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Homework Help: Really difficult question

  1. Jun 20, 2010 #1
    i am a junior physics teacher but i encounter a difficult problem to me

    200g zero degree celsius ice mixed with 100g 100 degree celsius steam

    what is the state of the mixture?

    and how to calculate?

    I have calculate using final temperature T as assumption

    but the result is T = 160 (approx.)

    but i think it is not logical. because that 100g 100 degree celsius steam should give out energy

    . i dont think the final mixture will become 160 degree cel which is totally not logical.
     
  2. jcsd
  3. Jun 20, 2010 #2

    phyzguy

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    How can the final result be hotter than either initial state? You have done something wrong. Why do you think that is the correct final result?
     
  4. Jun 20, 2010 #3

    K^2

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    He assumed all of the steam and ice turned to liquid. Obviously, that assumption is flawed. You can't just lump all the latent heats together simply because the states are mentioned in the problem.

    Final temperature will be 100°C, with the state being a mixture of liquid and steam. The amount of liquid water will be the 200g that came from ice + whatever quantity of steam condensed to provide energy to melt and heat the water from ice to 100°C.

    And this is the most basic thermodynamics question covered in any intro text. A Physics teacher being unable to solve it worries me deeply. If you are teaching the subject, make sure you actually learn it. You have a textbook for the class? I suggest starting with reading that first.

    Edit: Though it's good thing you asked, at least. I've known some teachers to come up with obviously nonsensical results and carry on explaining that nonsense to students.
     
    Last edited: Jun 20, 2010
  5. Jun 20, 2010 #4
    It's an interesting problem.....off hand, I don't know an elegant way to calculate the answer...but here is the "non mathematician" approach:

    Each gram of steam gives off about 539cal/gm as it condenses to water at 100 degrees, and each gram of ice absorbs about 80 cal/gram to cause a phase change to water at zero degrees. So the steam carries a lot more heat energy.

    From then on it takes about 1 calorie per gram to warm the cold water one degree C and the steam based hot water gives off the same amount as it approaches equilibrium. So far there is cold water, hot water and steam in these amounts:

    For all the ice to melt will take 200g x 80cal/g or 16,000 cal...while the steam could give off 100 x 539 or 53,900 cal ....so the assuming no loss of heat to exterior systems, all the ice appears to melt.....with some steam remaining:

    With all the ice melted, we have a mix of 200 grams of water (from the former ice) at zero degrees C, 16,000/539 or 29.7g of water at 100 C with 53,900 - 16,000 or 37,900 calories of steam to warm the water.

    I guess you can continue and eventually calculate T.
     
    Last edited: Jun 20, 2010
  6. Jun 20, 2010 #5
    I think Naty has the right idea but has approached it from the wrong end, so taking Naty's figures and reworking.

    Energy gained.
    200g of ice will melt to 200 g of water using 200 x 80 = 16000 cals at 0degC

    200g of water raised from 0 C to 100 C 200 x 1 x 100 = 20000 cals

    Total energy gained by ice 36,000 cals

    Energy lost

    36000/539 = 66.8 grams of steam will loose 36000 cals at 100degC

    So the final state is 266.6 grams of water in equilibrium with 33.2 grams of steam at 100degC
     
  7. Jun 21, 2010 #6
    that seems to be the all too common way I approach things!!!!!!!!!!!!!!!!!!!!!
     
  8. Jun 22, 2010 #7
    i have think a way to explain this question in a very logical step I am kenny1999 who post this question before.



    First. Two objects will not transfer energy when they have no temperature difference.

    Now, initially 200 gram of 0 degree cel ice in contact with 100 gram of 100 degree cel steam

    so there is no doubt that 200 gram of 0 degree cel ice will gain energy and that 100 gram of 100 degree cel steam will lose energy.

    that 100 degree cel steam will lose approx 226000J energy to turn all 100g 100 degree cel steam into 100 degree cel water. But this energy is far higher than that is required to turn
    that 200 gram of 0 degree cel ice into 100 degree cel water.

    As a result not all 100gram 100 degree cel steam will turn to become 100 degree cel water
    but only part of it, because 200gram of 0 degree cel ice cannot receive all energy if that 200 gram of 100 degree cel steam turns to 100 degree cel water. so only part of the 100 degree cel steam will turn to 100 degree cel water.

    Finally it will be the mixture of 100 degree cel water and 100 degree cel steam. Because their temperature is the same so there is no more NET energy transfer.

    But that 100 degree cel steam will somehow turn into 100 degree water but that 100 degree water will also gain back the energy to become 100 degree cel steam which reaches an equilibrium states.

    Do I explain correctly??
     
  9. Jun 22, 2010 #8
    There was no mention of what the ambient temperature was, or if it is in a container,
    whether it is an insulator or a conductor. That would influence your answer. If I was
    the student...and this was my test...The answer would be D) Not enough information.
     
  10. Jun 22, 2010 #9
    The shape and dimensions of the ice is crucial too. Is it flat? Is the steam below it? Above it?
     
  11. Jun 22, 2010 #10
    The latent heat of fusion for ice is 334 J/g and the latent heat for vaporization for steam is 2260 J/g. Also, the heat capacity of water is 4.18 J/(g K).

    Let us suppose that the whole mixture becomes a liquid with a temperature 0 oC < t < 100 oC. The released heat is from condensing the steam:

    100 g x 2260 J/g = 226,000 J

    and cooling the hot water

    100 g x 4.186 J/(g oC) x (100 - t) = 418.6 x (100 - t) J/(oC)

    The absorbed heat goes to melting the ice:

    100 g x 334 J/g = 33,400 J

    and warming the cold water:

    100 g x 4.186 J/(g oC) x t = 418.6 J/(g oC)

    These have to be equal:

    226,000 + 418.6 x (100 - t) = 33,400 + 418.6 t

    226,000 + 41,860 - 33,400 = 2 x 418.6 t

    234,460/(2 x 418.6) = t

    t = 280 oC

    But, this temperature is greater than the upper bound of the assumed interval (where water is liquid)!

    Can you conclude what is the equilibrium temperature of the system and what phases are in equilibrium? Can you find the masses of the phases?
     
  12. Jun 22, 2010 #11
    I like this problem. I wonder what Marylin Vos Savant would say. If your formula and
    calculations are correct, perhaps your answer is pointing to some of the water would
    then evaporate and the remainder would be something else not calculated by the
    equation? When we make presumptions and ignore constraints at the beginning of
    an equation, wouldn't the answer also be capable of carrying a "blind constraint" over
    to the answer?
     
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