temperature, Δt = 15°C
• Coefficient of volume expansion of mercury, γ = 1.82 × 10^-4 °C^-1

The increase in the volume of the mercury is

$$\begin{aligned}\Delta \text V = \text V\gamma\Delta \text t \\[1em] = 0.300 \times (1.82 \times 10^{-4}) \times 15 \\[1em] = 8.19 \times 10^{-4}\ \text{cm}^3\end{aligned}$$

This extra volume of mercury rises into the stem. The area of cross-section of the stem is

$$\begin{aligned}\text a = \pi \text r^2 = 3.14 \times (0.00500)^2 \\[1em] = 7.854 \times 10^{-5}\ \text{cm}^2\end{aligned}$$

If h be the rise of the mercury meniscus in the stem, then ΔV = a h, so

$$\begin{aligned}\text h = \frac{\Delta \text V}{\text a} = \frac{8.19 \times 10^{-4}}{7.854 \times 10^{-5}} \\[1em] = 10.43\ \text{cm}\end{aligned}$$

Hence, the mercury meniscus rises through 10.43 cm in the stem.

Given,

• Volume of the glass vessel, V = 256 cm³
• Rise in temperature, Δt = 100 − 20 = 80°C
• Coefficient of linear expansion of glass, α = 4 × 10^-6 °C^-1
• Coefficient of volume expansion of mercury, γ_m = 1.8 × 10^-4 °C^-1

The coefficient of volume expansion of glass is

$\gamma_\text g = 3\alpha = 3 \times (4 \times 10^{-6}) = 1.2 \times 10^{-5}\ ^\circ\text C^{-1}$

On heating, both the mercury and the vessel expand. The volume of mercury that overflows is the difference between the expansion of the mercury and the expansion of the vessel, that is, the apparent expansion of the mercury,

$\Delta \text V = \text V(\gamma_\text m - \gamma_\text g)\Delta \text t$

Substituting the values,

$$\begin{aligned}\Delta \text V = 256 \times (1.8 \times 10^{-4} - 1.2 \times 10^{-5}) \times 80 \\[1em] = 256 \times (1.68 \times 10^{-4}) \times 80 \\[1em] = 3.44\ \text{cm}^3\end{aligned}$$

Hence, 3.44 cm³ of mercury will overflow.

Given,

• Length of the glass tube, L = 133 cm
• Coefficient of volume expansion of glass, γ_g = 2.6 × 10^-5 °C^-1
• Coefficient of volume expansion of mercury, γ_m = 18.2 × 10^-5 °C^-1

Let a be the area of cross-section of the tube at a given temperature and l the length of the mercury column. Then the volume of the tube is L a and the volume of the mercury is l a.

The volume of the empty space above the mercury will remain the same at all temperatures only if the increase in the volume of the mercury is exactly equal to the increase in the volume of the tube. Hence, for a rise in temperature Δt,

$(\text l\text a)\gamma_\text m\Delta \text t = (\text L\text a)\gamma_\text g\Delta \text t$

Cancelling a and Δt,

$\text l\gamma_\text m = \text L\gamma_\text g$

$\text l = \frac{\text L\gamma_\text g}{\gamma_\text m} = \frac{133 \times (2.6 \times 10^{-5})}{18.2 \times 10^{-5}}$

$= \frac{133 \times 2.6}{18.2} = \frac{345.8}{18.2} = 19\ \text{cm}$

Hence, the length of the mercury column should be 19 cm.

Given,

• Diameter of the sphere = 7.0 cm, so radius r = 3.5 cm
• Mass of the sphere, m = 266.5 g
• Temperature at which the sphere just sinks, t = 35°C
• Density of the liquid at 0°C, ρ = 1.527 g cm^-3

The volume of the sphere is

$$\begin{aligned}\text V = \frac{4}{3}\pi \text r^3 = \frac{4}{3} \times 3.1416 \times (3.5)^3 \\[1em] = \frac{4}{3} \times 3.1416 \times 42.875 = 179.59\ \text{cm}^3\end{aligned}$$

The density of the sphere is

$\rho_\text s = \frac{\text m}{\text V} = \frac{266.5}{179.59} = 1.4839\ \text{g cm}^{-3}$

The sphere just begins to sink when the density of the liquid falls to the density of the sphere. Hence the density of the liquid at 35°C is

$\rho' = 1.4839\ \text{g cm}^{-3}$

When a liquid is heated its volume increases and its density decreases. Neglecting the smaller terms, the density of a liquid at temperature t is related to its density at 0°C by

$\rho' = \rho(1 - \gamma\text t)$

Substituting the values,

$1.4839 = 1.527(1 - 35\gamma)$

$1 - 35\gamma = \frac{1.4839}{1.527} = 0.9718$

$35\gamma = 1 - 0.9718 = 0.0282$

$\gamma = \frac{0.0282}{35} = 8.0 \times 10^{-4}\ ^\circ\text C^{-1}$

Hence, the coefficient of cubical expansion of the liquid is 8.0 × 10^-4 °C^-1.

Given,

• Mass of the aluminium sphere, m = 0.047 kg at 100°C
• Mass of the copper calorimeter, m_c = 0.14 kg
• Mass of water, m_w = 0.25 kg at 20°C
• Final steady temperature = 23°C
• Specific heat of water, c_w = 4.18 × 10³ J kg^-1 °C^-1
• Specific heat of copper, c_c = 0.386 × 10³ J kg^-1 °C^-1

Let c be the specific heat of aluminium.

The heat lost by the aluminium sphere in cooling from 100°C to 23°C is

$\text Q_1 = \text m\text c(100 - 23) = 0.047 \times \text c \times 77 = 3.619\text c$

The heat gained by the water and the calorimeter in warming from 20°C to 23°C is

$\text Q_2 = (\text m_\text w \text c_\text w + \text m_\text c \text c_\text c)(23 - 20)$

$= \left[(0.25 \times 4.18 \times 10^3) + (0.14 \times 0.386 \times 10^3)\right] \times 3$

$= (1045 + 54.04) \times 3 = 1099.04 \times 3 = 3297.1\ \text J$

By the principle of calorimetry, heat lost = heat gained,

$3.619\text c = 3297.1$

$\text c = \frac{3297.1}{3.619} = 911.1\ \text{J kg}^{-1}\space^\circ\text C^{-1}$

Hence, the specific heat of aluminium is about 0.911 × 10³ J kg^-1 °C^-1.

Given,

• Mass of water in the calorimeter, m₁ = 75 g at 15°C
• Mass of hot water poured in, m₂ = 50 g at 100°C
• Final temperature of the mixture = 25°C
• Specific heat of water, c = 1 cal g^-1 °C^-1

Let W be the water equivalent of the calorimeter.

The heat lost by the hot water in cooling from 100°C to 25°C is

$\text Q_1 = \text m_2\text c(100 - 25) = 50 \times 1 \times 75 = 3750\ \text{cal}$

The heat gained by the cold water and the calorimeter in warming from 15°C to 25°C is

$\text Q_2 = (\text m_1 + \text W)\text c(25 - 15) = (75 + \text W) \times 1 \times 10$

By the principle of calorimetry, heat lost = heat gained,

$10(75 + \text W) = 3750$

$75 + \text W = 375$

$\text W = 300\ \text g$

Hence, the water equivalent of the calorimeter is 300 g.

Given,

• Mass of ice, m₁ = 0.15 kg at 0°C
• Mass of water, m₂ = 0.30 kg at 50°C
• Final temperature of the mixture = 6.7°C
• Specific heat of water, c = 4.186 × 10³ J kg^-1 K^-1

Let L be the latent heat of melting of ice.

The heat lost by the water in cooling from 50°C to 6.7°C is

$$\begin{aligned}\text Q_1 = \text m_2\text c(50 - 6.7) \\[1em] = 0.30 \times (4.186 \times 10^3) \times 43.3 = 54376\ \text J\end{aligned}$$

The heat gained by the ice is used up first in melting the ice at 0°C and then in warming the water so formed from 0°C to 6.7°C,

$\text Q_2 = \text m_1\text L + \text m_1\text c(6.7 - 0)$

$$\begin{aligned}= 0.15\text L + \left[0.15 \times (4.186 \times 10^3) \times 6.7\right] \\[1em] = 0.15\text L + 4207\ \text J\end{aligned}$$

By the principle of calorimetry, heat lost = heat gained,

$0.15\text L + 4207 = 54376$

$0.15\text L = 50169$

$\text L = \frac{50169}{0.15} = 3.34 \times 10^5\ \text{J kg}^{-1}$

Hence, the latent heat of melting of ice is 3.34 × 10⁵ J kg^-1.

Given,

• Mass of ice, m = 3 kg at − 12°C
• Specific heat of ice, c_i = 2.100 × 10³ J kg^-1 K^-1
• Specific heat of water, c_w = 4.186 × 10³ J kg^-1 K^-1
• Latent heat of fusion of ice, L_f = 3.35 × 10⁵ J kg^-1
• Latent heat of steam, L_v = 2.256 × 10⁶ J kg^-1

The whole process takes place in four stages.

Stage 1 — Ice from − 12°C to 0°C :

$$\begin{aligned}\text Q_1 = \text m\text c_\text i\Delta \text T = 3 \times (2.100 \times 10^3) \times 12 \\[1em] = 75600\ \text J\end{aligned}$$

Stage 2 — Melting of ice at 0°C :

$$\begin{aligned}\text Q_2 = \text m\text L_\text f = 3 \times (3.35 \times 10^5) \\[1em] = 1.005 \times 10^6\ \text J\end{aligned}$$

Stage 3 — Water from 0°C to 100°C :

$$\begin{aligned}\text Q_3 = \text m\text c_\text w\Delta \text T = 3 \times (4.186 \times 10^3) \times 100 \\[1em] = 1.2558 \times 10^6\ \text J\end{aligned}$$

Stage 4 — Vaporisation of water at 100°C :

$$\begin{aligned}\text Q_4 = \text m\text L_\text v = 3 \times (2.256 \times 10^6) \\[1em] = 6.768 \times 10^6\ \text J\end{aligned}$$

The total heat required is

$\text Q = \text Q_1 + \text Q_2 + \text Q_3 + \text Q_4$

$= 75600 + (1.005 \times 10^6) + (1.2558 \times 10^6) + (6.768 \times 10^6)$

$= 9.10 \times 10^6\ \text J$

Hence, the heat required is 9.10 × 10⁶ J.

Note: The question states the initial temperature of the ice as 12°C, but ice cannot exist at a temperature above 0°C, so it has been taken as − 12°C. Also, the specific heat of ice is printed as 2.100 × 10⁵ J kg^-1 K^-1, which is not a physically correct value; the standard value 2.100 × 10³ J kg^-1 K^-1 has been used.

Given,

• Thermal conductivity of copper, K = 9.2 × 10^-2 kcal s^-1 m^-1 °C^-1
• Length of the rod, l = 25 cm = 0.25 m
• Area of cross-section, A = 1 cm² = 10^-4 m²
• Temperature difference, θ₁ − θ₂ = 125 − 0 = 125°C

The rate of transfer of heat in the steady state is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

Substituting the values,

$$\begin{aligned}\text H = \frac{(9.2 \times 10^{-2}) \times 10^{-4} \times 125}{0.25} \\[1em] = \frac{1.15 \times 10^{-3}}{0.25} \\[1em] = 4.6 \times 10^{-3}\ \text{kcal s}^{-1} = 4.6\ \text{cal s}^{-1}\end{aligned}$$

Hence, the rate of transfer of heat is 4.6 cal s^-1.

Given,

• Area of cross-section, A = 100 cm²
• Thickness of the plate, l = 2 cm
• Coefficient of thermal conductivity, K = 2 × 10^-4 cal s^-1 cm^-1 °C^-1
• Temperature difference, θ₁ − θ₂ = 50°C
• Time, t = 10 hours = 10 × 3600 = 36000 s

The quantity of heat flowing through the plate in time t is

$\text Q = \frac{\text{KA}(\theta_1 - \theta_2)\text t}{\text l}$

Substituting the values in the CGS system,

$$\begin{aligned}\text Q = \frac{(2 \times 10^{-4}) \times 100 \times 50 \times 36000}{2} \\[1em] = \frac{36000}{2} = 18000\ \text{cal}\end{aligned}$$

Hence, the amount of heat flowing through the plate in 10 hours is 18000 cal, that is, 18 kcal.

Given,

• Thickness of the plate, l = 2 cm = 0.02 m
• Area of cross-section, A = 5000 cm² = 0.5 m²
• Temperature difference, θ₁ − θ₂ = 150 − 140 = 10°C
• Coefficient of thermal conductivity of iron, K = 0.015 kcal s^-1 m^-1 °C^-1

The heat transmitted per second is

$\text H = \frac{\text{KA}(\theta_1 - \theta_2)}{\text l}$

Substituting the values,

$$\begin{aligned}\text H = \frac{0.015 \times 0.5 \times 10}{0.02} \\[1em] = \frac{0.075}{0.02} = 3.75\ \text{kcal s}^{-1}\end{aligned}$$

Hence, the rate of heat transmission through the plate is 3.75 kcal s^-1.