The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
AIEEE 2005, Diffcult
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${Q_1} = {T_0}{S_0} + \frac{1}{2}{T_0}{S_0} = \frac{3}{2}{T_0}{S_0}$
${Q_2} = {T_0}{S_0}$ and ${Q_3} = 0$
$\eta = \frac{W}{{{Q_1}}} = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}}$
$ = 1 - \frac{{{Q_2}}}{{{Q_1}}} = 1 - \frac{2}{3} = \frac{1}{3}$
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