SSC JE Mechanical Engineering 2018 Set 3 Paper

Question : 148 of 200

Marks: +1, -0

A heat engine performs a work of 100 kJ per cycle.
The efficiency of the engine is 50%. What will be the amount of heat rejected (in kJ) per cycle?

Solution:

For a heat engine efficiency =

{\work \done}/{\heat \supplied}

={Q_1-Q_2}/Q_1

0.50=100/Q_1⇒[Q_1=200]

Q_1-Q_2=100

⇒Q_2=200-100=100

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