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
$W = 1500 - 1050 = 450$ J. $\eta = \dfrac{450}{1500} = 0.30 = 30\%$.
Solution by Sreeraj P, M.Sc Physics