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The temperature-entropy diagram of a reversible engine cycle is given in figure. What is its efficiency ?

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$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{3}{4}$

answer is B.

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Detailed Solution

$Q_{1} = \frac{1}{2} S_{0} T_{0} + S_{0} T_{0} = \frac{3 S_{0} T_{0}}{2}, Q_{2} = S_{0} T_{0}; Q_{3} = 0$

$Efficiency : η = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{S_{0} T_{0}}{(3 \frac{S_{0} T_{0}}{2})} = 1 - \frac{2}{3} = \frac{1}{3}$

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