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Second Law and Reversibility

  1. Jan 27, 2012 #1
    I need a clarification about the reversibility criteria in classical thermodynamics.There are two criteria in it to my understanding.one is that there should be no dissipative forces i.e no friction.The second is that there is no heat flow across a definite temperature gap--only across an infinitesimal temperature difference.The apparent reason given for this is that all processes need to be reversible in a reversible machine and you can not possibly reverse the heat exchange if it occurs a definite temperature gap without causing any other change in the surroundings.I need a little more explanation about this....as in what kind of changes?

    Also in the definition of entropy,we talk about reversible heat.Yet when we do work against friction we say that the entropy of the thing we do work on(that gets heated up) rises by the heat supplied divided by the temperature of the body(provided its temperature does not rise much).Is the heat supplied here reversible?Another case is a hot stone being put in a cold reservoir.The heat exchanged here is not reversible;yet we use that heat to calculate the entropy change.

    Any help is appreciated.
     
  2. jcsd
  3. Jan 29, 2012 #2

    rude man

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    Sorry no one else has answerded you so far. I'll give it what I can.

    A definite temperature change can occur reversibly if the change is envisioned to be performed by an infinite series of heat reservoirs exchanging heat with the system. For any one reservoir, T is the same for the system as the reservoir, so dQ/T is + for one and - for the other, the two add to zero, and no change in entropy of the universe takes place. You don't need to add work to this process.

    The Carnot cycle is another, simpler example of moving a system from one finite temperature to another within the cycle, but this requires work. Since all sections of the cycle are performed reversibly, at all times the entropy change of the hot & cold reservoirs plus the system is zero.

    The question of your hot rock into a cooler liquid is a good one. As I see it, a reversible path can be found. What you do is take the rock and immerse it in a reservoir at T infinitesmally below the temperature of the rock. This is a reversible change since again T is "the same" for the rock and the reservoir, so again ΔS = 0 for both. You repeat this process on the rock an infinite number of times until its final temperature equals the final temperature of the rock + water, which you can easily compute ahead of time from heat lost = heat gained and the heat capacities of the rock & water.

    Then you perform the same stunt with the water, contacting it with an infinite series of reservoirs, each differentially higher in temperature than the previous, until the water too is brought up to the final temperature.

    For both processes, entropy of each system (rock and water separately) can be computed, so the total change in entropy of your universe (the rock and the water together) can be determined. It would be

    ΔS = Cr∫dT/T from Tir to Tf + Cw∫dT/T for Tiw to Tf
    where r stands for rock, w for water, i for initial and f for final.

    I'd appreciatye someone else competent looking at this!
    !
     
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