A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is:
Answer & explanation
Correct answer: option 2
$\text{Efficiency of a carnot engine is given by } \eta = 1 - \frac{T_2}{T_1} $
$\text{ Here }\eta = 0.5 , T_1 = 327^oC = 600 K$
$ 0.5 = 1 - \frac{T_2}{600} \Rightarrow \frac{T_2}{600} = 0.5 $
$\Rightarrow T_2 = 300K = 27^oC$