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Equilibrium Temperature with three substances

  1. Jan 26, 2017 #1
    1. The problem statement, all variables and given/known data
    If a block of 250 grams of lead (specific heat (c) =130) at 315 degrees celsius is placed in a 200 gram aluminum (c=900) calorimeter cup containing 900 grams of water (c=4.1), and the calorimeter and the water are both initially at 15 degrees celsius, what is the equilibrium temperature ?

    2. Relevant equations
    Q=mcΔt
    Tf=(mct+mct+mct)/(mc+mc+mc)
    3. The attempt at a solution
    Tf=((250x900x315)+(200x900x15)+(900x4.1x15))/((250x130)+(200x900)+(900x4.1)
    Tf=(70875000+2700000+55350)/(32500+180000+3690)
    Tf=40.1
    I'm not sure if I'm on the right track.
     
    Last edited: Jan 27, 2017
  2. jcsd
  3. Jan 27, 2017 #2

    Borek

    User Avatar

    Staff: Mentor

    Try to approach it in a systematic way - for a single substance Q=mcΔt, for the whole system

    [tex]\sum_i m_ic_i\Delta T_i = 0[/tex]

    where [itex]\Delta T_i = T_{final} -T_{initial(i)}[/itex] (Tfinal is common for all substances present).
     
  4. Jan 27, 2017 #3
    What are the units of specific heat, if lead is 130, aluminium is 900 and water is 4.1???
     
  5. Jan 28, 2017 #4
    The math is not consistent with your second Relevant equation. Try again.
     
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