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Heat flow and entropy

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Two bodies of temperature T and T+[tex]\Delta[/tex]T respectively (where [tex]\DeltaT>0[/tex]) are brought into direct thermal contact. Use a mathematical formulation of the second law to show:
    (i)Which way heat flows; is this heat flow reversible?


    2. Relevant equations
    2nd law of thermodynamics [tex]\int^{T+\Delta{T}}_{T}\frac{dQ}{T}dT[/tex]


    3. The attempt at a solution
    [tex]\int^{T+\Delta{T}}_{T}\frac{dQ}{T}dT=dQ\left[ln(T)\right]^{T+\Delta{T}}_{T}[/tex]

    [tex]=dQln\left(\frac{T+\Delta{T}}{T}\right)[/tex]

    heat flows from [tex]T+\Delta{T}[/tex] to T as the result is positive.

    Is this right? seems as if im missing something out here, does the result being positive show the direction of heat flow? or is it always from higher temps to lower? and how do i show whether the heat flow is reversible?
     
  2. jcsd
  3. Jan 28, 2010 #2

    Andrew Mason

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    The dT is incorrect.

    [tex]\Delta S_{total} = \Delta S_1 + \Delta S_2 = \int_{T_1}^{T_1+\Delta T} + dQ/T \int_{T_2}^{T_2-\Delta T}dQ/T[/tex]

    You can replace dQ with cmdT where c is the specific heat and m is the mass. This would result in:

    [tex]\Delta S_{total} = c_1m_1\int_{T_1}^{T_1+\Delta T} dT/T + c_2m_2\int_{T_2}^{T_2-\Delta T}dT/T = c_1m_1\ln{((T_1+\Delta T)/T_1)} + c_2m_2\ln{((T_2-\Delta T)/T_2)}[/tex]

    The second term is negative, of course.

    You can use the Clausius statement of the second law to show that the heat flow is from the higher temperature body to the colder one. You could also observe that if the hotter body got hotter and the colder got colder, the change in entropy of the universe would be negative, which would violate the second law.

    This is not a reversible process because the system is not in equilibrium at all times.

    AM
     
    Last edited: Jan 28, 2010
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