First slide

Heat engine

Question

A Carnot engine operates with source at $127^{0} C$ and sink at $27^{0} C$ . If the source supplies 40 kJ of heat energy, the work done by the engine is

Moderate

A

30 kJ
B

4 kJ
C

10 kJ
D

1 kJ

Solution

Efficiency of the Carnot cycle, $η = 1 - \frac{T_{2}}{T_{1}}$
$∴ η = 1 - \frac{(273 + 27)}{(273 + 127)} = 1 - \frac{3}{4} = \frac{1}{4}$
Also $η = \frac{Work done by engine}{Heat supplied by source} = \frac{W}{∆ Q} = \frac{W}{40 kJ}$
$W = 40 η = 40 \times \frac{1}{4} = 10 kJ$

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