Question

Помогите пожалуйста...тепловая машина с кпд 20% за цикл работы отдает холодильнику количество теплоты, равное 80 Дж. какую полезную работу машина совершает за цикл? должно получится 20 дж...но как это решить подскажить мне пожалуйста...физики ау

Answer 1

КПД=Апол./Qподв. , где A=Qподв - Qотв (т.е. A+Qотв=Qподв), где Qотв = 80 Дж. Получим:

20%=А/(А+Qотв)=А/(А+80)=0,2. Решим, получим А=20 Дж

Answer 2

Calculation of Useful Work Done by the Heat Engine

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

Efficiency (η) = Useful work output / Heat input

Given that the efficiency (η) is 20% and the heat input is 80 J, we can calculate the useful work done by the heat engine.

Let's denote the useful work done by the heat engine as W. We can rearrange the formula for efficiency to solve for W:

W = η * Heat input

Substituting the given values:

W = 0.20 * 80 J

Calculating the value:

W = 16 J

Therefore, the heat engine performs 16 J of useful work during the cycle.

Please note that the efficiency of a heat engine is defined as the ratio of useful work output to the heat input. In this case, the heat engine has an efficiency of 20%, which means that only 20% of the heat input is converted into useful work, while the remaining 80% is lost as waste heat.

I hope this helps! Let me know if you have any further questions.