Question

The efficiency of the heat engine working between ${327}^{o}C$ and ${27}^{o}C$ is to be increased by $10\%$. The temperature of the source should be increased by

• A. ${52}^{o}C$
• B. ${67}^{o}C$
• C. ${37}^{o}C$
• D. ${77}^{o}C$

Answer 1

The correct option is B ${67}^{o}C$

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. . Thus we get $\eta =1-\frac{300}{600}=50\%$. The efficiency is to be increased by $10\%$. Thus the new efficiency is 55%. Thus we get $0.55=1-\frac{300}{T_1}$
or
$T_1=\frac{300}{0.45}=667K$
Thus the increase in temperature of the source is $667-600={67}^{o}C$