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Homework Help: What is the entropy change of aluminum?

  1. Jan 23, 2007 #1
    1. The problem statement, all variables and given/known data
    In an experiment, 240 g of aluminum (with a specific heat of 900 J/kgK) at 100°C is mixed with 40.0 g of water (4186 J/kgK) at 20°C, with the mixture thermally isolated.
    a: What is the equilibrium temp in C?
    b: What is the entropy change of aluminum?
    c: What is the entropy change of the water?
    d: What is the entropy change of the aluminum-water system?

    2. Relevant equations
    Q=mC(Tf - Ti)
    MwCw(Tf - 20c) + MaCa(Tf - 100c)= 0
    S(entropy)= Q/T
    3. The attempt at a solution

    Basically I'm kinda stuff trying to solve for the final temp, I get 11.9C but that cannot be right since that means the water-Al system lost heat. :grumpy:
    any help is very welcome
  2. jcsd
  3. Jan 23, 2007 #2
    Nevermind, solved it just now. T(Final) = CaMa + CwMwTi/CwMw + CaMa
    Final temp came out to be 65.1*C.
    B: S = (MaCa)ln(Tf/Ti) = -21.2 J/K
    C: S = (MwCw)ln(Tf/Ti) = 24 J/K
    D: 24-21.2= 2.8 J/K

    Mod can close thread...
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