Q. The efficiency of a Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be
A) 100 K
B) 600 K
C) 400 K
D) 500 K
Explanation
(c) As $\eta = 1 = \frac{T_{2}}{T_{1}}\ $
$\therefore\ \frac{50}{100} = 1 = \frac{500}{T_{1}}\ or\ T_{1} = 1000K$
Again, $\frac{60}{100}\ = 1 - \frac{T_{2}}{1000}$
Or $T_{2} = 400\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.