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Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature ThT_{h} the other at Te,T_{e}, are brought into thermal contact and allowed to reach equilibrium. Evaluate the change in entropy for two blocks of copper, each of mass 500 g, with Cpm=24.4 JK1 mol1,C_{pm}=24.4\ \mathrm{JK}^{-1}\ \mathrm{mol}^{-1}, taking Th=500KT_{\mathrm{h}}=500 \mathrm{K} and Te=250K.T_{e}=250 \mathrm{K}.

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In this exercise we have to find an expression for the change in entropy when two block of the same substance and of equal mass are brought into thermal contact and allowed to reach equilibrium

We have to evaluate change in entropy for two blocks of copper

We will calculate the final temperature by using:

Heat lost == heat gained

m1(500KT)=m2(T250K)500g(500KT)=500g(T250K)2T=750T=375K\begin{align*} \mathrm{m}_{1}(500 \mathrm{K}-\mathrm{T})&=\mathrm{m}_{2}(\mathrm{T}-250 \mathrm{K})\\ 500_{\mathrm{g}}(500 \mathrm{K}-\mathrm{T})&=500_{\mathrm{g}}(\mathrm{T}-250 \mathrm{K})\\ 2 \mathrm{T}&=750\\ \mathrm{T}&=375 \mathrm{K}\\ \end{align*}

Firstly we will calculate ΔSC\Delta S_{C}:

ΔSC=nCP,mlnTfTi=500g63.55g mol24.4J K molln(375K250K)=7.86mol24.4J K mol(0.405)ΔSC=77.76J K\begin{align*} \Delta S_{C}&=nC_{P,m} \ln \frac{T_{f}}{T_{i}}\\ &=\frac{500 \mathrm{g}}{63.55 \mathrm{g}\ \mathrm{mol}} \cdot 24.4 \mathrm{J}\ \mathrm{K}\ \mathrm{mol} \cdot \ln \left(\frac{375 \mathrm{K}}{250 \mathrm{K}}\right)\\ &=7.86 \mathrm{mol} \cdot 24.4 \mathrm{J}\ \mathrm{K}\ \mathrm{mol} \cdot(0.405)\\ \Delta S_{C}&=77.76 \mathrm{J}\ \mathrm{K}\\ \end{align*}

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