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When 1 kg of ice at $0^\circ C$ melts to water at $0^\circ C$, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is
A
293 cal/K
B
273 cal/K
C
$8\times10^4$ cal/K
D
80 cal/K
Detailed Solution
Heat absorbed in melting: $Q = mL = 1000\ g\times80\ cal/g = 8\times10^4$ cal
Melting takes place at the constant temperature T = $0^\circ C$ = 273 K.
Change in entropy $\Delta S = \frac{Q}{T}$
$\Delta S = \frac{8\times10^4}{273}$
$\Delta S \approx 293$ cal/K
Note: the source prints the latent heat as '80 cal/°C'; the correct unit, cal/g, is used here.
Melting takes place at the constant temperature T = $0^\circ C$ = 273 K.
Change in entropy $\Delta S = \frac{Q}{T}$
$\Delta S = \frac{8\times10^4}{273}$
$\Delta S \approx 293$ cal/K
Note: the source prints the latent heat as '80 cal/°C'; the correct unit, cal/g, is used here.