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

Answer 1

Answer: (C), (D)

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

Initially $\eta =(1-\frac{T_2}{T_1})=\frac{W}{Q}=\frac{1}{6}$ ....(i)
Finally ${\eta }^{\prime }$=$(1-\frac{T_2^{\prime }}{T_1})=(1-\frac{(T_2-62)}{T_1})=1-\frac{T_2}{T_1}+\frac{62}{T_1}$
$=\eta +\frac{62}{T_1}$ ...(ii)
it is given that ${\eta }^{\prime }=2\eta$. Hence solving equation (i) and (ii)
$\Rightarrow T_1=372K={99}^{\circ }C$ and $T_2=310K={37}^{\circ }C$