A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :
Solution
$\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$
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$50 \%$ efficiency $\Rightarrow \frac{1}{2}=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$
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$\frac{1}{2}=1-\frac{T_{\text {sink }}}{600} \Rightarrow T_{\text {sink }}=300$
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Now, $70 \%$ efficiency $\Rightarrow \frac{7}{10}=1-\frac{T_{\text {sink }}}{T_{\text {source }}}$
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$\frac{300}{T_{\text {source }}}=\frac{3}{10}$
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$T_{\text {source }}=1000 \mathrm{~K}$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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