Question

A Carnot engine works between temperatures ${127}^{\circ }\text{C}$ and $-{73}^{\circ }\text{C}$ . In process $1$, by increasing the temperature of source by ${50}^{\circ }\text{C}$, efficiency of engine becomes ${\eta }_1$. In process $2$, by decreasing the temperature of sink by ${50}^{\circ }\text{C}$, efficiency of engine becomes ${\eta }_2$. Then which of the following option is correct?

• A. ${\eta }_1>{\eta }_2$
• B. ${\eta }_1={\eta }_2$
• C. ${\eta }_2>{\eta }_1$
• D. ${\eta }_2+2=2{\eta }_1$

Answer 1

The correct option is C ${\eta }_2>{\eta }_1$

First, let us find the efficiency of the engine initially,
$\eta =1-\frac{T_2}{T_1}$
Temperature of sink(colder body), $T_2=-\left(73\right)+273=200\text{K}$
Temperature of source(hotter body), $T_1=127+273=400\text{K}$
$\Rightarrow \eta =1-\frac{(273-73)}{(273+127)}=0.5$
When the temperature of the source is increased by ${50}^{\circ }C$
i.e., $T_1^{\prime }=(127+50)+273=450\text{K}$
$\Rightarrow {\eta }_1=1-\frac{200\text{K}}{450\text{K}}=0.555$
$\therefore {\eta }_1=0.555\times 100=55.5\%$
when the temperature of the sink is decreased by ${50}^{\circ }C$
i.e., $T_2^{\prime }=(-73-50)+273=150\text{K}$
$\Rightarrow {\eta }_2=1-\frac{T_2^{\prime }}{T_1}=1-\frac{150}{400}=0.625$
$\therefore {\eta }_2=0.625\times 100=62.5\%$
Hence $({\eta }_2>{\eta }_1)$ and reducing the temperature of the sink by ${50}^{\circ }\text{C}$ improves the efficiency of heat engine.