question_answer

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

A) \[\frac{1}{4}\]

B) \[\frac{1}{2}\]

C) \[\frac{2}{3}\]

D) \[\frac{1}{3}\]

Correct Answer: D

Solution :

[d]     \[{{Q}_{1}}={{T}_{0}}{{S}_{0}}+\frac{1}{2}{{T}_{0}}{{S}_{0}}=\frac{3}{2}{{T}_{0}}{{S}_{0}}\] \[{{Q}_{2}}={{T}_{0}}(2{{S}_{0}}-{{S}_{0}})={{T}_{0}}{{S}_{0}}\] and \[{{Q}_{3}}=0\] \[\eta =\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\]  \[=1-\frac{{{Q}_{1}}}{{{Q}_{2}}}=1-\frac{{{T}_{0}}{{S}_{0}}}{\frac{3}{2}{{T}_{0}}{{S}_{0}}}=\frac{1}{3}\]