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Difficulty: Very hardHeat Capacity and Specific Heat Capacity

An electric heater rated at 50 W50\text{ W} is used to heat a solid block of mass 1.5 kg1.5\text{ kg} for 4 minutes4\text{ minutes}. During this period, the temperature of the block increases from 30∘C30^\circ\text{C} to 70∘C70^\circ\text{C}. If 20%20\% of the heat energy supplied by the heater is lost to the surroundings, what is the heat capacity of the block?

  1. 240 J K−1240\text{ J K}^{-1}Answer
  2. B
    160 J K−1160\text{ J K}^{-1}
  3. C
    300 J K−1300\text{ J K}^{-1}
  4. D
    60 J K−160\text{ J K}^{-1}

Answer

240 J K−1240\text{ J K}^{-1}
The correct answer is 240 J K−1240\text{ J K}^{-1}. The heater delivers 12000 J12{}000\text{ J} of energy in 4 minutes4\text{ minutes}. Accounting for 20%20\% environmental heat loss leaves 9600 J9{}600\text{ J} absorbed by the block. Dividing this energy by the 40 K40\text{ K} temperature rise gives a total heat capacity of 240 J K−1240\text{ J K}^{-1}.

Step-by-Step Solution

1
Calculate total electrical energy supplied by heater
Qsupplied=P×t=50 W×(4×60 s)=12000 JQ_{\text{supplied}} = P \times t = 50\text{ W} \times (4 \times 60\text{ s}) = 12{}000\text{ J}
Electrical work converted to heat is given by power multiplied by time in seconds.
2
Determine useful heat absorbed by block after accounting for energy loss
Quseful=(1−0.20)×12000 J=0.80×12000 J=9600 JQ_{\text{useful}} = (1 - 0.20) \times 12{}000\text{ J} = 0.80 \times 12{}000\text{ J} = 9{}600\text{ J}
Since 20%20\% of supplied energy is lost, 80%80\% is retained to raise the temperature of the block.
3
Calculate temperature rise
ΔT=70∘C−30∘C=40 K\Delta T = 70^\circ\text{C} - 30^\circ\text{C} = 40\text{ K}
Temperature difference is the final temperature minus the initial temperature.
4
Calculate heat capacity of the block
C=QusefulΔT=9600 J40 K=240 J K−1C = \frac{Q_{\text{useful}}}{\Delta T} = \frac{9{}600\text{ J}}{40\text{ K}} = 240\text{ J K}^{-1}
Heat capacity CC is defined as total heat absorbed per unit temperature change (C=QΔTC = \frac{Q}{\Delta T}).

Key Concept

Heat Capacity (C=QΔTC = \frac{Q}{\Delta T}) represents total thermal capacity of a body, whereas Specific Heat Capacity (c=QmΔTc = \frac{Q}{m\Delta T}) is heat capacity per unit mass.
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Topics

Heat Capacity and Specific Heat Capacity

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