What is the efficiency of a lift with a gasoline engine, if 9 g of gasoline had to be burned to lift 150 kg of load to a height of 60 m?

Given: M = 150 kg, h = 60 m, m = 9g = 0.009 kg
Find: efficiency
Decision:
If the system were perfect, then the equation would look like this:
Мgh = qm, but it is not ideal, which means that not everything that burned out will go up, which means the equation looks like this:
Мgh = qmn, where n is efficiency, q is the specific heat of combustion of gasoline.
n = Mgh / qm
n = 150 kg * 10N / kg * 60m / 0.009 kg * 46 * 10 ^ 6J / kg = 90000/414000 = 0.217
n≈22%
Answer: 22%.