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 на вопрос
We can use the equations for the Carnot cycle to find the work and heat quantities for one cycle of the machine.
The efficiency of the Carnot cycle is given by:
η = 1 - T_cold/T_hot
where T_cold and T_hot are the temperatures of the cold and hot reservoirs, respectively. We can use this equation to find the efficiency of the machine:
η = 1 - 300/400 = 0.25
This means that 25% of the heat energy taken from the hot reservoir is converted into work, and the remaining 75% is rejected to the cold reservoir.
The work done by the machine is given by:
W = ηQ_hot = 0.25 x 2512 J = 628 J
where Q_hot is the heat energy taken from the hot reservoir.
The heat quantity given to the cold reservoir is given by:
Q_cold = (1-η)Q_hot = 0.75 x 2512 J = 1884 J
Therefore, for one cycle of the machine, the work done is 628 J and the heat given to the cold reservoir is 1884 J.