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Entropy - Thermal Equilibrium

  1. Jul 16, 2006 #1
    I stumbled across this question in one of the physics competition selection test but after thinking like 2 days i still cant figure out hw to solve it.
    I've been introduced to the equation: entropy, S=Q/T where Q is the heat energy and T is the temperature.
    Then i've been told that the entropy of a system always remains constant or increase.
    Then the question is to show that when a hotter object and a colder object are put together and isolated the hotter object always becomes colder and colder becomes hotter. We are expected to solve this with the information provided only.
    Help anyone ?? :biggrin: ?
  2. jcsd
  3. Jul 16, 2006 #2
    Well, I would suggest you look at the change in entropy. I.e. Delta S
  4. Jul 16, 2006 #3

    Andrew Mason

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    As Cyrus says, you must look at the change in entropy, which according to the second law of Thermodynamics, must be greater than or equal to zero. Assume the flow of heat is from the colder to the hotter object and determine whether the change in entropy of the system is greater than zero. See if it fits with the second law.

    The change entropy of the system is the sum of the entropy changes in each object. Use the convention: heat flow into an object is positive and heat flow out is negative. dS = dQ/T

  5. Jul 16, 2006 #4
    We already know the equation delta S = delta Q(reversible) / T

    Now we want to explain the way in which spontaneous processes occur using the entropy. Assume two spaces, one space has temperature T1 and the other T2 and we isolate this system.

    The difference between the two entropies delta S at equilibrium is given by:

    delta S = delta S(1) + delta S(2)

    (total entropy change of the system is the sum of the two separate entropy changes)

    Now we can write: delta S(1) = delta Q(reversible)(1) / T(1) and delta S(2) = delta Q(reversible)(2) / T2. Now because the system is isolated, Q(rev)1 = -Q(rev)2 (heat released by 1 must be taken up by 2). We say Q = Q(rev)1 = -Q(rev)2

    delta S = Q(1/T1 - 1/T2)

    Now we say that when T2 > T1 then delta S > 0 (spontaneous process) when Q > 0, so heat flows into system 1 . When T1 > T2, delta S > 0 when Q < 0 and thus heat flows into system 2.
    Last edited: Jul 16, 2006
  6. Jul 17, 2006 #5
    we see microsopic and macroscopic entropy differnetly ..
    microscopic is about statistical view of entropy?
    what can we say about the boltz man equation of entropy
  7. Jul 17, 2006 #6
    you mean S = k * ln W :p
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