Question

A Carnot engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is reduced by $62{}^{\circ }\text{C}$, its efficiency is doubled. The temperatures of the source and the sink are respectively

• A. $99{}^{\circ }\text{C},37{}^{\circ }\text{C}$
• B. $124{}^{\circ }\text{C},62{}^{\circ }\text{C}$
• C. $37{}^{\circ }\text{C},99{}^{\circ }\text{C}$
• D. $62{}^{\circ }\text{C},124{}^{\circ }\text{C}$

Answer 1

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

The efficiency of a Carnot engine is given by,
$\eta =1-\frac{T_2}{T_1}$
Here, $T_1$ and $T_2$ are temperatures of source and sink, respectively.

$\frac{1}{6}=1-\frac{T_2}{T_1}$

$\frac{1}{6}=\frac{T_1-T_2}{T_1}$

$T_1=6T_1-6T_2$

$\Rightarrow T_1=1.2T_2$

And, $\frac{1}{3}=1-\frac{T_2-62}{T_1}$

$\frac{1}{3}=\frac{T_1-T_2+62}{T_1}$

$\frac{1}{3}=\frac{1.2T_2-T_2+62}{1.2T_2}$

$1.2T_2=0.6T_2+186$

$\therefore T_2=310\text{K}=27{}^{\circ }\text{C}$

$\Rightarrow T_1=310\times 1.2=372\text{K}=99{}^{\circ }\text{C}$

Hence, $\left(B\right)$ is the correct answer.