Determine the efficiency of the tractor, which required 50MJ of heat from the heater to perform 15MJ work.

Given:

A = 15 MJ = 15 * 10 ^ 6 Joules – the work done by the tractor;

Qн = 50 MJ = 50 * 10 ^ 6 Joules – the amount of heat received by the tractor engine from the heater.

It is required to determine n (%) – the efficiency of the tractor engine.

Taking into account that the perfect work is equal to the difference between the amount of heat received from the heater and the amount of heat transferred to the refrigerator, we get:

n = A / Qн = 15 * 10 ^ 6 / (50 * 10 ^ 6) = 15/50 = 0.3 or 30%.

Answer: the efficiency of the tractor engine is 30%.