The heat engine lifts a weight of 58 kg to a height of 3.2 m. The engine efficiency is 29%;

The heat engine lifts a weight of 58 kg to a height of 3.2 m. The engine efficiency is 29%; the amount of heat received from the heater in one cycle is 64 J. How many cycles does the engine perform during the lifting of the load?

h = 3.2 m.

g = 9.8 m / s ^ 2.

m = 58 kg.

Efficiency = 29%.

Q1 = 64 J.

n -?

To lift a load, you need to spend E energy, the value of which is determined by the formula: E = m * g * h.

Let us write down the definition for efficiency: efficiency = E * 100% / Q, where Q is the amount of heat that the heat engine consumes to lift the load.

Q = n * Q1.

Efficiency = m * g * h * 100% / n * Q1.

n = m * g * h * 100% / efficiency * Q1.

n = 58 kg * 9.8 m / s ^ 2 * 3.2 m * 100% / 29% * 64 J = 98.

Answer: to lift the load, the heat engine must do n = 98 cycles.



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