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Change of Phase, Absolute Zero?

  1. Apr 3, 2004 #1
    Okay, getting some input on this question will finish up the last of the Confusing Questions of the latest Chapter Review.

    Q: Which involves the larger number of calories?
    (a) the condensation of 1 g of steam at 100 C to 100 C water.
    (b) the change of phase of 1 g of 100 C water to 1 g of ice AT ABSOLUTE ZERO (emphasis mine)
    (c) both involve the same number of calories.

    Well, from trying to figure it out, I thought (a) took approx 540 calories and (b) took (or rather releases?) 80 calories, and then I noticed the bit about absolute zero. I must really be confused, because I thought that was Kelvins. I thought the world would shrink out of existence if something went to absolute zero. I thought everything would stop and all the squirrels would fall out of the trees. Which, btw, is why I wish physicists would stop trying to get it down that far. I don't particularly want my last sight on this earth to be a poor dead squirrel falling onto my soon-to-be dead head. Or is absolute zero just another way of saying 0 degrees C in this case?

    Thanking you in advance for any help.
  2. jcsd
  3. Apr 3, 2004 #2
    Ah! Don't play with celsius! Convert it, right now! Kelvin. I want it in Kelvin.

    Then let's talk.

  4. Apr 3, 2004 #3
    cookiemonster, were I to make the ice really go to absolute zero, we'd all blip out of existence. Since we can't do that, can't I pick answer (a) and be correct? Or is this all merely metaphysical speculation, to paraphrase Oscar Wilde?

    If I WERE to convert it, wouldn't I have to convert it to Celsius? Absolute zero being -273 Celsius? Or do you mean convert the nice Celsius temperature to Kelvins? Okay, you taught me that conversion a week or so ago: 100 = 273 + 100 = 373 Kelvins. Now I am lost. Am I done now? I like answer (a).
  5. Apr 3, 2004 #4
    All right, the question reads after conversion:

    Q: Which involves the larger number of calories?
    (a) the condensation of 1 g of steam at 373 K to 373 K water.
    (b) the change of phase of 1 g of 373 K water to 1 g of ice at 0K
    (c) both involve the same number of calories.

    Now we just gotta calculate it. How do we do that?

  6. Apr 3, 2004 #5
    Well, calculations, they are not my strong point...maybe (a) equals 540 because it's at the same temp, and (b) is 80 x 373? Something like that? It takes 80 calories per degree, doesn't it?
  7. Apr 3, 2004 #6
    All right.

    Which ones go through a change of phase? How do we calculate the energy in a change of phase?

    Which ones go through a change of temperature? How do we calculate the energy in a change of temperature?

  8. Apr 3, 2004 #7
    And since I gotta go prevent starvation (with my only $13...), here's a few more hints.

    Energy for change of temperature, where c is the specific heat of H20 in one of the phases. c changes depending on the phase.
    [tex]Q = cm\Delta T[/tex]

    Energy for change of phase,
    [tex]Q = mH_{v\textrm{ or }f}[/tex]
    where H_v or f is either the heat of vaporization or the heat of fusion, depending on which phases you're going between.

  9. Apr 3, 2004 #8
    They are both going through a change of phase, one to liquid, one to ice. Then the ice gets colder.

    You are making my poor head hurt. Perhaps the ice does not get colder. Perhaps it sublimates or disappears. I do not know how we calculate energy in a change of temperature. I thought we subtracted, but somehow, I doubt any of my math-related thoughts are valid. I am frantically looking through the book, searching for something sensible to repeat, but finding...nothing. All I find is something about L fusion. I don't think that dog is in this fight.
  10. Apr 3, 2004 #9
    Thank you cookiemonster, enjoy your meal, you are out too late, however. I will work on those interesting equations you gave. Thanks again.
  11. Apr 3, 2004 #10
    Allrighty, I'm back and $2 poorer.

    H20 is in only one of three phases. Those are solid, liquid, or gas. Between 0 and 273K, H20 is in solid form. Between 273K and 373K, H20 is in liquid form. Above 373K, H20 is a gas.

    Now, part (a) asks for the energy to convert a gram of water vapor to a gram of water. We can calculate this with the equation

    [tex]Q = H_v m[/tex]

    where H_v is the heat of vaporization of water, 600 cal/g.

    Part (b) is a little more complicated. First you have to calculate the energy to reduce water at 373K to water at 273K using

    [tex]Q = cm\Delta T[/tex]
    where c is the specific heat of water, 1 cal/g/K.

    Then you have to calculate the energy in the change in phase of water to ice, using

    [tex]Q = H_f m[/tex]
    where H_f is the heat of fusion, 80 cal/g.

    Then you finally have to figure out the energy to reduce the ice's temperature, using

    [tex]Q = cm \Delta T[/tex]
    where this time c is the specific heat of ice, .49 cal/g/K.

    Now just add all those up and compare (a) to (b).

  12. Apr 4, 2004 #11
    I try to understand, I really do. I have it as answer (a), at the 600 calories. And answer (b) as only 313 calories total.

    Thank you very much. BTW, if you spent only $2 on your supper, you are not eating enough, you poor thing. I think you ought to get paid for all your help and remarks on the boards, that's what I think. You could amass points, and then once each month, the more decent people on the boards would have a chance to donate some money to the helpful people, and the board owner could divvy it up betweenst all of you. I'd rather donate some money to you than get my signature back. MUCH RATHER.
  13. Apr 4, 2004 #12
    And he could make a game, too, where people who help could have a chance to happen upon a question that has a dollar amount embedded in it. They happen to answer the question, they get the money. I would donate money to have a little pool of funds for the helpers.
  14. Apr 4, 2004 #13
    Haha, that'd be neat.

    But then I'd feel bad.

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