Q 11-11-033JEE MainMedium
A heat engine takes $1500$ J from a hot reservoir and gives out $1050$ J to a cold reservoir in each cycle. Find its efficiency as a percentage.
Numerical value type. Enter your answer.
Answer: 30
Concept. $W = Q_1 - Q_2$, $\eta = \dfrac{W}{Q_1}$.
Step 1. Work. $1500 - 1050 = 450$ J.
Step 2. Efficiency. $\dfrac{450}{1500} = 30\%$.
Answer. $30$.
Solution by Sreeraj P, M.Sc Physics