Question

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

• A. $1/2$
• B. $1/4$
• C. $1/3$
• D. $2/3$

Answer 1

Answer: (C)

$1/3$

Answer 2

According to the figure $Q_{1}=T_{0}\,S_{0}+\frac{1}{2} T_{0}\,S_{0}=\frac{3}{2}=T_{0}\,S_{0}$ $Q_{2}=T_{0}\left(2S_{0}-S_{0}\right)=T_{0}\,S_{0}$ $Q_{3}=0$ $\eta=\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}$ $=1-\frac{Q_{2}}{Q_{1}}=1-\frac{2}{3}$ $=\frac{1}{3}$