Q. A Carnot engine whose source is at 400 K take 200 cal of heat and rejects 150 cal to the sink. What is the temperature of the sink?
A) 800 K
B) 400 K
C) 300 K
D) Cannot say
Explanation
(c) As $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}\ \therefore T_{2} = \frac{Q_{2}}{Q_{1}} \times T_{1} = \frac{150}{200} \times 400 = 300\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.