In an ideal Carnot heat machine, the temperature of the refrigerator is t2 = 0, the temperature of the heater

In an ideal Carnot heat machine, the temperature of the refrigerator is t2 = 0, the temperature of the heater is t1 = 147. What amount of heat Q is transferred to the refrigerator in 5 cycles, if in one cycle the operating heat receives the amount of heat Q1 = 80 kJ from the heater?

Task data: Tx (abs. Refrigerator temperature) = 273 K (0 ºС); Тн (abs. Heater temperature) = 420 К (147 ºС); n (number of cycles) = 5 cycles; Qн1 (heat received by the working fluid for 1 cycle) = 80 kJ.

1) The efficiency of the specified Karnot heat engine: η = 1 – Тх / Тн = 1 – 273/420 = 0.35 (35%).

2) Heat transferred in 1 cycle: Qх1 = (1 – η) * Qн1 = (1 – 0.35) * 80 = 52 kJ.

3) Heat transferred in 5 cycles: Qx5 = Qx1 * n = 52 * 5 = 260 kJ.

Answer: For 5 cycles, the refrigerator of the specified Karnot heat engine will receive 260 kJ of heat.