Question

Идеальный тепловой двигатель совершает за один цикл работу 30 кДж. Известно, что температура нагревателя 127 °С, а температура холодильника 27 °С. Найти количество теплоты, отдаваемое за один цикл холодильнику.

Answer 1

Найдём КПД
η=(Τн-Тх)/Тн
Но также КПД определяется формулой
η=Qполуч/Qотд
Мы пользуемся соотношением Q=A(так как тепловыми потерями и прочим мы пренебрегаем)
Qотд-?
Приравняем:
Qпол/Qотд=(Тн-Тх)/Тн
Qотд=Qпол*Тн/(Тн-Тх)
Qотд=30*10^3*127/100
Qотд=38100 Дж
Ответ:38100 Дж

Answer 2

Calculation of Heat Transferred to the Refrigerator

To find the amount of heat transferred to the refrigerator in one cycle, we need to calculate the difference in heat between the hot reservoir (heater) and the cold reservoir (refrigerator).

Given: - Work done per cycle = 30 kJ - Temperature of the heater (hot reservoir) = 127 °C - Temperature of the refrigerator (cold reservoir) = 27 °C

We can use the formula for the efficiency of a heat engine to calculate the heat transferred to the refrigerator:

Efficiency = (Heat transferred to the refrigerator) / (Heat added to the heater)

The efficiency of an ideal heat engine is given by the Carnot efficiency formula:

Efficiency = 1 - (Temperature of the refrigerator / Temperature of the heater)

Rearranging the formula, we can solve for the heat transferred to the refrigerator:

Heat transferred to the refrigerator = Efficiency * Heat added to the heater

Let's calculate the efficiency first:

Efficiency = 1 - (27 °C / 127 °C)

Using the given temperatures, we can calculate the efficiency.

Calculation:

Efficiency = 1 - (27 / 127) = 0.7874

Now, we can calculate the heat transferred to the refrigerator:

Heat transferred to the refrigerator = Efficiency * Heat added to the heater

Heat transferred to the refrigerator = 0.7874 * 30 kJ

Answer:

The amount of heat transferred to the refrigerator in one cycle is approximately 23.622 kJ.

Please note that the calculations are based on the assumption of an ideal heat engine and the given temperatures.