SSC JE Mechanical Engineering 2018 Set 5 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:

Carnot efficiency

= {\Work \done}/{ \Heat \supplied (Q_1 )}

0.50 =100/Q_1

or,

Q_1 = 200

and, Work done =

Q_1 -Q_2

or,

Q_2 = 200 -100 = 100

kJ.

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