5.179. An ideal thermal machine, working according to Carnot cycle, gets from a heater 2512 J every cycle. Heater's temperature is 400 K, refrigerator one is 300 K. Find the work, done by machine, and heat quantity, given to the refrigerator, for one cycle.
Ответы
Ответ: The heat quantity given to the refrigerator for one cycle is 1884 J, and the work done by the machine is 628 J.
Объяснение:
η = 1 - T_Cold/T_Hot
where η is the efficiency of the machine, T_Cold is the temperature of the refrigerator, and T_Hot is the temperature of the heater. The maximum amount of work that the machine can do in one cycle is equal to the heat absorbed from the heater minus the heat released to the refrigerator. This can be expressed as:
W = Q_Hot - Q_Cold
where W is the work done by the machine, Q_Hot is the heat absorbed from the heater, and Q_Cold is the heat released to the refrigerator.
Given that the heater's temperature is 400 K and the refrigerator's temperature is 300 K, we can calculate the Carnot efficiency as:
η = 1 - T_Cold/T_Hot
η = 1 - 300/400
η = 0.25
This means that the maximum amount of work that the machine can do in one cycle is 25% of the heat absorbed from the heater. Therefore, the work done by the machine in one cycle is:
W = η * Q_Hot
W = 0.25 * 2512 J
W = 628 J
The heat released to the refrigerator can be calculated using the first law of thermodynamics:
Q_Cold = Q_Hot - W
Q_Cold = 2512 J - 628 J
Q_Cold = 1884 J