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A Carnot engine whose sink is at $300 \mathrm{~K}$ has an efficiency of $40 \%$. By how much should the temperature of source be increased so as to increase its efficiency by $50 \%$ of original efficiency?
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Verified Answer
The correct answer is:
$250 \mathrm{~K}$
Efficiency of carnot engine =
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1} \\
& \text { or } \quad \frac{T_2}{T_1}=1-\eta \\
& =1-\frac{40}{100}=\frac{3}{5} \\
& \therefore \quad T_1=\frac{50}{100} n=\frac{3}{2} n \\
& =500 \mathrm{~K} \\
& \begin{array}{l}
\text { New efficiency } \\
\eta^{\prime}=\eta+\frac{50}{100} \eta=\frac{3}{2} \eta=60 \%
\end{array} \\
& \therefore \quad \frac{T_2}{T_1}=1-\frac{60}{100}=\frac{2}{5} \\
& \therefore T_1^{\prime}=\frac{5}{2} \times T_2=\frac{5}{2} \times 300=750 \text {. } \\
&
\end{aligned}
$$
Increase in temperature of source
$$
\begin{aligned}
& =T_1^{\prime}-T_1 \\
& =750-300=250 \mathrm{~K} .
\end{aligned}
$$
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1} \\
& \text { or } \quad \frac{T_2}{T_1}=1-\eta \\
& =1-\frac{40}{100}=\frac{3}{5} \\
& \therefore \quad T_1=\frac{50}{100} n=\frac{3}{2} n \\
& =500 \mathrm{~K} \\
& \begin{array}{l}
\text { New efficiency } \\
\eta^{\prime}=\eta+\frac{50}{100} \eta=\frac{3}{2} \eta=60 \%
\end{array} \\
& \therefore \quad \frac{T_2}{T_1}=1-\frac{60}{100}=\frac{2}{5} \\
& \therefore T_1^{\prime}=\frac{5}{2} \times T_2=\frac{5}{2} \times 300=750 \text {. } \\
&
\end{aligned}
$$
Increase in temperature of source
$$
\begin{aligned}
& =T_1^{\prime}-T_1 \\
& =750-300=250 \mathrm{~K} .
\end{aligned}
$$
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