Question

Тепловая машина, работающая по циклу Карно, за один рабочий цикл получает от нагревателя количество теплоты Q=500 Дж и производит работу A=200 Дж. Каково при этом отношение Т1/Т2 температур нагревателя и холодильника у этой машины?

Answer 1

Ответ:

Объяснение:

A = Q нагревателя-Q холодильника

q холодильника = q нагревателя - A

Q2 = 300 Дж

500/300 = 5 к 3

(вроде бы так, но это не точно)

Answer 2

Calculation of the Temperature Ratio in a Carnot Cycle

To determine the temperature ratio T1/T2 of the heat source and the refrigerator in a Carnot cycle, we can use the equation:

T1/T2 = √(Q2/Q1)

Where: - T1 is the temperature of the heat source (nagrevatel) in Kelvin. - T2 is the temperature of the refrigerator (kholodilnik) in Kelvin. - Q1 is the amount of heat received from the heat source in Joules. - Q2 is the amount of heat rejected to the refrigerator in Joules.

Given that the Carnot engine receives Q = 500 J of heat from the heat source and produces A = 200 J of work, we need to determine the values of Q1 and Q2.

Since the Carnot cycle is a reversible cycle, the amount of heat rejected to the refrigerator is equal to the amount of heat received from the heat source. Therefore, Q2 = Q1 = 500 J.

Now we can calculate the temperature ratio:

T1/T2 = √(Q2/Q1) = √(500/500) = 1

Therefore, the temperature ratio T1/T2 for this Carnot engine is 1.

Please note that the provided search results did not contain relevant information for this specific calculation. The calculation is based on the principles of the Carnot cycle and the equation mentioned above.