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Heat Capacity and Entropy Textbook Definition - Quick Question

  1. Jun 7, 2012 #1
    Just a quick question of something I found in my textbook but cant get how they produced it.

    C_p =(∂Q/∂T)_p

    that is the definition of heat capacity at a constant pressure p. Q is heat and T is temperature. This equation is fine and I know how to derive it. Now it is the next line which worries me.

    C_p=(∂Q/∂T)_p = T(∂S/∂T)_p

    where S is the entropy

    why this bothers me is that this equation as I understand should only hold for C_V (heat capacity with constant volume)
    thats because dQ=TdS for a REVERSE-ABLE ONLY expansion (i.e dQ=0 , dS=0) ie adiabatic all of which occur as V, the volume is held constant for C_V. Hence dU=dQ and dQ(rev)=TdS = dU , which can then be simply substituted into the definition. OK

    Sorry for the spiel, but my question is how can this same line of reasoning be true for C_p where dV is not constant and dU is not equal to dQ and expansion isn't reversible?

  2. jcsd
  3. Jun 7, 2012 #2
    I think you are confusing U and H, have another look at your textbook.

    The partial differntial expressions you give for Cp are correct, but they are obtained by substituting Tds into the definition of H not U.

    Similar partials can be derived for Cv by substituting Tds into the definition of U.
  4. Jun 7, 2012 #3
    The definition of specific heat, C, under any condition is:
    C= ∂Q/∂T
    The definition of S, under any condition is:
    where Q is the heat energy being transferred.
    So regardless of whether the process is isobaric or isovolumeteric,
    C= T(∂S/∂T)

    You have made an incorrect hypothesis. The definition of specific heat, C, is not:
    C= (∂U/∂T)
    where U is the internal energy. The definition of specific heat is:
    C= (∂Q/∂T)

    One can easily prove that:
    C_V= (∂U/∂T)_V
    C_P= (∂H/∂T)_P
    where H is the enthalpy, not the internal energy. Your logic worked
    for C_V, but it didn't work for C_P because of the incorrect hypothesis.
  5. Jun 7, 2012 #4

    Andrew Mason

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    Your question is why:

    [tex]C_p=\left(\frac{\partial{Q}}{\partial{T}}\right)_p = \left(\frac{\partial{Q_{rev}}}{\partial{T}}\right)_p[/tex]

    The answer is that it is implicit in the left side that Q is Qrev. Cp is determined by a reversible process:

    Cp = dQ/dT = dU/dT + dW/dT

    Now if dW = PdV and if P is constant,

    Cp = dU/dT + PdV/dT = Cv + PdV/dT = Cv + P(RdT/P)dT = Cv+R

    But dW = PdV ONLY in a quasi-static process.

  6. Jun 7, 2012 #5
    Ok very good Andrew, makes sense though seems a rather large implication that C_p is determined by a reversible process only. Oh and Darwin as I understand it entropy S is not defined by dQ=TdS for all conditions but dQ(rev)=TdS where (res) is for a reversible process only. Huge difference.

    Thanks for the fast responses, I have an exam next week where I have to use that substitution and am much happier now I know where It comes from. Thanks again
    Last edited: Jun 7, 2012
  7. Jun 7, 2012 #6
    I'm new to this forum, is there some sort of thanks button / points system to show appreciation ?
  8. Jun 7, 2012 #7
    your 100% right.
    since dQ(rev)=TdS
    dH=dQ(rev) at at constant pressure
    since C_p=∂Q/∂T we have

    thanks a bunch
  9. Jun 8, 2012 #8
    The fact that you have come back to the thread to let us know how you got on is great, keep it up.

    It is quite dispiriting if you put in a deal of effort to write something and you never hear further. Others have put in more the I have here.

    go well

  10. Jun 8, 2012 #9

    Andrew Mason

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    You seem to be missing a step at the beginning. Start with:

    H = U + PV

    dH = dU + PdV + VdP

    For constant pressure dP = 0 so:

    dH = dU + PdV = dQrev (first law, substituting PdV for ∂W).

    Since dQrev = TdS by definition (ie dS = dQrev/T), then

    [tex]\left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{T\partial S}{\partial T}\right)_P = \left(\frac{\partial Q_{rev}}{\partial T}\right)_P = C_p[/tex]

    Similarly, since

    dU = TdS - PdV , for a constant volume process (dV = 0):

    [tex]\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{T\partial S}{\partial T}\right)_V = \left(\frac{\partial Q_{rev}}{\partial T}\right)_V = C_V[/tex]

    Last edited: Jun 8, 2012
  11. Jun 8, 2012 #10
    This hypothesis is not valid for conditions under which the process is irreversible. Once example would be if a Carnot engine where modified by having kinetic friction between the movable piston and the cylinder. I would call this the "the rusty Carnot problem".
    This probably won't appear on any test. However, I am now interested for its
    own sake. I am curious how one would take into account kinetic friction. I think I have an idea, but I am not sure. I also wonder if a solution to "the rusty Carnot problem"
    could be published anywhere.
  12. Jun 9, 2012 #11

    Andrew Mason

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    The work done by the rusty Carnot would include the friction. There would be no change the efficiency of the engine if you include that friction as part of the output.

    The Carnot cycle is reversible in the sense that if you were able to store the output mechanical work you could reverse the heat flow and restore the original state of the reservoirs using that stored energy. So in the forward Carnot engine cycle if the output work was used to lift a weight, you could then lower that weight to reverse the direction of the cycle so that it acts as a Carnot refrigerator and flow all that heat back to the hot reservoir.

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