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dybincka [34]
3 years ago
13

In one cycle, a freezer uses 585 J of electrical energy in order to remove 1750 J of heat from its freezer compartment at 10.0 F

.a. What is the coefficient of performance of this freezer?b. How much heat does it expel into the room during this cycle?
Physics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

(a) 0.525

(b) 1750J

Explanation:

Coefficient of Performance (COP) = Q2/(Q1 - Q2)

Q1 = heat rejected to the surrounding = 1750J

Q2 = heat absorbed in the environment = 585K

COP = 585/(1700 - 585) = 585/1115 = 0.525

(b). Heat expelled to the surrounding (room) = 1750J

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Answer:

39.05°

Explanation:

As we know that, the diffraction is the phenomena of bending of light when it passes through  an obstacle.

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dsin\theta=n\lambda

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Given that, d is 84 cm, n is 2, and the wavelength can be calculated as,

\lambda=\frac{c}{f}

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Here, c is 343 m/s and f is 1300 Hz,

Therefore,

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Recall diffraction equation in term of sin\theta.

sin\theta=\frac{n\lambda}{d}

Put all the variables.

sin\theta=\frac{2\times 0.264 m}{84 cm}\\sin\theta=\frac{2\times 0.264 m}{0.84 m}\\\theta=39.05^{\circ}

Therefore, it is the required angle between the first 2 order of diffraction.

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