In a heat engine, the temperature of the source and sink are $500\, K$ and $375\, K$. If the engine consumes $25\times10^5\, J$ per cycle, the work done per cycle is
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Here, $T_{1}=500 \mathrm{K}, T_{2}=375 \mathrm{K}$
$\mathrm{Q}_{1}=25 \times 10^{5} \mathrm{J}$
$\therefore \eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=1-\frac{375}{500}=0.25$
$\mathrm{W}=\eta \mathrm{Q}=0.25 \times 25 \times 10^{5}=6.25 \times 10^{5} \mathrm{J}$
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