Question

в тепловом двигателе количество теплоты, переданное холодильнику, составляет 15 кДж, а теплота нагревателя равна 0,020 МДж. Определите КПД такого двигателя (в %) ​

Answer 1

Ответ:

25%

Объяснение:

Q1=20 кДж    Q2=15 кДж    n=?

===

n=(Q1-Q2)/Q1=(20-15)/20=0.25

======================

Answer 2

Calculation of Efficiency (КПД) of a Heat Engine

To determine the efficiency (КПД) of a heat engine, we need to calculate the ratio of the useful work output to the heat input. In this case, the heat input is the amount of heat supplied by the heater, and the useful work output is the amount of heat transferred to the refrigerator.

Given: - Heat transferred to the refrigerator (Q2) = 15 kJ - Heat supplied by the heater (Q1) = 0.020 MJ

To calculate the efficiency, we can use the formula:

Efficiency (КПД) = (Q2 / Q1) * 100%

Let's substitute the given values into the formula:

Efficiency (КПД) = (15 kJ / 0.020 MJ) * 100%

Converting the units to match:

- 1 MJ = 1000 kJ

Efficiency (КПД) = (15 kJ / (0.020 MJ * 1000 kJ/MJ)) * 100%

Simplifying the calculation:

Efficiency (КПД) = (15 / 20) * 100%

Efficiency (КПД) = 75%

Therefore, the efficiency (КПД) of the heat engine is 75%.