Q 11-11-142JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
A heat engine is involved with exchange of heat of $1915\ \text{J}$, $-40\ \text{J}$, $+125\ \text{J}$ and $-Q\ \text{J}$, during one cycle achieving an efficiency of $50.0\%$. The value of $Q$ is:
Answer: (C) $980\ \text{J}$
Heat absorbed: $1915 + 125 = 2040$ J. Net work $W = 1915 - 40 + 125 - Q = 2000 - Q$.
$$\eta = \frac{W}{2040} = 0.5 \Rightarrow 2000 - Q = 1020 \Rightarrow Q = 980\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics