Question

A Carnot's engine has an efficiency of $50$% at sink temperature ${50}^{\circ }$. Calculate the temperature of source :

• A. ${133}^{\circ }C$
• B. ${143}^{\circ }C$
• C. ${100}^{\circ }C$
• D. ${373}^{\circ }C$

Answer 1

Answer: (D)

${373}^{\circ }C$

$\eta =\frac{\text{Net work done per cycle}}{\text{Total amount of heat absorbed per cycle}}$
or $\eta =\frac{W}{Q_1}$
$\therefore \eta =\frac{Q_1-Q_2}{Q_1}$
or $\eta =1-\frac{Q_2}{Q_1}$
as $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$
$\therefore \eta =1-\frac{T_2}{T_1}....\left(i\right)$
where, $T_2$ is temperature of sink and $T_1$ is the temperature of the sources.
Here, $\eta =50$%
$T_2={50}^{\circ }C=273+50$
$=323K$
Putting these values in Eq. (i), we get
$\frac{50}{100}=1-\frac{323}{T_1}$
$T=646K={373}^{\circ }C$.