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An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s efficiency? b. How much work is done in each cycle? c. What is the power output of the engine if each cycle lasts for 0.305 s?

Respuesta :

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

[tex]\eta=\dfrac{W}{Q_1}[/tex]

[tex]\eta=\dfrac{1210}{1749}[/tex]

[tex]\eta=0.6918[/tex]

η = 69.18 %

We know that rate of work done is known as power

[tex]P=\dfrac{W}{t}[/tex]

[tex]P=\dfrac{1210}{0.305}\ W[/tex]

P=3967.21 W

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