Determine the efficiency of the tractor engine, which required 2 kg of fuel with

Determine the efficiency of the tractor engine, which required 2 kg of fuel with a specific heat of combustion of 4.2 MJ / kg to perform work of 2.3 MJ.

These tasks: Ap (useful work performed by the considered tractor) = 2.3 MJ (2.3 * 10 ^ 6 J); m (mass of the required fuel) = 2 kg; q (specific heat of combustion of the fuel taken) = 4.2 MJ / kg (4.2 * 10 ^ 6 J / kg).

The efficiency of the engine of the tractor under consideration can be determined by the formula: η = Ap / Q = Ap / (q * m).

Calculation: η = 2.3 * 10 ^ 6 / (4.2 * 10 ^ 6 * 2) = 0.274 or 27.4%.

Answer: The considered tractor has an engine efficiency of 27.4%.