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A Carnot engine whose efficiency is $40 \%$, receives heat at $500 \mathrm{~K}$. If the efficiency is to be $50 \%$, the source temperature for the same exhaust temperature is
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The correct answer is:
$600 \mathrm{~K}$
We know that, efficiency of a Carnot engine is given by
$\eta=\left(1-\frac{T_2}{T_1}\right)$
Given, $\eta_1=0.4$ and $T_1=500 \mathrm{~K}$ (temperature of source)
So, $0.4=1-\frac{T_2}{500} \Rightarrow \frac{T_2}{500}=1-0.4=0.6$
$\Rightarrow \quad T_2=0.6 \times 500=300 \mathrm{~K}$ (temperature of sink)
Now, in the second gas,
$\eta=0.5 \text { and } T_2=300 \mathrm{~K}$
$\therefore \quad 0.5=\left(1-\frac{300}{T_1}\right)$
$\Rightarrow \quad \frac{300}{T_1}=1-0.5=0.5 \Rightarrow T_1=\frac{300}{0.5}=600 \mathrm{~K}$
$\eta=\left(1-\frac{T_2}{T_1}\right)$
Given, $\eta_1=0.4$ and $T_1=500 \mathrm{~K}$ (temperature of source)
So, $0.4=1-\frac{T_2}{500} \Rightarrow \frac{T_2}{500}=1-0.4=0.6$
$\Rightarrow \quad T_2=0.6 \times 500=300 \mathrm{~K}$ (temperature of sink)
Now, in the second gas,
$\eta=0.5 \text { and } T_2=300 \mathrm{~K}$
$\therefore \quad 0.5=\left(1-\frac{300}{T_1}\right)$
$\Rightarrow \quad \frac{300}{T_1}=1-0.5=0.5 \Rightarrow T_1=\frac{300}{0.5}=600 \mathrm{~K}$
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