An ideal heat engine gives the refrigerator 30% of the energy it receives from heating. Find the efficiency of this machine.

Given:

Qx = 30% of Qn = 0.3 * Qn – the heat engine gives the refrigerator 30% of the energy received from the heater.

It is required to determine n – the coefficient of performance (COP) of the heat engine.

Let Q be the amount of energy received by the machine from the heater. Then, according to the condition of the problem, the machine gives the refrigerator an amount of energy equal to Q1 = 0.3 * Q.

The useful work done by the machine will be equal to:

A = Q – Q1 = Q – 0.3 * Q = 0.7 * Q.

Then the efficiency of the heat engine will be equal to:

n = A / Q = 0.7 * Q / Q = 0.7 = 70%.

Answer: the efficiency of the heat engine is 70%.