500 g of water and 100 g of ice at 0^° C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100°C is added to it. Then water in…

$\displaystyle 500$ g of water and $\displaystyle 100$ g of ice at $\displaystyle 0^{\circ} \mathrm{C}$ are in a calorimeter whose water equivalent is $\displaystyle 40$ g. $\displaystyle 10$ g of steam at $\displaystyle 100$°C is added to it. Then water in the calorimeter is : (Latent heat of ice $\displaystyle =80 \mathrm{cal} / \mathrm{g}$, Latent heat of steam $\displaystyle =540 \mathrm{cal} / \mathrm{g}$ )