An engine is used to lift a 2700-kg truck to a height of 3.0 m at constant speed. In the lifting process, the engine received 4.18 × 10⁵ J of heat from the fuel burned in its interior. What is the efficiency of the engine?

Question

Quizplus · Accepted Answer

The efficiency of the engine is given by the ratio of the work done by the engine to the heat received by the engine. In this case, the work done by the engine is equal to the potential energy gained by the truck, which is mgh = (2700 kg)(9.8 m/s^2)(3.0 m) = 79,860 J. The heat received by the engine is 4.18 × 10^5 J. Therefore, the efficiency of the engine is 79,860 J / 4.18 × 10^5 J = 0.19.

An Engine Is Used to Lift a 2700-Kg Truck to a Height