Question

4. Термический коэффициент полезного действия теплового дви гателя п. =20%. Определите работу, совершенную двигателем за цикл. если количество теплоты, переданное холодильнику, Q 2 =-1.6 Если можно на листок ​

Answer 1

Ответ:

400 Дж

Объяснение:

n=0.2 (20%)    Q2=1600 Дж    A=?

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n=A/Q1

Q1=Q2+A

Q2+A=A/n

A=Q2/(1/n-1)=1600/(1/0.2-1)=400 Дж

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Answer 2

Calculation of Work Done by the Engine

To calculate the work done by the engine during a cycle, we need to use the formula:

Work = Q1 - Q2

Where: - Q1 is the heat input to the engine - Q2 is the heat rejected by the engine to the refrigerator

Given that the thermal efficiency of the engine (η) is 20% (0.2), we can calculate the heat input (Q1) using the formula:

Q1 = Q2 / η

Substituting the given values, we have:

Q1 = -1.6 / 0.2 = -8

Therefore, the heat input to the engine is -8 units.

Now, we can calculate the work done by the engine:

Work = Q1 - Q2 = -8 - (-1.6) = -6.4 units

Therefore, the work done by the engine during the cycle is -6.4 units.

Please note that the negative sign indicates that work is being done on the engine, rather than by the engine.