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Difficulty: MediumHeat Capacity and Specific Heat Capacity

An electric heater rated at 100 W100\text{ W} is used to heat a metal block of mass 2.5 kg2.5\text{ kg} for 5 minutes5\text{ minutes}. If the temperature of the block increases from 25∘C25^\circ\text{C} to 45∘C45^\circ\text{C} and no heat energy is lost to the surroundings, what is the specific heat capacity of the metal?

Answer: 600 J kg^-1 K^-1

Answer

600 J kg−1 K−1600\text{ J kg}^{-1}\text{ K}^{-1}
The energy transferred by the 100 W100\text{ W} heater over 300 seconds300\text{ seconds} is Q=30,000 JQ = 30,000\text{ J}. Dividing this by the product of mass (2.5 kg2.5\text{ kg}) and temperature rise (20 K20\text{ K}) yields the specific heat capacity c=600 J kg−1 K−1c = 600\text{ J kg}^{-1}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate total electrical heat energy supplied to the metal block
Q=P×t=100 W×(5×60 s)=30,000 JQ = P \times t = 100\text{ W} \times (5 \times 60\text{ s}) = 30,000\text{ J}
Heat energy supplied by an electric heater is given by power multiplied by heating time in seconds.
2
Calculate the change in temperature
ΔT=45∘C−25∘C=20 K\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20\text{ K}
Temperature difference is calculated by subtracting initial temperature from final temperature.
3
Solve for the specific heat capacity
c=QmΔT=30,000 J2.5 kg×20 K=600 J kg−1 K−1c = \frac{Q}{m \Delta T} = \frac{30,000\text{ J}}{2.5\text{ kg} \times 20\text{ K}} = 600\text{ J kg}^{-1}\text{ K}^{-1}
Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass by one kelvin.

Key Concept

Specific Heat Capacity and Electrical Energy
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Topics

Heat Capacity and Specific Heat Capacity

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