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Sear's and Zemansky Ex. 17.10: What's Cooking?

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A heavy copper pot of mass 2.0 kg (including the copper lid) is at a temperature of 150°C. You pour 0.10 kg of water at 25°C into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.

    m(copper) = 2.0 kg
    T(copper) = 150°C
    c(copper) = 390 J/kg*K

    m(water) = 0.10 kg
    T(water) = 25°C
    c(water) = 4190 J/kg*K
    Lv(water) = 2256x10^3 J/kg

    2. Relevant equations
    Heat transfer: Q = mc(delta)T
    Heat transfer in Phase Change: Q = m*(Lv)
    Conservation of energy: Q(pot) + Q(water) = 0

    3. The attempt at a solution
    I calculated the final T to be 106°C for the pot/water system, that was the easy part. The next part says the real final Temp is 100°C because the energy that would have gone into heating the water, actually went into changing its phase from liquid to gas. I don't understand how to do this part at all. I tried calculating a second Q for a (delta)T of 6°C and got a value in Joules. Then I plugged this energy into Q = m*(Lv) and got a mass of 1.1g water vapor which is wrong. So my approach is flawed, why?

    4. The actual solution
    The final temperature is 100°C, and of the 0.10kg of water in the pot, 0.0034kg or 3.4g of the water was changed into vapor.
  2. jcsd
  3. Feb 2, 2009 #2


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    Homework Helper

    The "heat of vaporization" for water is 2260000 J/kg according to
    I guess modify the first stage of your calc to just go up to 100 degrees and see how much heat is leftover. Put that into heat of fusion = m*2260000 J/kg and see what mass of water is vaporized.

    Kind of contrived: if the lid prevents steam from escaping, the pressure will go up and the boiling temp will rise . . . so quite possibly your first answer is better than the one in the back of the book!
  4. Feb 3, 2009 #3

    5. The actual solution... again!
    The solution was simple. I applied the heat transfer equation Q = mc(delta)T for both the water and the copper pot. I assumed the final temperature to be 100°C and calculated both values separately. Since energy is conserved, I knew that Q(pot) + Q(water) = 0 but then I immediately saw there was a difference that did NOT equal 0. That difference is the energy the copper pot lost to convert water into steam! In my case it was -7575 J, then it was a matter of figuring out how much steam was converted using Q = m*Lv. Regardless of which Lv I used for water, the magic of sig figs resulted in a m = 0.0034 kg or 3.4g! Turns out the book was right!
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