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.
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%.
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