Question

A reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is reduced by ${62}^{\circ }C$, the efficiency of the engine is doubled. The temperature of the source and sink are:

• A. ${99}^{\circ }C,{37}^{\circ }C$
• B. ${80}^{\circ }C,{37}^{\circ }C$
• C. ${95}^{\circ }C,{37}^{\circ }C$
• D. ${90}^{\circ }C,{37}^{\circ }C$

Answer 1

The correct option is A ${99}^{\circ }C,{37}^{\circ }C$

Given : Work $W=\frac{1}{6}{Q}_H$

$\therefore$ Efficiency of engine $\eta =\frac{W}{Q_H}=\frac{1}{6}$ .........(1)

Let the initial source and sink temperatures be $T_H$ and $T_L$ respectively.

$\therefore$ $\eta =1-\frac{T_L}{T_H}$

or, $\frac{1}{6}=1-\frac{T_L}{T_H}$ $⟹T_H=1.2T_L$ ........(2)

Now the sink temperature is reduced by ${62}^{o}C$ i.e $T_L^{\prime }=T_L-62$ ........(3)

New efficiency is doubled i.e ${\eta }^{\prime }=\frac{2}{6}=\frac{1}{3}$

Using ${\eta }^{\prime }=1-\frac{T_L^{\prime }}{T_H}$

$\therefore$ $\frac{1}{3}=1-\frac{T_L-62}{1.2T_L}$ (using (2) and (3))

or, $\frac{T_L-62}{1.2T_L}=\frac{2}{3}$ $⟹T_L=310K$ or, $T_L=320-273={37}^{o}C$

$\therefore$ $T_H=1.2\times 310=372K$

$⟹$ $T_H=372-273={99}^{o}C$