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Question
50 g of metal piece at 27°C requires 2400 J of heat energy so as to attain a temperature of 327°C . Calculate the specific heat capacity of the metal.
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Solution
Given: Mass of metal, \(m=50\,\text{g}\)
Initial temperature, \(T_1=27^\circ\text{C}\)
Final temperature, \(T_2=327^\circ\text{C}\)
Heat energy supplied, \(Q=2400\,\text{J}\)
Using the formula,
\[ Q=mc\Delta T \]
\[ c=\frac{Q}{m(T_2-T_1)} \]
The rise in temperature.
\[ \Delta T=327-27=300^\circ\text{C} \]
The specific heat capacity.
\[ c=\frac{2400}{50\times300} \]
\[=\frac{2400}{15000} \]
\[{=0.16\,\text{J g}^{-1}\,^\circ\text{C}^{-1}} \]
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