Question

Тепловой двигатель за один цикл получает от нагревателя 150 кДж теплоты, а холодильнику отдает 60000 Дж.Определите КПД этого двигателя. Хеееееееелпппп физика

Answer 1

Ответ:

25%

Объяснение:

Дано :

Ап = 150 кДж = 150000 Дж

Аз = 60000 Дж

_____________

КПД - ?

Решение :

КПД = Ап / Аз • 100% = 150000 Дж : 60000 Дж • 100% = 25%

Answer 2

Ответ:

Ответ:

25%

Объяснение:

Дано :

Ап = 150 кДж = 150000 Дж

Аз = 60000 Дж

_____________

КПД - ?

Решение :

КПД = Ап / Аз • 100% = 150000 Дж : 60000 Дж • 100% = 25%

Объяснение:

Answer 3

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.

Given: - Heat received by the heat engine (нагреватель): 150 kJ - Heat rejected by the refrigerator (холодильник): 60,000 J

To calculate the efficiency, we can use the following formula:

Efficiency (КПД) = (Useful work output / Heat input) x 100%

In this case, the useful work output is the difference between the heat received and the heat rejected:

Useful work output = Heat received - Heat rejected

Substituting the given values into the formula:

Useful work output = 150 kJ - 60,000 J

Now, we can calculate the efficiency:

Efficiency (КПД) = (Useful work output / Heat input) x 100%

Please note that we need to convert the units to ensure consistency. Let's convert 150 kJ to J:

150 kJ = 150,000 J

Substituting the values into the formula:

Efficiency (КПД) = ((150,000 J - 60,000 J) / 150,000 J) x 100%

Simplifying the equation:

Efficiency (КПД) = (90,000 J / 150,000 J) x 100%

Efficiency (КПД) = 0.6 x 100%

Efficiency (КПД) = 60%

Therefore, the efficiency (КПД) of this heat engine is 60%.