Project ForgeThermodynamics
Entropy & Second Law
GATE 2014
Medium
A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K. It rejects 1 kJ of heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is
- A0.8
- B1.0
- 1.4
- D2.0
Explanation
For a reversible cyclic engine, the net entropy exchanged with all reservoirs is zero: Q1/T1+Q2/T2=Qr/Tr. So 2/1000+Q2/800=1/400 ⟹ Q2/800=0.0025−0.002=0.0005 ⟹ Q2=0.4 kJ. Net work = (Q1+Q2)−Qr = (2+0.4)−1 = 1.4 kJ.
Source: GATE ME 2014 (Session 1 / Set-1) (2014)
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