Question

КПД тепловой машины равен 20%20%. Средняя мощность передачи теплоты холодильнику в ходе её работы составляет 8 кВт8 кВт. Какую работу совершает тепловая машина за минуту? Ответ выразить в кДжкДж, округлив до целых.

Answer 1

Из формулы кпд=(P1-P2)/P1 найдем среднюю мощность передачи теплоты нагревателю:
P1=P2/(1-кпд)=8/(1-0,2)=10 кВт
A=Q1-Q2=(P1-P2)*t=(10000-8000)*60=120000 Дж=120 кДж

Answer 2

Calculation of Work Done by the Heat Engine

To calculate the work done by a heat engine, we can use the formula:

Work = Efficiency * Heat Input

Given that the efficiency of the heat engine is 20% (0.2) and the average heat transfer power to the refrigerator during its operation is 8 kW, we can calculate the work done by the heat engine per minute.

Work = Efficiency * Heat Input = 0.2 * 8 kW = 1.6 kW

To convert this value to kilojoules (kJ), we can multiply it by 60 (since there are 60 seconds in a minute) and divide by 1000 (since 1 kilowatt is equal to 1000 watts).

Work = 1.6 kW * 60 / 1000 = 0.096 kJ

Therefore, the work done by the heat engine per minute is approximately 0.096 kilojoules (rounded to the nearest whole number).

Please note that the provided sources did not contain specific information related to the calculation of work done by a heat engine. The calculation was performed based on the given information and general knowledge of heat engines.