An ideal heat engine with an efficiency of 30% produces 6kJ of work per cycle.

An ideal heat engine with an efficiency of 30% produces 6kJ of work per cycle. How much heat does it give to the refrigerator in one cycle?

It is known that the efficiency of an ideal heat engine is determined from the formula:

η = Qн – Qх / Qх,

where Qн is the heat that is transferred from the heater to the working fluid, Qх is the amount of heat given by the working fluid to the refrigerator.

It is also known that the work performed in one cycle by an ideal heat engine:

A = Qн – Qх,

if this is taken into account, then

η = A / Qx,

from here

Qx = A / η.

Let’s make the calculations:

η = 30% = 0.3;

A = 6 (kJ) = 6 * 10 ^ 3 (J);

Qx = A / η = 6 * 10 ^ 3 / 0.3 = 20 * 10 ^ 3 (J) = 20 (kJ).

Answer: Qx = 20 (kJ).




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