Q. An ideal heat engine working between temperature $T_{1}$ and $T_{2}$ has an efficiency $\eta$, the new efficiency if both the source and sink temperature are doubled, will be
A) $\frac{\eta}{2}$
B) $\eta$
C) $2\eta$
D) $3\eta$
Explanation
(b) In first case $\eta_{1} = \frac{T_{1} - T_{2}}{T_{1}}$
In second case $\eta_{2} = \frac{2T_{1} - 2T_{2}}{2T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} = \eta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.