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KIM [24]
3 years ago
14

How much energy becomes unavailable for work in an isothermal process at 440 k, if the entropy increase is 25 j/k?

Physics
1 answer:
just olya [345]3 years ago
8 0
Answer: 11,000 J

Explanation:

In an isothermal process,

\text{entropy increase} =  \frac{\text{amount of energy in heat transfer}}{\text{temperature} }  (1)

Note that, the energy used in heat transfer is not available for work. So, the amount of energy unavailable for work is equal to the energy used in heat transfer.

To obtain the amount of energy in heat transfer, we multiply both sides of equation (1) by the denominator of the right side of (1) so that 

amount of energy in heat transfer = (entropy increase)(temperature)
                                                      = (25 J/K)(440 K)
                                                      = 11,000 J

Since the amount of energy unavailable for work is equal to the amount of energy in the heat transfer, therefore the amount of energy unavailable for work is 11,000 J.
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Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, \theta_1=47.7^{\circ}

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Using Snell's law for air liquid interface as :

n_1\ sin\theta_1=n_2\ sin(90)

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n_1=\dfrac{1}{sin(47.7)}

n_1=1.35

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

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3 years ago
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Answer:

0.015A

Explanation:

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0.015A

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Answer:

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Explanation:

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