Question:
A heat engine is involved with exchange of heat of $1915 \mathrm{~J},-40 \mathrm{~J},+125 \mathrm{~J}$ and $\mathrm{QJ}$, during one cycle achieving an efficiency of $50.0 \%$. The value of $Q$ is:
Correct Option: , 3
Solution:
$\eta=\frac{\text { Work done }}{\text { Heat supplied }}$
$\frac{1}{2}=\eta=\frac{1915-40+125-Q}{1915+125}$
$\frac{1}{2}=\frac{2000-\mathrm{Q}}{2040}$
$2040=4000-2 Q$
$2 Q=1960$
$Q=980 \mathrm{~J}$
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