Question

A system S receives heat continuously from an electrical heater of power 10W. The temperature of S becomes constant at 50°C when the surrounding temperature is 20°C. After the heater is switched off, S cools from 35.1°C to 34.9°C in 1 minute. The heat capacity of S is

• A. 100J/°C
• B. 300J/°C
• C. 750J/°C
• D. 1500J/°C

Answer 1

Answer: (D)

1500J/°C

Answer 2

In steady state, the rate of heat supplied equals the rate of heat lost. $P = k(T_S - T_{surr})$ $10\text{ W} = k(50^\circ\text{C} - 20^\circ\text{C})$ $10 = 30k \implies k = \frac{1}{3} \text{ W/}^\circ\text{C}$.

When the heater is switched off, the heat lost is $dQ = C (T_1 - T_2) = C (35.1^\circ\text{C} - 34.9^\circ\text{C}) = 0.2C$. The average temperature during cooling is $T_{avg} = \frac{35.1 + 34.9}{2} = 35^\circ\text{C}$. The average temperature difference is $T_{avg} - T_{surr} = 35^\circ\text{C} - 20^\circ\text{C} = 15^\circ\text{C}$. The average rate of heat loss is $k(T_{avg} - T_{surr}) = \frac{1}{3} \times 15 = 5$ W. The time taken is $\Delta t = 1$ minute $= 60$ seconds. The total heat lost is approximately (Average rate of heat loss) $\times$ (time). $dQ \approx (5 \text{ W}) \times (60 \text{ s}) = 300$ J. Equating the two expressions for $dQ$: $0.2C = 300$ J $C = \frac{300}{0.2} = 1500 \text{ J/}^\circ\text{C}$.