The heat engine consumed 6 kg of kerosene in 2 hours. What is the power of the motor if its efficiency is 25%?

Initial data: t (operating time of the heat engine) = 2 h; m (mass of consumed kerosene) = 6 kg; η (heat engine efficiency) = 25% or 0.25.

Reference data: q (specific heat of combustion of kerosene) = 46 * 10 ^ 6 J / kg.

SI system: t = 2 h = 2 * 3600 s = 7200 s.

Engine power can be expressed from the following formula:

η = N * t / (q * m) and N = q * m * η / t.

Let’s do the calculation:

N = 46 * 10 ^ 6 * 6 * 0.25 / 7200 = 9583 W.

Answer: The motor power is 9583 W.