ThermodynamicsHard

Question

A Carnot engine operates between 327°C and 117°C. If it absorbs 120 cal heat per cycle from the source, then the heat rejected per cycle to the sink is

Options

A.120 cal
B.42 cal
C.78 cal
D.90 cal

Solution

$\frac{\left| q_{rej} \right|}{q_{abs}} = \frac{T_{C}}{T_{H}} \Rightarrow \left| q_{rej} \right| = \frac{390}{600} \times 120 = 78\text{ cal}$

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