Question

Figure below represents a Carnot engine that works between temperatures $T_1$ 400 K and $T_2$ 150 K and drives aCarnot refrigerator that works between temperatures $T_3$ 325 K and $T_4$ 225 K.What is the ratio $\frac{Q3}{Q1}$?

• A. 104/85
• B. 104/95
• C. 85/104
• D. 95/104

Answer 1

The correct option is C

85/104

Let us consider the efficiency of the heat engine first.

${\eta }_{H.E}=1-\frac{T_2}{T_1}$

$=1-\frac{150}{400}=\frac{5}{8}$

$\therefore W={\eta }_{H.E}\times Q_1$

$=\frac{5}{8}{Q}_1$.......................(1)

Now this work is utilized by the refrigerator to add heat to the hot body from the cold body. The efficiency of this refrigerator is calculated using.

${\eta }_r=1-\frac{T_3}{T_4}$

$=1-\frac{225}{325}$

$=\frac{4}{13}$...........(2)

We know that ${\eta }_r=\frac{Q_2}{w}$

$\Rightarrow Q_2=w\times {\eta }_r$ ............(3)

From $1^{st}$ law of thermodynamics we know

$Q_3=Q_2+w$

$=w+w\times {\eta }_r$ - - - - - - (4)

Substituting for w and $et{a}_r$ in (4) we get

$Q_3=\frac{5}{8}{Q}_1+\frac{5}{8}{Q}_1\times \frac{4}{13}$

$=Q_1\left(\frac{65+20}{8\times 13}\right)$

$=Q_1\times \frac{85}{104}$

$\frac{Q_3}{Q_1}=\frac{85}{104}$