Question
A Carnot engine has an efficiency of $50 \%$ when its source is at a temperature $327^{\circ}\,C$. The temperature of the sink is $.........^{\circ} C$
$\% \eta=\left(1-\frac{T_{\text {sink }}}{T_{\text {source }}}\right) \times 100$
$T_{\text {source }}=327^{\circ}\,C =600\,K$
$50=\left(1-\frac{T_{\text {sink }}}{600}\right) \times 100$
$\frac{1}{2}=1-\frac{T_{\text {sink }}}{600}$
$T _{\text {Sink }}=300\,K$
So temp. of sink is ${ }^{\circ} C =300-2763=27^{\circ}\,C$
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