Question

Two Carnot engines $A$ and $B$ are operated in series. Engine $A$ receives heat from a reservoir at $600K$ and rejects heat to a reservoir at temperature $T$. Engine $B$ receives heat rejected by engine $A$ and in turn rejects it to a reservoir at $100K$. If the efficiencies of the two engines $A$ and $B$ are represented by ${\eta }_A$ and ${\eta }_B$, respectively, then what is the value of $\frac{{\eta }_A}{{\eta }_B}$?

• A. $\frac{12}{7}$
• B. $\frac{7}{12}$
• C. $\frac{12}{5}$
• D. $\frac{5}{12}$

Answer 1

The correct option is B $\frac{7}{12}$

Let the output of both engines be W

Let the first engine take $Q_1$ heat as input at temperature $T_1=600K$ and gives out heat $Q_2$ at temperature $T$ the second engine receives $Q_2$ as input ang give out $Q_3$ at temperature $T_3=100K$ to the sink.

Work done by first engine $W=Q_1-Q_2$

Work done by secondt engine $W=Q_2-Q_3$

Thus,

$Q_1-Q_2=Q_2-Q_3$

Dividing both sides by $Q_1$

$1-\frac{Q_2}{Q_1}=\frac{Q_2}{Q_1}-\frac{Q_3}{Q_1}$

$\Rightarrow 1-\frac{T}{T_1}=\frac{Q_2}{Q_1}(1-\frac{Q_3}{Q_2})$

$\Rightarrow 1-\frac{T}{T_1}=\frac{Q_2}{Q_1}(1-\frac{T_3}{T})$

$\Rightarrow 1-\frac{T}{T_1}=\frac{T}{T_1}(1-\frac{T_3}{T})$

$\Rightarrow \frac{T_1}{T}-1=1-\frac{T_3}{T}$

$\Rightarrow \frac{T_1}{T}+\frac{T_3}{T}=2$

$\Rightarrow \frac{1}{T}(T_1+T_3)=2$

$\Rightarrow T=\frac{(T_1+T_3)}{2}$

$\Rightarrow T=\frac{600+100}{2}=350K$

Now using Carnot engine efficiency formula

${\eta }_A=\frac{600-T}{600}$

${\eta }_B=\frac{T-100}{T}$

$\frac{{\eta }_A}{{\eta }_B}=\frac{\frac{600-350}{600}}{\frac{350-100}{350}}=\frac{\frac{250}{600}}{\frac{250}{350}}=\frac{350}{600}=\frac{7}{12}$