Find the specific heat of the body, when 0.2 kg of a body at 1000C is dropped into 0.5 kg of
water at 100C, the resulting temperature is 160C. Given: Specific heat of water is 4.2 x 10³
J/kg/0
C.

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Solution

$\begin{array}{l} Let the specific heat of the body be x \\ m = 0.2 kg \\ m_{w} = 0.5 kg \\ T_{1} = 100^{0} C \\ T_{2} = 160^{0} C \\ s = 4.2 \times 10^{3} {Jkg}^{- 1} C^{- 1} \\ T_{b o d y} {= 1000}^{0} C \\ Heat lost by the body = m x (T_{b o d y} - T_{2}) \\ = 0.2 x \times (1000 - 160) \\ = 168 x J \\ Heat gained by the water = m_{w} (T_{2} - T_{1}) s \\ = 0.5 \times (160 - 100) \times 4.2 \times 10^{3} \\ = 126 \times 10^{3} J \\ Heat gained by water will be equal to heat lost by the body \\ 168 x = 126 \times 10^{3} \\ = > x = \frac{126 \times 10^{3}}{168} = 750 {Jkg}^{- 1} C^{- 1} \end{array}$

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