When $1 \mathrm{~kg}$ of ice at $0^{\circ} \mathrm{C}$ melts to water at $0^{\circ} \mathrm{C}$, the…
When $1 \mathrm{~kg}$ of ice at $0^{\circ} \mathrm{C}$ melts to water at $0^{\circ} \mathrm{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 \mathrm{cal} /{ }^{\circ} \mathrm{C}$, is
$273 \mathrm{cal} / \mathrm{K}$
$253 \mathrm{cal} / \mathrm{K}$
$263 \mathrm{cal} / \mathrm{K}$
$293 \mathrm{cal} / \mathrm{K}$
Solution
Change in entropy is given by $\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}$ or $\Delta \mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{mL}_{\mathrm{f}}}{273}$
$\Delta \mathrm{S}=\frac{1000 \times 80}{273}=293 \mathrm{cal} / \mathrm{K}$
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