Determine the efficiency of a tractor that consumed 1.5 kg of fuel with a specific heat of combustion

Determine the efficiency of a tractor that consumed 1.5 kg of fuel with a specific heat of combustion of 42 MJ / kg to perform work of 18.9 MJ.

Given:

Useful = 18.9 MJ = 18.9 * 10 ^ 6 Joules – the work done by the tractor;

m = 1.5 kilograms is the mass of fuel consumed by the tractor;

q = 42 MJ / kg = 42 * 10 ^ 6 Joule / kilogram – specific heat of combustion of fuel.

It is required to determine N – tractor efficiency.

Let’s find the amount of energy that was released during the complete combustion of the fuel:

Q = q * m = 42 * 10 ^ 6 * 1.5 = 63 * 10 ^ 6 Joules.

Then the efficiency of the tractor will be equal to:

N = A useful / Q = 18.9 * 10 ^ 6 / (63 * 10 ^ 6) = 18.9 / 63 = 0.3 = 30%.

Answer: The efficiency of the tractor is 30%.

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