The power output of a car engine running at 2500 rpm is 500 kW . You may want to review (Pages 574 - 577) . Part A How much work is done per cycle if the engine's thermal efficiency is 20.0 %

Explanation:

2500 rpm = 2500 / 60 rps ( revolution per second )

= 41.67 rps .

power output = 500 kW

thermal efficiency = 20 %

power input = (100 /20) x 500 kW

= 2500 kW .

work done per second = 2500 kJ

work done per cycle = 2500 / 41.67

= 60 kJ .

they should be similar or the same

the answer is b, have negligible mass