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Dahasolnce [82]
3 years ago
15

Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and

Y?
Physics
2 answers:
Thepotemich [5.8K]3 years ago
3 0

<u>First law of thermodynamics:</u>

  • It states that <em>"Energy neither be created nor it can be destroyed". </em>simply it converts one form of energy into another form.
  • It is also known as<em> "law of conservation of energy"</em>

<u>Limitations of First law</u>

  1. It doesn't provide a clear idea about the direction of transfer of heat.
  2. It doesn't provide the information that how much heat energy converted inti work.
  3. Its not given any practical applications.

<u>II law of thermodynamics:</u>

It states that <em>"the total entropy of the system can never decrease over time"</em>

It is strongly proved by two laws, they are

<em>1. Kelvin-plank statement:</em>

      He stated that "any engine does not give 100% efficiency". It violates the Perpetual motion of machine II kind<em>(PMM-II).</em>

<em>2. Classius statement: </em>

<em>    </em><em>    It states that "Heat always flows from high temperature body to low temperature body, without aid of external energy". </em>

<em>          Also it stated that " Heat can also be transferred from low temperature body to high temperature body, by the aid of an external energy".</em>

<em>Applications of II law: </em>

<em>Refrigeration &Air conditioning, Heat transfer, I.C. engines, etc.</em>

KatRina [158]3 years ago
3 0

Answer:

The answer is D.

Explanation:

I just took the quiz.

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Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi
meriva

Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Explanation:

So we are told that it is a RC circuit. We are told Q = C V [1 - e^(-t/RC)] = 12.0 V, R =  1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

Q = CV[1-e^{-t/RC} ]

The capacitor is fully charge when t approaches infinity, therefore:

Q =  \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.

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