A) - 4.5 cal/K
B) + 4.5 cal/K
C) + 5.4 cal/K
D) - 5.4 cal /K
Correct Answer: B
Solution :
Gain of entropy of ice \[{{S}_{1}}=\frac{\Delta Q}{T}=\frac{mL}{T}=\frac{80\times 100}{(0+273)}\] \[=\frac{8\times {{10}^{3}}}{273}cal/K\] Loss of entropy of water \[{{S}_{2}}=\frac{\Delta Q}{T}=-\frac{mL}{T}\] \[=-\frac{80\times 100}{(273+50)}=-\frac{8\times {{10}^{3}}}{323}cal/K\] Total change of entropy \[{{S}_{1}}+{{S}_{2}}=\frac{8\times {{10}^{3}}}{273}-\frac{8\times {{10}^{3}}}{323}=+4.5\,\,cal/K\]
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