$100 \, g$ of water is supercooled to $-10 \, ^\circ C$. At this point,due to some disturbance,some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $[S_W = 1 \, cal \, g^{-1} \, ^\circ C^{-1}$ and $L_{fusion} = 80 \, cal \, g^{-1}]$

(C) Mass of water $m = 100 \, g$.
When supercooled water at $-10 \, ^\circ C$ is disturbed,it releases heat as it warms up to $0 \, ^\circ C$. This heat is used to freeze a portion of the water into ice.
Heat released by $100 \, g$ of water to reach $0 \, ^\circ C$:
$Q = m \cdot S_W \cdot \Delta T = 100 \times 1 \times (0 - (-10)) = 1000 \, cal$.
Let $m'$ be the mass of water that freezes into ice at $0 \, ^\circ C$. The heat released by the water is absorbed by the freezing process:
$Q = m' \cdot L_{fusion}$.
$1000 = m' \times 80$.
$m' = \frac{1000}{80} = 12.5 \, g$.
Since the final state is a mixture of ice and water at equilibrium,the temperature of the resultant mixture will be $0 \, ^\circ C$ and the mass of ice formed is $12.5 \, g$.