Question

$5gm$ of water at ${30}^{\circ }C$ and $5gm$ of ice at $-{20}^{\circ }C$ are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice $=0.5cal/g{m}^{\circ }C$ and latent heat of ice $=80cal/gm$.

Answer 1

In this case heat is given by water and taken by ice
Heat available with water to cool from ${30}^{\circ }C$ to $0^{\circ }C$
$=ms\theta =5\times 1\times 30=150cal$
Heat required by $5gm$ ice to increase its temperature up to $0^{\circ }C$ is
$ms\theta =5\times 0.5\times 20=50cal$
Out of $150cal$ heat available, $50cal$ is used for increasing temperature of ice from $-{20}^{\circ }C$ to $0^{\circ }C$. The remaining heat $100cal$ is used for melting the ice.
If mass of ice melted is $mgm$ then
$m\times 80=100\Rightarrow m=1.25gm$
Thus, $1.25gm$ ice out of $5gm$ melts and mixture of ice and water is at $0^{\circ }C$.