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)

According to the figure
$Q_1=T_0{S}_0+\frac{1}{2}{T}_0{S}_0=\frac{3}{2}=T_0{S}_0$
$Q_2=T_0\left(2S_0-S_0\right)=T_0{S}_0$
$Q_3=0$
$\eta =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}$
$=1-\frac{Q_2}{Q_1}=1-\frac{2}{3}$
$=\frac{1}{3}$