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Three Carnot engines operate in series between a heat source at temperature $T_1$ and heat sink at a temperature $T_4$. There are two other reservoirs at temperatures $T_2$ and $T_3$. The three engines are equally efficient, if (given that, $\left.T_1>T_2>T_3>T_4\right)$
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Verified Answer
The correct answer is:
$T_2=\left(T_1^2 \cdot T_4\right)^{1 / 3}$ and $T_3=\left(T_1 \cdot T_4^2\right)^{1 / 3}$
Given, efficiencies
$$
\eta_1=\eta_2=\eta_3=\eta
$$
and temperature,
$$
T_1>T_2>T_3>T_4
$$
As we know that, in case of Carnot engine,
$$
\begin{aligned}
& \operatorname{Efficiency}(\eta)=1-\frac{T_{\text {sink }}}{T_{\text {source }}} \\
& \therefore \quad \eta_1=\eta=1-\frac{T_2}{T_1} \\
& \Rightarrow \quad T_2=(1-\eta) T_1 \\
& \text { Similarly, } T_3=(1-\eta) T_2 \\
& T_4=(1-\eta) T_3 \\
&
\end{aligned}
$$
Now, by using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
\left(T_1^2 \cdot T_4\right)^{1 / 3} & =\left[\left(\frac{T_2}{1-\eta}\right)^2 \cdot(1-\eta) T_3\right]^{1 / 3} \\
& =\left[\frac{T_2^2}{(1-\eta)} \cdot T_3\right]^{1 / 3} \\
& =\left[\left(\frac{T_2^2}{1-\eta}\right)(1-\eta) T_2\right]^{1 / 3} \\
& =\left[T_2^3\right]^{1 / 3}=T_2
\end{aligned}
$$
Again, by using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
\left(T_1 \cdot T_4^2\right)^{1 / 3} & =\left\{\left(\frac{T_2}{1-\eta}\right)\left[(1-\eta) T_3\right]^2\right\}^{1 / 3} \\
& =\left[T_2(1-\eta) T_3^2\right]^{1 / 3} \\
& =\left[\frac{T_3}{1-\eta}(1-\eta) T_3^2\right]^{1 / 3}=\left[T_3^3\right]^{1 / 3} \\
& =T_3
\end{aligned}
$$
$$
\eta_1=\eta_2=\eta_3=\eta
$$
and temperature,
$$
T_1>T_2>T_3>T_4
$$
As we know that, in case of Carnot engine,
$$
\begin{aligned}
& \operatorname{Efficiency}(\eta)=1-\frac{T_{\text {sink }}}{T_{\text {source }}} \\
& \therefore \quad \eta_1=\eta=1-\frac{T_2}{T_1} \\
& \Rightarrow \quad T_2=(1-\eta) T_1 \\
& \text { Similarly, } T_3=(1-\eta) T_2 \\
& T_4=(1-\eta) T_3 \\
&
\end{aligned}
$$
Now, by using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
\left(T_1^2 \cdot T_4\right)^{1 / 3} & =\left[\left(\frac{T_2}{1-\eta}\right)^2 \cdot(1-\eta) T_3\right]^{1 / 3} \\
& =\left[\frac{T_2^2}{(1-\eta)} \cdot T_3\right]^{1 / 3} \\
& =\left[\left(\frac{T_2^2}{1-\eta}\right)(1-\eta) T_2\right]^{1 / 3} \\
& =\left[T_2^3\right]^{1 / 3}=T_2
\end{aligned}
$$
Again, by using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
\left(T_1 \cdot T_4^2\right)^{1 / 3} & =\left\{\left(\frac{T_2}{1-\eta}\right)\left[(1-\eta) T_3\right]^2\right\}^{1 / 3} \\
& =\left[T_2(1-\eta) T_3^2\right]^{1 / 3} \\
& =\left[\frac{T_3}{1-\eta}(1-\eta) T_3^2\right]^{1 / 3}=\left[T_3^3\right]^{1 / 3} \\
& =T_3
\end{aligned}
$$
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