A Carnot engine has efficiency $25\%$ . It operates between reservoirs of constant temperature with temperature difference of $80\,K$ . What is the temperature of low temperature reservoir ...... $^oC$
Medium
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$\eta=25 \% \quad \eta=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}} \times 100 \%$
$\therefore \frac{80}{\mathrm{T}_{1}} \times 100=25 \quad \Rightarrow \mathrm{T}_{1}=320 \mathrm{K}$
$\therefore \mathrm{T}_{2}=320-80=240 \mathrm{K}$
$=-33^{\circ} \mathrm{C}$
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