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-${ }^0\mathrm{C}$)

Let ‘m’  be the required mass of water.

Then heat lost by hot water

${\mathrm{Q}}_{\mathrm{lost}} = \left(m\right)\left(1\frac{\mathrm{cal}}{\mathrm{g}-{ }^0\mathrm{C}}\right){(50-40)}^0\mathrm{C}------\left(\mathrm{i}\right)$

Heat gained by the calorimeter

    ${\mathrm{Q}}_1 = \left[\right(1500 \mathrm{g}\left)\right(\frac{390}{4.2 \times {10}^3}\frac{\mathrm{cal}}{\mathrm{g}-{ }^0\mathrm{C})}\left)\right]{(40-25)}^0\mathrm{C}$

Heat gained by the water present in it

${\mathrm{Q}}_2 = \left[\right(200 \mathrm{g}\left)\right(1\frac{\mathrm{cal}}{\mathrm{g}-{ }^0\mathrm{C}}\left)\right]{(40-25)}^0\mathrm{C}$

Total heat gained by the calorimeter and the water present in it

     $\mathrm{Q}{ }_{\mathrm{gain}} = {\mathrm{Q}}_1+{\mathrm{Q}}_2-------\left(\mathrm{ii}\right)$

From principle of calorimetry

heat lost by hot body = heat gained by cold body

So, equating Eqs. (i) and (ii) and solving for m, we get

m = 508.93 gram