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Equilibrium temperature of three different substances

  1. Oct 13, 2015 #1
    I've been asked to find the final temperature of different in an insulated container.

    A combination of 0.25kg water at 20C, 0.4kg aluminum at 26C, and 0.1kg copper at 100C are combined, and allowed to come to thermal equilibrium.

    Finding the equilibrium temperature for two substances is easy enough, but throws me off when another one is involved...
    I approached the problem by finding the equilibrium temperature between aluminum and water separately (21.5 C), and copper and water (22.8 C).
    Q(Al) = Q(H2O) --> Tfinal = 21.53C
    Q(Cu) = Q(H2O) --> Tfinal = 22.85C

    My rationale is that both of the metals will heat the water separately, and then the water at two different temperatures will mix.

    But now that I have the final temperatures of each metal in equilibrium with water, I'm not quite sure how to proceed. I don't think I would just average the two values, as the amount of water heated by both metals is different.
     
  2. jcsd
  3. Oct 13, 2015 #2
    Why don't you just write the balance of the heat transfer for all three and solve the equation for the final temperature?
     
  4. Oct 13, 2015 #3
    Okay, so I just tried that: Q(Al) = Q(Cu) = Q(H2O). Solving for temperature, I got 38C, which makes sense. This problem is much easier than I thought -- the principles are exactly the same as two objects that are in thermal equilibrium with each other! Thank you.
     
  5. Oct 13, 2015 #4
    No, the heats don't have to be equal.
    The heat given must equal the heat taken. It may be that the two metals "give" and the water "accepts" heat.
     
  6. Oct 13, 2015 #5
    So I suppose then, that the proper relationship to solve for thermal equilibrium would be -Q(Al) = -Q(Cu) = +Q(H2O) then? Since aluminum and copper are the warmer substances and "give away" heat, and water is the cooler substance, so it "accepts" heat.
     
  7. Oct 13, 2015 #6
    No.

    Q(Al)+Q(Cu)+Q(H2O) = 0

    Chet
     
  8. Oct 13, 2015 #7
    Why is this so? The temperature of Cu and Al are greater than water, so wouldn't copper and aluminum be giving heat, while water will be receiving it -- eventually meeting at some final temperature. I understand that the net exchange is 0, but in that form, why wouldn't the relationship be: Q(H2O) - Q(Al) - Q(Cu) = 0 ?
     
  9. Oct 13, 2015 #8
    Q may be negative. They way Chet wrote it, you don't assume who gives and who takes. It comes out from the calculation.
    It's not always easy to know which one does what.
     
  10. Oct 13, 2015 #9
    Without having the (average) specific heat of Cu an Al? You're a genius, believe me!

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  11. Oct 13, 2015 #10
    Are you saying that a sufficiently accurate answer to this problem cannot be obtained without including the temperature dependence of the heat capacities? Why don't you try it both ways, with- and without the temperature dependence (say, in the later case, using the heat capacities at the initial temperatures) and report back to us on the comparison of your final results?

    Chet
     
  12. Oct 14, 2015 #11
    No, I'm saying that the specific heats of those substances are not included in the data. Certainly the OP has a table with values, but, apart from water's specific heat which is known, she could write those of Al and Cu, at least so that we could control her computations (furthermore, not all tables give the same exact values).

    (P.S. Curious you intended this because, actually, the need of temperature dependence of specific heats is what I was thinking for another thread, that of supercooled water I started some weeks ago :smile:)

    Regards.

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