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Calorimetry Promblem

  1. Feb 14, 2007 #1
    1. What is the final temperature when a 3.0 kg gold bar at 99 degrees celsius is dropped into 0.22 kg of water at 25 degrees celsius.

    H20 Heat Capacity (CpH20)= 4186
    H20M (mass) = .22 kg
    H20Ti (initial temperature) = 25 degrees celsius
    Au (CpAu)= 129
    AuM (mass) = 3.0 kg
    AuTi (initial temperature) = 99 degrees celsius


    2. CpH20 * H20M * ∆TH20 = CpAu * AuM * ∆TAu
    (where ∆T is change in temperature)


    3. I *think* Im missing two variables, but Im not sure if I am correct. I am missing the change in temperature for H20, and the final temperature. In order to get either answer, I need another variable. However, I do know that I could get ∆TAu by (Temperature F - Temperature initial), but again, I would need the final temperature.

    Im not sure what I should do in order to find the change in temperature of H20, or maybe Im just reading the problem wrong. I do know that the answer should come out to 47 degrees celsius.

    Any advice or help?

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 14, 2007 #2

    hage567

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    The final temperature is going to be the same for both, since the system goes to equilibrium. That is the only unknown variable. You have to remember to conserve energy, so the heat lost by the gold bar will be gained by the water. This means your equation isn't quite right.

    ∆T= (Temperature F - Temperature initial) is useful.
    Put this expression for each substance in your equation instead of just ∆T, since "Temperature F" is what you're solving for.
     
  4. Feb 14, 2007 #3

    Dick

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    How about final temperature as another variable? What does your equation 2. look like in terms of that variable?
     
  5. Feb 14, 2007 #4
    ok, im not sure how to set up the equation if its wrong. I can change the equation to CpH20 * H20M * (Tf - Ti) = CpAu * AuM * (Tf - Ti). That would mean that energy is conserved. Is that enough to solve the problem. I could try to get Tf by itself, and get the answer that way.

    Well, I plugged in the numbers, with the answer, and here is what I got;

    4186 * .22 * (47 - 25) = 129 * 3 * (47 - 99)

    This comes to 20260 = -20124. There seems to be a large margin of error, so Im not sure its correct.
     
    Last edited: Feb 14, 2007
  6. Feb 14, 2007 #5

    hage567

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    You are still missing one important point. The energy must be conserved, as I mentioned before. So think of it like
    (heat lost by bar) + (heat gained by water) = 0.
    Do you see what is wrong with your equation now?
     
  7. Feb 14, 2007 #6
    Ive been messing around with the problem for a while, but have had no luck thus far. I'll try to figure it out tomorrow, but thanks for the help guys.
     
  8. Feb 14, 2007 #7

    hage567

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    So close, yet so far...

    You're almost there.

    "I can change the equation to CpH20 * H20M * (Tf - Ti) = CpAu * AuM * (Tf - Ti). That would mean that energy is conserved."

    This is almost correct, you are just missing a negative sign on one side of your equation.

    "Heat lost by bar must be the same as heat gained by water"

    Do you see where it should go? Once you have that in the right place, carefully solve your equation for Tf.
     
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