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)

99$^{\circ}$C, 37$^{\circ}$C

Answer 2

$\eta_{1}=1-\frac{T_{L}}{T_{H}}=\frac{W}{Q_{1}}=\frac{1}{6}$ or $5T_{H}-6T_{L}=0 ...\left(i\right)$ $\eta_{2}=1-\frac{T_{L}-62}{T_{H}}=2\eta_{1}=\frac{1}{3} \left(Given\right)$ $\Rightarrow 1-\frac{1}{3}=\frac{T_{L}-62}{T_{H}}$ or $2T_{H}-3T_{L}=-186 ...\left(ii\right)$ Solving $\left(i\right)$ and $\left(ii\right)$, we get $\therefore T_{H} = 372 \,K = 99^{\circ}C$ $T_{L}=\frac{5}{6}T_{H}=\frac{5}{6}\times 372\, K=310\, K =37^{\circ}C$