A) 2.4 J/K
B) 3.6 J/K
C) 4.2 J/K
D) 5.2 J/K
Correct Answer: A
Solution :
[a] : The entropy change of the block L is \[\Delta {{S}_{L}}=(1.5kg)(386J/kgK)\int\limits_{333K}^{313K}{\frac{dT}{T}}\] \[=(1.5kg)(386J/kg\,K)ln\frac{313K}{333K}=-35.8J/K\] The entropy change of the block R is \[\Delta {{S}_{R}}=(1.5kg)(386J/kg\,K)ln\int\limits_{293K}^{313K}{\frac{dT}{T}}\] \[=(1.5kg)(386J/kg\,K)ln\left( \frac{313K}{293K} \right)=+38.2J/K\] Thus, the net entropy change for the two block system during irreversible process is \[\Delta {{S}_{irrev}}=\Delta {{S}_{L}}+\Delta {{S}_{R}}=-35.8J/K+38.2J/K=2.4J/K\]
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