Question

A Carnot engine has an efficiency of $50\%$ when its sink is at temperature of ${27}^{o}C$. The temperature of the source is

• A. ${300}^{o}C$
• B. ${327}^{o}C$
• C. ${373}^{o}C$
• D. ${273}^{o}C$

Answer 1

The correct option is B ${327}^{o}C$

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. The temperature of the sink is given as ${27}^{o}C=300K$. Thus we get $0.5=1-\frac{300}{T_1}$
or
$T_1=600K={327}^{o}C$