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Temperature of a mixture

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the final temperature of a mixture made from 3kg of ice at -5ºC and 10.8kg o fcopper at 2000ºC. The melting point of copper is 1083ºC.

    2. Relevant equations

    Q=mct(delta)t
    mc(delta)t + mc(delta)t = 0J

    m=mass
    c=specific heat
    delta t = temperature change
    Q=energy

    3. The attempt at a solution
    I have no attempt at such a question because I don't know how to attempt it with there being a phase change.
     
  2. jcsd
  3. Jan 16, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I didn't know you could mix copper and water! Just take the phase change into account by using the "heat of melting" and/or the "temperature of vaporization".

    Here's what I would do. First calculate the heat necessary to raise the temp of the ice to 0º C using the specific heat of the ice. Then add the "heat of melting" of ice. Calculate the heat of the copper using the specific heat of copper and subtract the total heat neccessary to melt the ice. Calculate the temperature of the copper with the heat left. That puts you in the situation in which both water and copper are liquid, the water at 0º C and the copper at whatever temperature (slightly less than 2000º C) you have calculated.

    It seems to me very likely that the water will vaporize when you drop it on copper at that temperature! Calculate the heat necessary to raise the temperature of the water to 100º C and the "heat of vaporisation" of water. If that is still less than the heat of copper (which you have already calculated), then the water will vaporize. Subtract that from the heat of the copper. Now calculate the heat contained in steam at temperature "x" and the heat of copper at temperature "x". Add those, set their sum equal to the heat contained in copper after the phase changes etc. and solve for x.

    (If there is not enough heat in the copper to vaporize the water, ignore the heat of vaporization and do that last step, the heat in water and copper at temperature "x" for liquid water.)
     
  4. Jan 16, 2008 #3
    for these types of questions, put the final temperature equal to x degrees. then consider the energy involved for both ice and copper to reach that temperature.

    ice will increase from -5 to 0. then it will melt. it will then reach x degrees. hopefully x is not higher than 100 degrees, else you would have to take into considerations the temperature rise till 100 degrees, then boiling, then, increase in temperature till x degrees.

    now for copper, it will experience a decrease in temperature from 2000 degrees to it's melting point, then you have to take into consideration its latent heat of fusion, then a decrease from melting point to x.

    you will have two equations in x. equate them to obtain x. since the energy lost by copper in cooling is equal to the energy gained by ice in getting hotter.
     
  5. Jan 16, 2008 #4
    awww i didn't read the post by HallsofIvy, it's the same stuff..... sorry
     
  6. Jan 16, 2008 #5
    I just did the calculation, and it turns out all the water vaporizes. From then on, do I disclude water from the calculation or what? What I did was found how much energy it took for the water to vaporize, and whatever was left I used to find what the temperature of the copper would be. Or would I continue to include water in the equation, regardless of the fact that it is a gas?
     
  7. Jan 17, 2008 #6
    the assumption we make is that whatever energy is taken from copper, it is totally transfered to the ice/water/water vapour.

    for water:

    you calculate the energy for:
    1. increase in temperature of ice from -5 to 0 degrees.
    2. melting of ice at 0 degrees.
    3. increase in temperature of water from 0 to 100 degrees.
    4. boiling of water
    5. increase in temperature of water from 100 degrees to x degrees.

    you add these up to obtain an expression in x, which gives the energy taken from copper by ice/water/water vapour.

    use the same logic to find the energy given by copper and obtain another expression in x.

    equate both expressions to find x.

    i didn't understand your question, but i hope this makes everything clear.
     
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