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Mass, Temperature, and Relativity?

  1. Sep 5, 2015 #1
    By how much (in picograms) does the mass of 1 mol of water at 0° c differ from the mass of 1 mol of ice at 0° c?

    1 mol of water = (assuming) 1 mol of ice = 18 g = 1.8E13 pg

    So, is this a trick question? The logic of a trickster would tell you that there would be no difference, if one assumes ice to be water ice, and that water is typically frozen at 0 c (further how do I determine the mass of ice, if I'm not told the chemical makeup). However my question extends more to, does special relativity say anything about mass effecting temperature? Does 1 mol of water weigh more at 100 c vs 0 c?

    I have no attempt at solving this question, because I don't know where to start. My text book gives a basic example, that heating a balloon increases the mass of the balloon as a system, but that the balloon's atoms themselves do not increase in mass. However, it does not provide a way to determine how the mass has changed.
     
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  3. Sep 5, 2015 #2
    Well, water has latent energy stored in it (I forgot the proper use of those words, but I think you can figure them out) When water freezes, heat is expelled, deducing with the conservation of energy, it's only logical to think that this energy is stored in the water where it is absent in the ice. Thinking about it from another point of view, there are no bonds in a body of water where they are everywhere in ice, to form bonds, energy is released, so the amount of energy that the ice lacked would be the amount of energy that was released from the formation of those bonds.

    EDIT:
    Oh, by the way, use the equation that we all know but rarely use (at least I rarely use): E = mc^2
     
  4. Sep 5, 2015 #3
    Thanks for the reply!

    I thought about using E=mc^2, but I don't know how to account for temp, since I am assuming this question is asking indirectly about kinetic energy as technically the atoms are in motion. Plus, since its asking for the same chemical, at the same temp, even if somehow one sample was still liquid while the other was frozen, would they not have the same kinetic energy?
     
  5. Sep 5, 2015 #4

    SteamKing

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    When liquid water freezes, there is no change in temperature. The latent heat of fusion is the amount of energy which must be removed from the liquid at 0°C to convert it to ice at 0°C.

    You should brush up on the physics of phase changes ...
     
  6. Sep 5, 2015 #5
    interesting, so there is an overall change in energy between the two samples of matter of different phases, not reflected in their temperatures? I am not familiar with what you are talking about, perhaps that should be where I start looking then?

    EDIT: would the equation Q=mL work, where:
    Q is the amount of energy released or absorbed during the change of phase, m is the mass of the substance, and L is the specific latent heat for a particular substance (334 kJ-kgm−1 according to Wikipedia)[1]
     
    Last edited: Sep 5, 2015
  7. Sep 5, 2015 #6
    Yes, latent heat of fusion is the amount of energy a sample of material needs to absorb or release in order to change phase, and that is the change in energy not reflected by temperature. Speaking of which, to calculate the amount of energy that IS shown by temperature, you should take a look at "Boltzmann constant". I am not very familiar with the specifics of that either, but it should be relevant.
     
  8. Sep 5, 2015 #7

    SteamKing

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    Usually, this is discussed in physical science classes in high school, if not earlier. It is important to know if you are working out problems in calorimetry.

    See the Link below "phasechange.pdf"

    Yes, this is how you determine how much latent heat to remove to freeze a given mass of water.
     

    Attached Files:

  9. Sep 5, 2015 #8
    This is a note on my current (and likely very wrong) attempt at this problem.

    Based on what I have learned so far, there is an energy drop between two samples of matter at the same temperature, but that are in different phases of matter (I very vaguely remember this from chemistry now). This can be calculated using Q=mL, where Q is the amount of energy gained or lost in a phase change, m is mass, and L is the specific latent heat (which is 334~336 kJ-kg−1 depending on the source). Using that I get:

    .018kg * 334 kJ/kg = 6.012 kJ = 6012J


    Then using the equation to calculate internal energy, E=mc^2, we can find out the change in mass, E/c^2=m


    6012J/ (3*10^8 m/s)^2= 6.68*10^-14 kg = 66.8 pg


    This seems like an awful lot of mass though. Any pointers?
     
  10. Sep 5, 2015 #9

    SteamKing

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    It takes the removal of a relatively large amount of heat from water to produce ice. It requires an order of magnitude greater addition of heat to turn water into steam, not counting the heat required to raise its temperature to 100°C first.

    https://en.wikipedia.org/wiki/Latent_heat

    This is one reason why we are able to make and enjoy a cold drink by adding ice, and why steam power is often used to make electricity fairly economically.
     
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