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A Carnot's engine has an efficiency of $25 \%$ when its sink is at $27^{\circ} \mathrm{C}$. If it has to be increased to $40 \%$, what should be the temperature of the sink keeping the temperature of the source constant?
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The correct answer is:
$240 \mathrm{~K}$
Efficiency of Carnot's engine,
$$
\eta=25 \%=0.25
$$
Temperature of sink,
$$
T_2=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}
$$
Temperature of source $=T_1$
$$
\begin{array}{ll}
\therefore & \eta=1-\frac{T_2}{T_1} \\
\Rightarrow & 0.25=1-\frac{T_2}{T_1} \\
\Rightarrow & \frac{T_2}{T_1}=1-0.25=0.75 \\
\Rightarrow & T_1=\frac{T_2}{0.75}=\frac{300}{0.75}=400 \mathrm{~K} \\
\Rightarrow & T_1=400 \mathrm{~K}
\end{array}
$$
When efficiency, $\eta^{\prime}=40 \%=0.4$
$$
\begin{array}{ll}
\text { then, } & 1-\frac{T_2^{\prime}}{T_1}=0.4 \Rightarrow \frac{T_2^{\prime}}{T_1}=1-0.4=0.6 \\
\Rightarrow & T_2^{\prime}=0.6 T_1=0.6 \times 400=240 \mathrm{~K}
\end{array}
$$
$$
\eta=25 \%=0.25
$$
Temperature of sink,
$$
T_2=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}
$$
Temperature of source $=T_1$
$$
\begin{array}{ll}
\therefore & \eta=1-\frac{T_2}{T_1} \\
\Rightarrow & 0.25=1-\frac{T_2}{T_1} \\
\Rightarrow & \frac{T_2}{T_1}=1-0.25=0.75 \\
\Rightarrow & T_1=\frac{T_2}{0.75}=\frac{300}{0.75}=400 \mathrm{~K} \\
\Rightarrow & T_1=400 \mathrm{~K}
\end{array}
$$
When efficiency, $\eta^{\prime}=40 \%=0.4$
$$
\begin{array}{ll}
\text { then, } & 1-\frac{T_2^{\prime}}{T_1}=0.4 \Rightarrow \frac{T_2^{\prime}}{T_1}=1-0.4=0.6 \\
\Rightarrow & T_2^{\prime}=0.6 T_1=0.6 \times 400=240 \mathrm{~K}
\end{array}
$$
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