A copper calorimeter of a mass 300g contains 500g of water at a temperature of 20°C. A 500 g of copper block at 100°C is dropped into the calorimeter. If the resultant temperature is 25°C , then find the specific heat of copper in Jkg^-1K^-1 (nearly)

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390

190

490

290

answer is B.

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Detailed Solution

Heat lost by copper block = heat gained by water + Heat gained by calorimeter
$\begin{array}{l} m_{c} s_{c} (t_{h} - t_{r}) = m_{w} s_{w} (t_{r} - t_{c}) + m_{c} s_{c} (t_{r} - t_{c}) \\ 500 \times s_{c} (100 - 25) = 500 \times 1 \times (25 - 20) + 300 \times s_{c} (25 - 20) \\ ∴ s_{c} = 0 .0694 cal / {gm}^{\circ} C \\ ∴ s_{c} = 0 .0694 \times 4200 ≃ 292 J / kg \cdot K \end{array}$