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 $$80\,cal{/^ \circ }C,$$ is
A.
$$8 \times {10^4}cal/K$$
B.
$$80\,cal/K$$
C.
$$293\,cal/K$$
D.
$$273\,cal/K$$
Answer :
$$293\,cal/K$$
Solution :
Change in entropy is given by
$$\Delta S = \frac{{ml}}{T} = \frac{{1000 \times 80}}{{273}} = 293\,cal\,{K^{ - 1}}$$
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