Q. A Carnot engine works between 600 K and 300 K. In each cycle of operation, the engine draws 1000 J of heat energy from the source. The efficiency of the engine is

A) 50%
B) 70%
C) 20%
D) 80%
Explanation

(a) $\eta = 1 - \frac{T_{2}}{T} = 1 - \frac{300}{600} = \frac{1}{2} = 50\%$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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