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Difficulty: MediumHeat Capacity and Specific Heat Capacity

A container of heat capacity 80 J K−180\text{ J K}^{-1} holds 0.4 kg0.4\text{ kg} of a liquid. When 9.6 kJ9.6\text{ kJ} of heat energy is supplied to the system, the temperature of the container and liquid rises by 20 K20\text{ K}. What is the specific heat capacity of the liquid?

  1. 1000 J kg−1 K−11000\text{ J kg}^{-1}\text{ K}^{-1}Answer
  2. B
    1200 J kg−1 K−11200\text{ J kg}^{-1}\text{ K}^{-1}
  3. C
    400 J kg−1 K−1400\text{ J kg}^{-1}\text{ K}^{-1}
  4. D
    480 J kg−1 K−1480\text{ J kg}^{-1}\text{ K}^{-1}

Answer

1000 J kg−1 K−11000\text{ J kg}^{-1}\text{ K}^{-1}
The option specifying 1000 J kg−1 K−11000\text{ J kg}^{-1}\text{ K}^{-1} is correct because out of the 9600 J9600\text{ J} supplied, 1600 J1600\text{ J} (80×2080 \times 20) is absorbed by the container. The remaining 8000 J8000\text{ J} raises the temperature of 0.4 kg0.4\text{ kg} of liquid by 20 K20\text{ K}, yielding c=80000.4×20=1000 J kg−1 K−1c = \frac{8000}{0.4 \times 20} = 1000\text{ J kg}^{-1}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the thermal energy absorbed by the container
Qcontainer=1600 JQ_{\text{container}} = 1600\text{ J}
The heat absorbed by a body of heat capacity CC during a temperature change ΔT\Delta T is given by Qcontainer=CΔT=80 J K−1×20 K=1600 JQ_{\text{container}} = C \Delta T = 80\text{ J K}^{-1} \times 20\text{ K} = 1600\text{ J}.
2
Determine the net thermal energy absorbed by the liquid
Qliquid=8000 JQ_{\text{liquid}} = 8000\text{ J}
By energy conservation, Qtotal=Qcontainer+QliquidQ_{\text{total}} = Q_{\text{container}} + Q_{\text{liquid}}, so Qliquid=9600 J−1600 J=8000 JQ_{\text{liquid}} = 9600\text{ J} - 1600\text{ J} = 8000\text{ J}.
3
Calculate the specific heat capacity of the liquid
cliquid=1000 J kg−1 K−1c_{\text{liquid}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}
Using the specific heat capacity relationship Q=mcΔTQ = m c \Delta T, cliquid=QliquidmΔT=8000 J0.4 kg×20 K=1000 J kg−1 K−1c_{\text{liquid}} = \frac{Q_{\text{liquid}}}{m \Delta T} = \frac{8000\text{ J}}{0.4\text{ kg} \times 20\text{ K}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}.

Key Concept

Distinction and calculation involving heat capacity of a container and specific heat capacity of a substance
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Topics

Heat Capacity and Specific Heat Capacity

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