Question

Figure below shows a reversible heat engine $E_R$ having heat interactions with three constant temperature systems. Calculate the thermal efficiency of the heat engine

• A. 60

Answer 1

The correct option is A 60
For reversible cycle, $\oint \frac{\delta Q}{T}=0$
$\frac{100}{1000}+\frac{50}{500}-\frac{Q_3}{300}=0$
$0.2-\frac{Q_3}{300}=0$ or
$Q_3=60kJ$
Now, $Q_1+Q_2=W+Q_3$
$100+50=W+60$
$W=90kJ$
$\eta =\frac{W}{Q_{added}}=\frac{90}{150}=0.6=60$