First slide

Heat engine

Question

The temperature-entropy diagram of a reversible engine cycle is given in fig. Its efficiency is

Moderate

A

2/3
B

1/3
C

1/4
D

1/2

Solution

We know that change in entropy $∆ S$ is given by $ΔS = \frac{ΔQ}{T} or ΔQ = TΔS$
For calculation of Q₁ and Q₂, see fig.
Here,
$\begin{array}{l} Q_{1} = T_{0} S_{0} + \frac{1}{2} T_{0} S_{0} = \frac{3}{2} T_{0} S_{0} \\ Q_{2} = T_{0} S_{0} and Q_{3} = 0 \end{array}$
Now, $η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{2}{3} = \frac{1}{3}$

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