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Difficulty: HardHeat Capacity and Specific Heat Capacity

An electric immersion heater rated at 200 W200\text{ W} is used to heat 0.8 kg0.8\text{ kg} of a liquid contained in a vessel of heat capacity 120 J K−1120\text{ J K}^{-1}. The initial temperature of the liquid and vessel is 20∘C20^\circ\text{C}. If the heater is operated for 3.5 minutes3.5\text{ minutes} and the final temperature reaches 50∘C50^\circ\text{C}, calculate the specific heat capacity of the liquid in J kg−1 K−1\text{J kg}^{-1}\text{ K}^{-1}, assuming no thermal energy is lost to the surroundings.

Answer: 1600 J kg^-1 K^-1

Answer

The specific heat capacity of the liquid is 1600 J kg−1 K−11600\text{ J kg}^{-1}\text{ K}^{-1}.
Total energy supplied by the heater is Q=200 W×210 s=42,000 JQ = 200\text{ W} \times 210\text{ s} = 42,000\text{ J}. The temperature increase is ΔT=30 K\Delta T = 30\text{ K}. The energy absorbed by the container is Qvessel=CΔT=120×30=3,600 JQ_{\text{vessel}} = C \Delta T = 120 \times 30 = 3,600\text{ J}. The remaining energy 38,400 J38,400\text{ J} is absorbed by the liquid. Dividing this value by the product of the mass of liquid (0.8 kg0.8\text{ kg}) and temperature rise (30 K30\text{ K}) gives the specific heat capacity 1600 J kg−1 K−11600\text{ J kg}^{-1}\text{ K}^{-1}.

Step-by-Step Solution

1
Convert the heating time to seconds and compute total heat energy supplied by the heater.
Q=P×t=200 W×(3.5×60 s)=42,000 JQ = P \times t = 200\text{ W} \times (3.5 \times 60\text{ s}) = 42,000\text{ J}
Heat energy supplied by an electric source equals electrical power multiplied by duration in seconds.
2
Determine the change in temperature of the system.
ΔT=50∘C−20∘C=30 K\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}
Both the vessel and liquid experience the same initial and final temperatures.
3
Calculate the heat energy absorbed by the vessel.
Qvessel=C×ΔT=120 J K−1×30 K=3,600 JQ_{\text{vessel}} = C \times \Delta T = 120\text{ J K}^{-1} \times 30\text{ K} = 3,600\text{ J}
Heat capacity CC represents heat required per unit temperature rise for the container as a whole.
4
Subtract vessel absorption from total heat supplied to find heat absorbed by the liquid.
Qliquid=42,000 J−3,600 J=38,400 JQ_{\text{liquid}} = 42,000\text{ J} - 3,600\text{ J} = 38,400\text{ J}
Conservation of energy dictates Qtotal=Qvessel+QliquidQ_{\text{total}} = Q_{\text{vessel}} + Q_{\text{liquid}}.
5
Calculate the specific heat capacity cc of the liquid.
c=QliquidmΔT=38,4000.8×30=1600 J kg−1 K−1c = \frac{Q_{\text{liquid}}}{m \Delta T} = \frac{38,400}{0.8 \times 30} = 1600\text{ J kg}^{-1}\text{ K}^{-1}
Specific heat capacity isolates heat absorbed per unit mass per Kelvin.

Key Concept

Principle of conservation of thermal energy combining heat capacity of a vessel (CC) and specific heat capacity of a liquid (cc).
Estimated Time:1m 30s
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Topics

Heat Capacity and Specific Heat Capacity

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