A copper calorimeter of mass 1.5 kg has 200 g of water at ${25}^0\mathrm{C}$. How much water (approx.) at ${50}^0\mathrm{C}$ is required to be poured in the calorimeter, so that the equilibrium temperature attained is ${40}^0\mathrm{C}?$ (Sp. heat capacity of copper = 390 J/kg-${ }^0\mathrm{C}$)

detailed solution

Correct option is A

Let ‘m’  be the required mass of water.Then heat lost by hot waterQlost = (m)(1calg- 0C)(50-40)0C------(i)Heat gained by the calorimeter    Q1 = [(1500 g)(3904.2 ×103calg- 0C))](40-25)0CHeat gained by the water present in itQ2 = [(200 g)(1calg- 0C)](40-25)0CTotal heat gained by the calorimeter and the water present in it     Q gain = Q1+Q2-------(ii)From principle of calorimetryheat lost by hot body = heat gained by cold bodySo, equating Eqs. (i) and (ii) and solving for m, we getm = 508.93 gram