# When burning gasoline in a car engine for 2 seconds, energy losses amounted to 300 kJ, while the engine

When burning gasoline in a car engine for 2 seconds, energy losses amounted to 300 kJ, while the engine developed a power of 50 kW. Determine the efficiency of the motor.

t = 2 s.

Wpot = 300 kJ = 3 * 10 ^ 5 J.

Np = 50 kW = 5 * 10 ^ 4 W.

Efficiency -?

The formula for determining the efficiency coefficient of the engine efficiency will be as follows: efficiency = Ap * 100% / Az.

We will express the useful work Ap by the formula: Ap = Az – Wpot.

Az = An + Wpot.

The expended work Az is expressed through the power: An = Nz * t.

Ap = 5 * 10 ^ 4 W * 2 s = 10 * 10 ^ 4 J = 1 * 10 ^ 5 J.

Az = 1 * 10 ^ 5 J +

Az = 1 * 10 ^ 5 J + 3 * 10 ^ 5 J = 4 * 10 ^ 5 J /

Efficiency = 1 * 10 ^ 5 J * 100% / 4 * 10 ^ 5 J = 25%.

Answer: a gasoline engine has an efficiency of 25%.

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