Q.100 g of water is supercooled to $-10^\circ$C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $\left[ S_w = 1 \text{ cal/g/}^\circ\text{C and } L^w_{Fusion} = 80 \text{ cal/g} \right]$

The key idea is that the supercooled water is unstable. The disturbance triggers freezing, and the latent heat released by the freezing portion warms the entire system to the equilibrium temperature.

1. The mixture can only stabilise at $0^\circ$C (the freezing point). Any ice formed will be at $0^\circ$C, and the remaining water will also be at $0^\circ$C. …

Supercooled water at $-10^\circ\text{C}$ is metastable. Once freezing is triggered, the latent heat released warms the sample. Since ice and water coexist at $0^\circ\text{C}$, the temperature rises no further than $0^\circ\text{C}$, so the final state is an ice-water mixture at $0^\circ\text{C}$.

Using energy conservation for the isolated system: the heat absorbed in warming all $100\ \text{g}$ from $-10^\circ\text{C}$ to $0^\circ\text{C}$ is supplied by the latent heat released when a mass $m$ freezes.

$m\,L_{\text{Fusion}} = m_{\text{total}}\,S_w\,\Delta T$ …

Write the energy balance directly as a ratio rather than as two separate 'heat released' / 'heat absorbed' equations: the fraction of water that must freeze is simply $\dfrac{m}{M}=\dfrac{S_w\Delta T}{L_f}=\dfrac{1\times10}{80}=\dfrac18$, so $m=\dfrac{100}{8}=12.5$ g in one step. Sanity check by limiting cases: if the water were supercooled by only $1°C$, almost none would freeze ($m\approx1. …