Question

Температура нагревателя идеальной тепловой машины Карно 427 С, а температура холодильника 27 С. за 1 цикл рабочее тело машины совершает работу 20 кДж. какое количество теплоты отдаст рабочее тело за 1 цикл холодильнику?

Answer 1

Вот так вот тоже долго решала

Answer 2

Calculation of Heat Transfer in a Carnot Cycle

To calculate the amount of heat transferred to the refrigerator in one cycle, we can use the formula for the efficiency of a Carnot cycle:

Efficiency (η) = 1 - (Tc/Th)

Where: - η is the efficiency of the Carnot cycle, - Tc is the temperature of the cold reservoir (in Kelvin), - Th is the temperature of the hot reservoir (in Kelvin).

In this case, the temperature of the hot reservoir (nagrevatel) is 427°C, which is equivalent to 700 Kelvin, and the temperature of the cold reservoir (holodilnik) is 27°C, which is equivalent to 300 Kelvin.

Using these values, we can calculate the efficiency:

η = 1 - (300/700) = 1 - 0.4286 = 0.5714

The efficiency of the Carnot cycle is 0.5714, which means that 57.14% of the heat energy is converted into work, and the remaining 42.86% is transferred to the cold reservoir (holodilnik).

To calculate the amount of heat transferred to the refrigerator (Qc), we can use the equation:

Qc = η * Qh

Where: - Qc is the amount of heat transferred to the refrigerator, - η is the efficiency of the Carnot cycle, - Qh is the amount of heat absorbed from the hot reservoir.

In this case, the work done by the working substance (rabocheye telo) in one cycle is given as 20 kJ.

Using the equation, we can calculate the amount of heat transferred to the refrigerator:

Qc = 0.5714 * 20 kJ = 11.43 kJ

Therefore, the amount of heat transferred to the refrigerator in one cycle is 11.43 kJ.

Please note that the calculations are based on the idealized assumptions of a Carnot cycle and may not account for all real-world factors.