Question

У тепловой машины, работающей по циклу Карно, температура нагревателя в 1,6 раза больше температуры холодильника. за один цикл машина производит работу 12,0 кДж. какая работа за цикл затрачивается на изотермическое сжатие рабочего тела.

Answer 1

очень простая задача. кпд любой тепловой машины равно (А1-А2)/А1=1-А2/А1. для машины Карно А1 - работа при адиабатическом расширении, а А2 - при изотермическом сжатии. но у неё кпд так же равен (Т1-Т2)/Т1=1-Т1/Т2. сравнивая формулы получаем А2/А1=Т2/Т1, отсюда А2=А1(Т2/Т1)=(1/1,6)*12 кДж=7,5 кДж

Answer 2

Calculation of Work for the Carnot Cycle

To calculate the work done during the isothermal compression phase of the Carnot cycle, we need to use the given information. Let's break down the problem step by step:

1. The temperature of the heater is 1.6 times greater than the temperature of the refrigerator. 2. The machine produces 12.0 kJ of work per cycle.

Now, let's calculate the work done during the isothermal compression phase.

Solution:

1. Let's assume the temperature of the refrigerator is Tc. 2. The temperature of the heater will be 1.6Tc.

According to the Carnot cycle, the work done during the isothermal compression phase is given by the formula:

W = Qh - Qc

Where: - W is the work done - Qh is the heat absorbed from the heater - Qc is the heat rejected to the refrigerator

In the Carnot cycle, the heat absorbed from the heater is equal to the heat rejected to the refrigerator. Therefore:

Qh = Qc

Now, let's substitute the values into the equation:

W = Qh - Qc = Qh - Qh = 0

Therefore, the work done during the isothermal compression phase of the Carnot cycle is 0 kJ.

Please let me know if there's anything else I can help you with!