Question

A heat engine receives half of its heat supply at 1000 k and half at 500 k, while rejecting heat to the sink at 300 k. What is the maximum thermal efficiency of the engine ?

• A. 90 %
• B. 45 %
• C. 55 %
• D. 60 %

Answer 1

The correct option is C 55 %

Maximum efficiency is obtained when engine works on reversible cycle,



For maximum efficiency

$\Delta S_{\text{isolated system}}=0$

$\text{or}\oint \frac{dQ}{T}=0$

$\Rightarrow \Delta S=\sum \frac{Q}{T}+\left[\frac{-Q_3}{T_3}\right]$

$\Rightarrow 0=\frac{Q_1}{1000}+\frac{Q_2}{500}-\frac{Q_3}{300}$

According to question,

$Q_1=Q_2=Q$

$\Rightarrow \frac{Q_3}{300}=Q[\frac{1}{1000}+\frac{1}{500}]$

$\Rightarrow \frac{Q_3}{Q}=\frac{300}{1000}+\frac{300}{500}=\mathrm{0.9.....}\left(1\right)$

Now, Efficiency,

$\eta =\frac{W}{Q_1+Q_2}=\frac{W}{2Q}$

${\eta }_{max}=\frac{2Q-Q_3}{2Q}=1-\frac{Q_3}{2Q}$.

[Using equation (1)]

${\eta }_{max}=1-\frac{0.9}{2}=0.55$