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Heat energy in a mixture to get final temperature

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A glass mug has mass 125g empty. It contains 180g of
    coffee. Both are at 70 C. I add 15g of creme at 5 C.
    Find final temperature. Assume creme has specific
    heat of 2900 J/kg.C .

    2. Relevant equations

    Q = m C (delta)T in joules or Kcalories

    m = mass in kg

    C is specific heat where C-coffee 1 kcal/kg.C
    C-glass .2 kcal/kg.C
    C-creme 2900 J/kg.C

    delta T is *change* in temperature

    3. The attempt at a solution

    Attempt at solution: First I don't see a "change"
    in temperature. But I'm somehow suppose to apply
    the above formula for Q-heat energy. There are 3
    Qs .. 2 at 70 C .. and 1 at 5 C. I am somehow
    suppose to weight these separate energies, calculate
    a change in temperature. Then show the final equilibrium

    I can only guess that I can do something like

    Q-coffee = mC-coffee x 70 to get an absolute ?? energy

    Q-glass = mC-glass x 70

    Q-creme = mC-creme x 5

    and somehow weight the absolute thermal energies to
    get delta-T .. and then T ???????
  2. jcsd
  3. Apr 30, 2013 #2
    Two hints: First, get your constants into the same units.
    Second: Remember that the heat lost by the glass and coffee is the heat gained by the creme.
  4. Apr 30, 2013 #3
    Thanks Barryj .. I got it. I did the heat gained = heat lost Conservation of Energy ( heat )

    m1C1(70 - T) + m2C2(70 - T) = m3C3( T - 5 ) with the m3 conversion going back to kcal/kg.C

    and I got a very reasonable answer. Amazing! Not one word of how to balance mixtures is anywhere in the
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