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Question 2.44: An engine working on Carnot cycle receives 1200 kJ from a he...

An engine working on Carnot cycle receives 1200 kJ from a heat source at a constant temperature of 1000°C and rejects to a heat sink at a constant temperature of 35°C. Calculate (a) thermal efficiency of the engine, and (b) work done.

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\begin{aligned}&T_1=1000+273=1273 \ K \\&T_2=35+273=308 \ K\end{aligned}

(a) \eta_{\text {th }}=1-\frac{T_2}{T_1}=1-\frac{308}{1273}=0.758 \text { or } 75.8 \%

(b) \eta_{ th }=\frac{W}{Q_1}

W = 0.758 × 1200 = 909.6 kJ

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