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Question

A piece of iron of mass $0.05 k g$ is heated to a temperature of $200^{\circ} C$ and dropped into a beaker containing $0.4 k g$ of water at $20^{\circ} C$ . If the final temperature of water and iron is ${22.4}^{\circ} C$ , find the specific heat capacity of iron.

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Solution

By the energy conservation, thermal energy lost by hot iron $=$ thermal energy gained by cold water
If $T_{f}$ be the final temperature of mixture, $m_{i} S_{i} (T_{i} - T_{f}) = m_{w} S_{w} (T_{f} - T_{w})$ where $S_{i} =$ specific of iron and $S_{w} = 4180 J / k g^{o} C$ , specific of water.

or $S_{i} = \frac{m_{w} S_{w} (T_{f} - T_{w})}{m_{i} (T_{i} - T_{f})} = \frac{(0.4) (4180) [(22.4 - 20)]}{0.05 (200 - 22.4)} = \frac{4012.8}{8.88} = 451.89 J / k g^{o} C$

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