Q. $200\ cal$ of heat is given to a heat engine so that it rejects $150\ cal$ of heat, if source temperature is $400\ K$, then the sink temperature is
A) $300\ K$
B) $200\ K$
C) $100\ K$
D) $50\ K$
Explanation
(a) Here, $Q_{1} = 200cal,\ Q_{2} = 150cal,\ T_{1} = 400K$
As $\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}}$
$\therefore T_{2} = \frac{Q_{2}}{Q_{1}} \times T_{1} = \frac{150}{200} \times 400 = 300K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.