First slide

Heat engine

Question

A heat engine receives 50 kcal of heat from the source per cycle, and operates with an efficiency of 20%. The heat rejected by engine to the sink per cycle is

Moderate

A

40 kcal
B

25 kcal
C

30 kcal
D

50 kcal

Solution

Given $Q_{1} = 50 kcal and η = 20 %$
Using $η = \frac{W}{Q_{1}}, we have W = η \times Q_{1}$
i.e., work obtained per cycle W = $20 % \times 50 kcal$
= 10 kcal = 42kJ
Since, $Q_{1} = Q_{2} + W so, Q_{2} = Q - W$
i.e., heat rejected to the sink per cycle $Q_{2}$
= 50 kcal -10 kcal = 40 kcal

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