A solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature is 25°C. Calculate the heat capacity of solid.

Given,

Mass of solid (m_s) = 60 g

Mass of water (m_w) = 150 g

Fall in temperature of solid (△t_s) = (100 - 25) = 75°C

Rise in temperature of water (△t_w) = (25 - 20) = 5°C

Specific heat capacity of water (c_w) = 4.2 Jg^-1K^-1

Let specific heat capacity of solid be c_s.

Then,

Heat energy given by solid = m_sc_s△t_s = 60 x c_s x 75 = 4500 x c_s ......(1)

Heat energy taken by water = m_wc_w△t_w = 150 × 4.2 × 5 = 3150 ......(2)

Assuming that there is no loss of heat energy,

Heat energy given by solid = Heat energy taken by water.

Equating equations 1 & 2, we get,

4500 × c_s = 3150

$⇒ \text c_\text s = \frac{3150}{4500}$

⇒ c_s = 0.7 J g^-1 K^-1

Now,

Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K^-1

Hence, Heat capacity of solid = 42 J K^-1.