Question

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

• A. ${80}^{\circ }C,{37}^{\circ }C$
• B. ${95}^{\circ }C,{28}^{\circ }C$
• C. ${90}^{\circ }C,{37}^{\circ }C$
• D. ${99}^{\circ }C,{37}^{\circ }C$

Answer 1

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

The correct answer is option(D).

WE know,

Efficiency,

$\eta =\frac{(T_2–T_1)}{T_2}=\frac{1}{6}⟹6T_2-6T_1=T_2⟹T_2=1.2T_1.......\left(i\right)$

Where $T_2$ is the source temperature and $T_1$ is the sink temperature.

If the sink temperature is reduced by 62°C, the efficiency gets doubled i.e.

$\eta =\frac{T_2-(T_1-62)}{T_2}=2\times \frac{1}{6}=\frac{2}{6}⟹6T_2-6T_1+372=2T_2⟹6T_2-2T_2-6T_1+372=0⟹4T_2-6T_1+372=0$

Substituting the value of $T_2$ from equation (i), we will get,

$4\left(1.2T_1\right)–6T_1+372=0⟹4.8T_1-6T_1+372=0⟹372=1.2T_1⟹T_1=310K=37°C$

Therefore, $T_2=1.2\times 310=372K=99°C$

Hence, the temperature of the source and the sink is 99°C and 37°C respectively.