Question 84 of 126

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