# What is the efficiency of the engine if a 55 kW car per 1 kW / hour consumes 0.31 kg of gasoline?

What is the efficiency of the engine if a 55 kW car per 1 kW / hour consumes 0.31 kg of gasoline?

Np = 55 kW = 55000 W.

t = 1 h = 3600 s.

q = 4.6 * 107 J / kg.

m = 0.31 kg.

Efficiency -?

The efficiency of a car of the “Volga” brand will be expressed by the formula, according to the definition: efficiency = Ap * 100% / Az, where Ap, Az is the useful and expended work of the car engine.

The useful work of the engine for t = 1 hour An will be expressed by the formula: An = Np * t.

Ap = 55000 J * 3600 s = 198000000 J.

During this time t = 1 hour, the engine burns mb = 55 * m kg of gasoline. From it the expended amount of heat is released Az = q * mb = q * 55 * m.

Az = 4.6 * 107 J / kg * 55 * 0.31 kg = 784300000 J.

Efficiency = 198,000,000 J * 100% / 784,300,000 J = 25%.

Answer: the engine of a car “Volga” has an efficiency of 25%. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.