Q. What is the temperature of source in Carnot cycle of 10% efficiency when heat exhausts at 270 K?
A) 400 K
B) 500 K
C) 300 K
D) 600 K
Explanation
(c) As $\eta = 1 - \frac{T_{2}}{T_{1}}\ \ \therefore\ \frac{T_{2}}{T_{1}} = 1 - \eta = 1 - \frac{10}{100} = \frac{90}{100}$
or $T_{1} = \frac{100T_{2}}{90}$
$= \frac{100}{90} \times 270 = 300\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.