Question

The temperature of sink of Carnot engine is ${27}^{\circ }$C. Efficiency of engine is 25%. Then temperature of source is

• A. 227^o C
• B. 327^o C
• C. 127^o C
• D. 27^o C

Answer 1

The correct option is C 127^o C

$\eta =1-\frac{T_2}{T_1}\Rightarrow \frac{25}{100}=1-\frac{300}{T_1}\Rightarrow \frac{1}{4}=1-\frac{300}{T_1}$

$T_1=400K={127}^{\circ }C$