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Homework Help: Clarifications of entropy and heat transfer between reservoirs.

  1. Apr 22, 2006 #1
    Hello all, I just need some clarifications on my approach to this question.

    Heat is absorbed along a metal bar from a hot thermal reservoir Th to a cold thermal reservoir at temperature Tc. Apart from the ends, which are in contact with the thermal reservoirs, the bar is perfectly insulated. The bar is made of a material having a temperature-independant thermal conductivity of 406W/(m.K), Th = 400K and Tc = 200K. The cross sectional area of the bar is 5cm^2 and the length of the bar is 25cm.

    a)calculate the total rate of heat absorbtion by the bar at the hot and cold ends. Calculate the nett rate of heat absorbtion by the bar over it's whole surface.

    Heat absorbed(hot):
    dQ(hot)/dt = lambda . dT/dx . Area
    = 406.200.0.0005/0.25 = 162.4J/s
    Heat absorbed(cold):
    dQ(cold)/dt = -lambda . dT/dx . Area
    = -162.4J/s
    Cold reservoir <-----<-----< Hot reservoir

    .'. dQ(nett)/dt = Sinflow - Soutflow = 162.4-(-162.4) = 324.8J/s

    b)calculate the rate of entropy transfer into the bar at the hot and cold ends.

    dS/dt = 1/Th . dQ(reversible)/dt
    = 1/Th . 324.8 = 324.8/400 = 0.812J/K
    dS/dt = -1/Tc . dQ(reversible)/dt
    = -1/200 . 324.8 = -1.624J/K

    c)calculate the rate of entropy production inside the bar.
    dS/dt = lambda . dT/dx. A (1/Tc - 1/Th)
    = 0.406J/K

    d)calculate the rate of change of entropy of the universe (the bar and 2 reservoirs) and show that it satisfies the second law of thermodynamics.

    dS/dt = dS(external)/dt + dS(internal)/dt
    dS(external) = lambda . dT/dx . A (1/Th - 1/Tc)
    = -0.406J/K
    dS(internal) = entropy production inside the bar = 0.406J/K
    .'. dS/dt = -0.406 + 0.406 = 0 J/K

    Now, I thought that the second law of thermodynamics says that that should be greater than 0 in a irreversible process(Is this irreversible?)... Why did I get 0? Any one that can explain it to me would be kindly appreciated.
  2. jcsd
  3. Apr 23, 2006 #2

    Andrew Mason

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    I think this should be 0. The heat in = heat out so there is no net rate of heat absorption.

    The heat flow out of the hot reservoir is negative and the heat flow into the cold reservoir is positive.

    As you correctly state, the rate of change in entropy of the universe is the sum of the rates of change of entropy of the hot and reservoirs and bar.

    [tex]\Delta S_{hot}/dt = \frac{d}{dt}Q_{hot}/T_h = -162.4/400 = -.406 J/Ksec[/tex]

    [tex]\Delta S_{cold}/dt = \frac{d}{dt}Q_{cold}/T_c = +162.4/200 = +.812 J/Ksec[/tex]

    [tex]\Delta S_{bar}/dt = .406 J/Ksec[/tex]

    [tex]\Delta S_{universe}/dt = -.406 + .812 + .406 = .812 J/Ksec[/tex]

  4. Apr 23, 2006 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    I don't think this is correct.

    The rate of entropy 'production' in the bar is the integral of ds along the length of the bar. For a 'slice' of width dx, the entropy change is:

    [tex]dS = dQ_{in}/T - dQ_{out}/(T-dT)[/tex]

    Since both heats are the same:

    [tex]ds = (Q(T-dT)/T(T-dT)) - QT/T(T-dT) = -QdT/T(T-dT) = -QdT/T^2[/tex]

    [tex]dS/dt = d/dt\int_{T_c}^{T_h} -QT^{-2} dT[/tex]

    So work that integral out to find the rate of entropy production in the bar.

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