Question

Тепловая машина совершает за цикл работу А=320 кДж и передает холодильнику количество теплоты Q2=320 кДж. Определите КПД цикла.

Answer 1

Дано A=320 кДж Q2=320 кДж КПД - ?

КПД=(Q1-Q2)/Q1=A/Q1

Q1=A+Q2=640 кДж

КПД=320*100%/640=50%

Answer 2

Calculating the Efficiency of the Cycle

To determine the efficiency of the cycle, we can use the formula for the efficiency of a heat engine:

Efficiency (η) = 1 - (Q2 / A)

Where: - Q2 is the amount of heat transferred to the refrigerator - A is the work done by the heat engine during the cycle

Given: - A = 320 kJ - Q2 = 320 kJ

We can now calculate the efficiency using the provided values:

Efficiency (η) = 1 - (320 kJ / 320 kJ)

Efficiency (η) = 1 - 1

Efficiency (η) = 0

Therefore, the efficiency of the cycle is 0.