When $1 ext{ kg}$ of ice at $0^{\circ} ext{C}$ melts to water at $0^{\circ} ext{C}$, the resulting change i

Question

When $1	ext{ kg}$ of ice at $0^{\circ}	ext{C}$ melts to water at $0^{\circ}	ext{C}$, the resulting change in its entropy, taking latent heat of ice to be $80	ext{ cal/g}$, is

Accepted Answer

The change in entropy ($\Delta S$) during a phase transition at a constant temperature is given by the formula $\Delta S = \frac{Q}{T}$ [1], where $Q$ is the heat absorbed and $T$ is the absolute temperature.
Given:
Mass of ice, $m = 1	ext{ kg} = 1000	ext{ g}$
Latent heat of fusion of ice, $L = 80	ext{ cal/g}$
Absolute temperature, $T = 0^{\circ}	ext{C} = 273	ext{ K}$
The total heat absorbed to melt the ice is $Q = m 	imes L = 1000	ext{ g} 	imes 80	ext{ cal/g} = 80000	ext{ cal}$.
Now, calculating the change in entropy:
$\Delta S = \frac{80000	ext{ cal}}{273	ext{ K}} \approx 293	ext{ cal/K}$.

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