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Maximum entropy and thermal equilibrium

  1. Oct 13, 2006 #1
    3) An object of mass m1, specific heat c, and temperature T1 is placed in contact with a second object of mass m2, specific heat c2 and temperature T2>T1. As a result, the temperature of the first object increases to T and the temperature of the second object decreases to T'.

    a) Show that the entropy increase of the system is

    deltaS = m1c1 ln(T/T1) + m2c2ln (T'/T2)

    b) Show that energy conservation requires that

    m1c1(T-T1) = m2c2 (T2-T')

    c)Show that the entropy change in S, considered as a function of T, is a maximum if T' = T, which is just the condition of thermodynamic equilibrium.
    Last edited: Oct 13, 2006
  2. jcsd
  3. Oct 13, 2006 #2


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    step by step with the variables...
  4. Oct 13, 2006 #3
    This is what I get

    a) This is a calculation of an entropy change for an irreversible process. Since entropy is a state function, ΔS is independent of path. All we have to do is imagine a reversible path which will effect the same change and calculate the entropy change for the reversibly path.

    ΔS_total = ΔS_cold + ΔS_hot

    ΔS_total = m1c1 integral T1 to T (dT/T) + m2c2 integral T2 to T' (dT/T)

    ΔS_total = m1c1 ln(T/T1) + m2c2 ln(T'/T2)

    b) C = Q/dT

    therefore m1(Q/dT)(T-T1)=m2(Q_2/dT)(T2-T')

    Due to conservation of energy, these two equations must be equal. Because the Heat capacities differ, T and T' are different as well.
    I know the specific heat is somehow supposed to cancel out the temp change but Im not sure what dt = for each side. somebody help :confused:

    C) no clue what do do here, maybe integrate from T to T'? :confused:
  5. Oct 13, 2006 #4


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    c) you must get the total differential of S considered as a function of T (i.e. an equation of the form dS = f(T)dT +...

    And since a max in entropy means dS=0, what condition does this set on f(T)?
  6. Oct 13, 2006 #5
    total differential would be f(t)dt+f(t1)dt+f(t2)dt+f(t')dt?

    sorry, I'm very bad at calculus.
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