Three Carnot engines operate in series betwee | vedclass

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Question

(C) For a Carnot engine,efficiency $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given that the three engines are equally efficient,let $\eta

Vedclass · Accepted Answer

$T_2 = (T_1^2 \cdot T_4)^{1/3}$ and $T_3 = (T_1 \cdot T_4^2)^{1/3}$

Vedclass · Answer

(C) For a Carnot engine,efficiency $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given that the three engines are equally efficient,let $\eta_1 = \eta_2 = \eta_3 = \eta$.For the first engine: $\eta = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = 1 - \eta$.For the second engine: $\eta = 1 - \frac{T_3}{T_2} \Rightarrow \frac{T_3}{T_2} = 1 - \eta$.For the third engine: $\eta = 1 - \frac{T_4}{T_3} \Rightarrow \frac{T_4}{T_3} = 1 - \eta$.Since $1 - \eta$ is constant,we have $\frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} = k$,where $k = 1 - \eta$.From this,$T_2 = T_1 k$,$T_3 = T_2 k = T_1 k^2$,and $T_4 = T_3 k = T_1 k^3$.Thus,$k^3 = \frac{T_4}{T_1} \Rightarrow k = \left(\frac{T_4}{T_1}ight)^{1/3}$.Substituting $k$ back:$T_2 = T_1 \left(\frac{T_4}{T_1}ight)^{1/3} = T_1^{2/3} T_4^{1/3} = (T_1^2 T_4)^{1/3}$.$T_3 = T_1 \left(\frac{T_4}{T_1}ight)^{2/3} = T_1^{1/3} T_4^{2/3} = (T_1 T_4^2)^{1/3}$.Therefore,the correct option is $C$.

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