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Specific heat/equilibrium in three component system (Thermo)

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  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Rather than retyping the problem, I've uploaded a screenshot here >> http://imgur.com/EB5MrtP

    2. Relevant equations
    [itex]Q_1 + Q_2 + Q_3 ... = 0[/itex]
    [itex] mc \Delta T = Q[/itex]
    3. The attempt at a solution
    a. [itex] Q_{tea} + Q_{met} + Q_{cr} = 0 [/itex]

    I converted the temperatures to K, so [itex]T_{cr}[/itex] and [itex]T_{met}[/itex] = 293.15K, while [itex]T_{tea}[/itex] = 373.15K

    Constructing the equation, I got:

    [itex] 0 = (m_{met} * c_{met}* (T_{equil}-293.15)) + (m_{cr} * c_{cr} * (T_{equil}-293.15)) + (m_{tea} * c_{tea} * (T_{equil} - 373.15)) [/itex]

    I'm stuck here now, because the problem doesn't give any of the masses, it only relates them to each other. The specific heats I can find because the specific heat of water is a known value, but I'm stuck on the masses. What am I missing?
     
    Last edited: Sep 15, 2015
  2. jcsd
  3. Sep 15, 2015 #2

    TSny

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    I believe the problem is asking you to express the equilibrium temperature in terms of the symbols for the masses and specific heats. If so, you do not need to plug in numerical values for these quantities.
     
  4. Sep 15, 2015 #3
    Oh, there's a small part at the bottom that got cropped out, it specifically asks for a numerical value. It also mentions something about using a binomial expansion to solve it, which I didn't understand how it applied to this problem. I've amended the link in the OP with the fixed version.
     
  5. Sep 15, 2015 #4

    TSny

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    OK. Sorry, I did not see part (c) when I first looked at the problem.

    Try expressing all masses in terms of one of the masses, say ##m_{met}##. Likewise, express all specific heats in terms of one of the specific heats, say ##c_{met}##.

    You will then find that the equation simplifies such that you do not need any numerical values for the masses or specific heats.

    Also, you do not need to express the temperatures in K. You can keep them in oC because you are dealing with temperature differences here.
     
  6. Sep 15, 2015 #5
    Oh, duh, I see it now. Thank you! I probably should have tried that first, but I just didn't think of it.
     
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