JEE Main 2014 · 12 Apr · Shift 1 · Physics Q12

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A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be :

20% and −43°C

40% and −33°C

50% and −20°C

70% and −10°C

Answer and worked solution

Official answer: B

$\eta=1-\frac{600}{1000}=40\%$ . For a Carnot engine $\frac{T_2}{T_1}=\frac{Q_2}{Q_1}$ :
$T_2=0.6\times400=240\text{ K}=-33^\circ\text{C}.$

Topics:Carnot cycle

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 12