Question

Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by ${62}^{o}C$, its efficiency gets doubled. The temperature of the source and sink respectively are

• A. $372K$ and $310K$
• B. $273K$ and $300K$
• C. ${99}^{o}C$ and ${10}^{o}C$
• D. ${200}^{o}C$ and ${37}^{o}C$

Answer 1

Answer: (A)

$372K$ and $310K$

Initial efficiency ${\eta }_1=\frac{T_1-T_2}{T_1}=\frac{1}{6}$
New efficiency ${\eta }_2=\frac{1}{3}=\frac{T_1-(T_2-62)}{T_1}=\frac{t_1-T_2}{T_1}+\frac{62}{T_1}$
$\therefore$ $\frac{1}{3}=\frac{1}{6}-\frac{62}{T_1}$
$⟹T_1=372K$
Now $\frac{1}{6}=\frac{T_1-T_2}{T_1}=\frac{372-T_2}{372}$
$⟹T_2=310K$