Solution:
Entropy change $Delta S = frac{Delta Q}{T} = frac{m cdot L}{T} = frac{1000 cdot 80}{273} = 293 text{ cal/K}$
When $1 text{ kg}$ of ice at $0^circ text{C}$ melts to water at $0^circ text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 text{ Cal}/^circ text{C}$, is:
(2011 Pre)
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