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Thermal physics - entropy change (latent heat, specific heat etc)

  1. Oct 30, 2013 #1
    Hi all, could somebody have a look over my answers for this question please? The value I got for the second part seems quite feeble.

    1. The problem statement, all variables and given/known data

    VolcanistQ_zpsce3a1e3b.jpg


    3. The attempt at a solution

    Part a)

    Key
    m = mass
    cs = specific heat bronze
    cm = specific heat molten bronze
    Tfus = melting temperature bronze
    T1 = initial temperature (of the bronze coin)
    T2 = final temperature (of the bronze coin)
    Hfus = latent heat fusion bronze

    Ok so to find the entropy change I used the following equation:

    ΔS = m [csln(Tfus /T1) + (Hfus /Tfus) + cmln(T2 /Tfus)]

    plugging the numbers in:

    ΔS = 0.007 [377ln(1223/300) + (168000/1223) + 325ln(1473/1223)]
    ∴ ΔS = 5.09 JK-1



    Part b)

    Key
    mw = mass of water (the beer)
    mm = mass of molten bronze coin
    cw = specific heat water
    cm = specific heat molten bronze
    Qw = heat transfer water
    Qm = heat transfer bronze coin

    For this question I used the equation Q = mcΔT

    Qw = mw x cw x ΔT
    Qw = 0.5 x 4200 x (T - 288)

    Qm = mm x cm x ΔT
    Qm = 0.007 x 325 x (1473 - T)

    At equilibrium Qw = Qm

    So 0.5 x 4200 x (T - 288) = 0.007 x 325 x (1473 - T)

    ∴ T = 289.4 K which is about 16 °c, a mere 1 °c rise.
     
  2. jcsd
  3. Oct 30, 2013 #2
    It looks like part A was done correctly. In part B, why didn't you take into account the latent heat to freeze the bronze, and the sensible heat to cool the solid bronze which forms? Also, in part b, you have to also calculate the change in entropy for the beer between its initial temperature and its final temperature.

    Chet
     
  4. Nov 1, 2013 #3
    Hi Chet thanks for pointing that out, missed it completely
     
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