WBCS Preliminary Examination 2023 — General Studies · Question 173 of 200

Answer: (C) 40%

Carnot efficiency is $\eta=1-\dfrac{T_2}{T_1}$, with temperatures in kelvin. Here $T_1=227+273=500$ K and $T_2=27+273=300$ K, so $\eta=1-\dfrac{300}{500}=0.4$, i.e. 40%. The usual mistake is to use Celsius values, which gives a meaningless 88%. No heat engine can be 100% efficient, since that would need the sink to be at absolute zero (second law of thermodynamics), and the Carnot value is the upper limit for any engine working between the same two temperatures.