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 engine is doubled. The temperatures 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

Answer: (A)

${\eta }_1=1-\frac{T_L}{T_H}=\frac{W}{Q_1}=\frac{1}{6}$
or $5T_H-6T_L=\mathrm{0...}\left(i\right)$
${\eta }_2=1-\frac{T_L-62}{T_H}=2{\eta }_1=\frac{1}{3}\left(Given\right)$
$\Rightarrow 1-\frac{1}{3}=\frac{T_L-62}{T_H}$
or $2T_H-3T_L=-\mathrm{186...}\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get
$\therefore T_H=372K=99^{\circ }C$
$T_L=\frac{5}{6}{T}_H=\frac{5}{6}\times 372K=310K=37^{\circ }C$