Question

Two Carnot heat engines are put in series such that heat rejected by one engine is used in the second engine as heat input. If efficiency of first engine is $30\text{\%}$, and overall combination has efficiency of $65\text{\%}$, then the efficiency of second engine is

• A. $20\text{\%}$
• B. $35\text{\%}$
• C. $40\text{\%}$
• D. $50\text{\%}$

Answer 1

The correct option is D $50\text{\%}$

Let the first heat engine absorbs heat at $T_1$ and releases heat at $T_2$.

So according to the question, second heat engine absorbs heat at $T_2$. And let it release the heat at $T_3$.

${\eta }_1=1-\frac{T_2}{T_1}\Rightarrow 1-{\eta }_1=\frac{T_2}{T_1}....\left(1\right)$

${\eta }_2=1-\frac{T_3}{T_2}\Rightarrow 1-{\eta }_2=\frac{T_3}{T_2}....\left(2\right)$

For combination,

$\eta =1-\frac{T_3}{T_1}\Rightarrow 1-\eta =\frac{T_3}{T_1}....\left(3\right)$

Using equations $\left(1\right),\left(2\right)$ and $\left(3\right)$,

$\Rightarrow (1-\eta )=(1-{\eta }_1)(1-{\eta }_2)$

$\Rightarrow (1-0.65)=(1-0.3)(1-{\eta }_2)$

$\Rightarrow 0.35=\left(0.7\right)(1-{\eta }_2)$

$\Rightarrow 1-{\eta }_2=0.5$

$\therefore {\eta }_2=0.5=50\text{\%}$

Hence, $\left(D\right)$ is the correct answer.