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Carnot Cycle

A cyclic heat...

Question

A cyclic heat engine does 50kJ of work per cycle. If efficiency of engine is 75%, the heat rejected per cycle will be:

A

60.6

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B

16.6

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C

200

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D

600

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Solution

The correct option is B 16.6
Carnot efficiency = Work done/Heat supplied(Q1)

$0.75 = 50 / Q 1$

or, $Q_{1} = \frac{200}{3}$

and, $W o r k = Q_{1} - Q_{2}$

or, $Q_{2} = \frac{200}{3} - 50 = \frac{50}{3} = 16.6 k J$

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