$100\,g$ of water is supercooled to $-\,10\,^oC$. 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 ? $[S_W = 1\,cal\,g^{-1}\,^oC^{-1}$ and ${L^W}_{{\text{fussion}}}$ $= 80\,cal\,g^{-1}]$
$100\,g$ of water is supercooled to $-\,10\,^oC$. 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 ? $[S_W = 1\,cal\,g^{-1}\,^oC^{-1}$ and ${L^W}_{{\text{fussion}}}$ $= 80\,cal\,g^{-1}]$
Mass of water $m=100 \mathrm{~g}$
Change in temperature $\Delta \mathrm{T}=0-(-10)=10^{\circ} \mathrm{C}$
Specific heat of water $\mathrm{s}_{\mathrm{w}}=1$ calg $^{-1}{ }^{\circ} \mathrm{C}^{-1}$
Latent heat of melting of water $\mathrm{L}_{f}=80$ calg $^{-1}$
Heat required to convert ice at $-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ water.
$\mathrm{Q}=m \mathrm{~s}_{\mathrm{w}} \Delta \mathrm{T}$ $=100 \times 1 \times 10$ $=1000 \text { cal }$
Suppose, ' $m$ ' gram ice melt,
$\therefore \quad \mathrm{Q}=m \mathrm{~L}$
$m=\frac{\mathrm{Q}}{\mathrm{L}}$
$=\frac{1000}{80}$
$=12.5 \mathrm{~g}$
As only $12.5 \mathrm{~g}$ ice is melted from $100 \mathrm{~g}$, the temperature of mixture will be $0^{\circ} \mathrm{C}$.
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