In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper of mass 0.12 kg is heated on the flame for a long time. The copper then is quickly transferred into a beaker of negligible heat capacity containing 0.84 kg of water and the temperature of the water rose from 15°C to 35°C. Calculate the temperature of the flame.
[Given sp. Heat capacity of copper = 0.4 Jg-1 oC-1, Sp. Heat capacity of water = 4.2 Jg-1 oC-1.]
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Given,
- Mass of copper (mC) = 0.12 kg = 0.12 x 1000 g = 120 g
- Mass of water (mW) = 0.84 kg = 0.84 x 1000 = 840 g
- Initial Temperature of water (T1) = 15 °C
- Final Temperature of water = Final Temperature of copper = (T2) = 35 °C
- Specific heat capacity of copper (cC) = 0.4 J g-1 °C-1
- Specific heat capacity of water (cW) = 4.2 J g-1 °C-1
Let,
Temperature of the flame = Initial Temperature of copper = T
Now,
Heat lost by copper = mC x cC x (T - T2)
= 120 x 0.4 x (T - 35)
= 48 x (T - 35)
And
Heat gained by water = mW x cW x (T2 - T1)
= 840 x 4.2 x (35 - 15)
= 840 x 4.2 x 20
= 70560 J
As,
Heat lost by copper = Heat gained by water
⟹ 48 x (T - 35) = 70560
⟹ T - 35 = $\frac{70560}{48}$ = 1470
⟹ T = 1470 + 35 = 1505 °C
Hence, temperature of the flame is 1505 °C.