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$100 \mathrm{~g}$ of water is supercooled to $-10^{\circ} \mathrm{C}$. 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? $\left[S_w=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\right.$ and $\left.L^{w_{\text {Fusion }}}=80 \mathrm{cal} / \mathrm{g}\right]$
PhysicsThermal Properties of Matter
Solution:
1808 Upvotes Verified Answer
As given that, mass of water $(\mathrm{m})=100 \mathrm{~g}$
Change in temperature $\Delta T=0-(-10)=10^{\circ} \mathrm{C}$
Specific heat of water $\left(S_w\right)=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Latent heat of fusion of water $L_{\text {fusion }}^w=80 \mathrm{cal} / \mathrm{g}$
Heat required to bring water in super cooling from $-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$
$$
Q=m s_w \Delta T=100 \times 1 \times 10=1000 \mathrm{cal}
$$
Let $m$ gram of ice be melted, so, $Q=m L$
or $m=\frac{Q}{L}=\frac{1000}{80}=12.5 \mathrm{~g}$
So, there is $m=12.5 \mathrm{~g}$ water and $100-12.5 \mathrm{~g}$ ice in mixture. As small mass of ice is melted. Hence the temperature of the mixture will remain $0^{\circ} \mathrm{C}$.

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