$500 \mathrm{~g}$ of water and $100 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ are in a calorimeter whose…
$500 \mathrm{~g}$ 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 $=540 \mathrm{cal} / \mathrm{g}$ )
$580 \mathrm{~g}$
$590 \mathrm{~g}$
$600 \mathrm{~g}$
$610 \mathrm{~g}$
Solution
As $1 \mathrm{~g}$ of steam at $100^{\circ} \mathrm{C}$ melts $8 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$.
$10 \mathrm{~g}$ of steam will melt $8 \times 10 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ Water in calorimeter $=500+80+10 \mathrm{~g}=590 \mathrm{~g}$