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JEE Main 2020 (2 Sep, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (2 Sep, Shift 2)

Q52JEE Main 2020 · 2 SEP · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A heat engine is involved with exchange of heat of 1915 J, – 40J, + 125 J and –Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is

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Answer: (a)

Heat absorbed $Q_{in} = 1915 + 125 = 2040\text{ J}$ .
Net work $W = 1915 - 40 + 125 - Q = 2000 - Q$ .
$\eta = \frac{W}{Q_{in}}$ : $0.5 \times 2040 = 2000 - Q \Rightarrow Q = 980\text{ J}$
So the answer is (a) 980 J.

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