Determine the efficiency of the tractor engine, which required fuel weighing 4 kg with a specific heat

Determine the efficiency of the tractor engine, which required fuel weighing 4 kg with a specific heat of combustion of 42 MJ / kg to perform an effective action of 39 mJ.

Ap = 39 MJ = 39 * 10 ^ 6 J.

m = 4 kg.

q = 42 MJ / kg = 42 * 10 ^ 6 J / kg.

Efficiency -?

The efficiency factor of the tractor engine efficiency shows what part of the expended work of the engine Az goes into the useful work Ap of the tractor: Efficiency = Ap * 100% / Az.

The expended work A is the amount of thermal energy Qs, which is released during the combustion of fuel in the engine.

Qz = q * m, where q is the specific heat of combustion of the fuel, m is the mass of the fuel that has burned.

Efficiency = Ap * 100% / q * m.

Efficiency = 39 * 10 ^ 6 J * 100% / 42 * 10 ^ 6 J / kg * 4 kg = 23.3%.

Answer: the tractor engine has an efficiency of 23.2%.