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A Carnot engine whose low temperature reservoir is at $7^{\circ} \mathrm{C}$ has an efficiency of $50 \%$. It is desired to increase the efficiency to $70 \%$. By how many degrees should the temperature of the high temperature reservoir be increased
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The correct answer is:
$380 \mathrm{~K}$
\(\begin{aligned}
& \text {Initially } \eta=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1} \\
& \Rightarrow 0.5=\frac{\mathrm{T}_1-(273+7)}{\mathrm{T}_1} \\
& \Rightarrow \frac{1}{2}=\frac{\mathrm{T}_1-280}{\mathrm{~T}_1} \Rightarrow \mathrm{T}_1=560 \mathrm{~K}
\end{aligned}\)
Finally \(\eta_1^{\prime}=\frac{\mathrm{T}_1^{\prime}-\mathrm{T}_2}{\mathrm{~T}_1^{\prime}} \Rightarrow 0.7=\frac{\mathrm{T}_1^{\prime}-(273+7)}{\mathrm{T}_1^{\prime}}\)
\(\Rightarrow T_1^{\prime}=933 \mathrm{~K}\)
\(\therefore\) increase in temperature
\(=933-560=373 \mathrm{~K} \approx 380 \mathrm{~K}\)
& \text {Initially } \eta=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1} \\
& \Rightarrow 0.5=\frac{\mathrm{T}_1-(273+7)}{\mathrm{T}_1} \\
& \Rightarrow \frac{1}{2}=\frac{\mathrm{T}_1-280}{\mathrm{~T}_1} \Rightarrow \mathrm{T}_1=560 \mathrm{~K}
\end{aligned}\)
Finally \(\eta_1^{\prime}=\frac{\mathrm{T}_1^{\prime}-\mathrm{T}_2}{\mathrm{~T}_1^{\prime}} \Rightarrow 0.7=\frac{\mathrm{T}_1^{\prime}-(273+7)}{\mathrm{T}_1^{\prime}}\)
\(\Rightarrow T_1^{\prime}=933 \mathrm{~K}\)
\(\therefore\) increase in temperature
\(=933-560=373 \mathrm{~K} \approx 380 \mathrm{~K}\)
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