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Vaporisation of coffee in a sealed cup

  1. Jul 30, 2012 #1
    Just wondering - if you have one of those cups of coffee with a lid on it, how much of the liquid would evaporate (assuming reasonable figures for the size of the cup and temperature of the liquid or whatever other variables are relevant)? Since it's essentially a sealed container, I'd assume there'd have to be some kind of equilibrium reached. What kind of figure would you be looking at in terms of the heat that the liquid loses to the air? Is this a big factor in the overall loss of heat from the coffee compared to conduction and radiation? Thanks in advance!
     
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  3. Jul 30, 2012 #2

    CWatters

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  4. Jul 31, 2012 #3
    I saw this page. But how do we calculate the mass or moles of particles in the vapour state? How do we equate pressure to actual amount of substance?
     
  5. Jul 31, 2012 #4

    NascentOxygen

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    Hi Musimatician. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
    Knowing the temperature and pressure, and recognizing the relative humidity will be 100%, you consult a chart relating those parameters to the maximum mass of water vapour that can be held per unit volume.

    This will be a useful start: http://www.engineeringtoolbox.com/relative-humidity-air-d_687.html

    You will have to approximate your data, because there will be a thermal gradient in the space above your coffee—near the coffee will be hotter than adjacent to the colder lid.

    It is surprising just how much water is held as gas in our ambient air. Even in the middle of the Sahara and its low relative humidity, there is ample water for any household, and it's nearing feasiblity to extract it using a solar powered compressor (think of it collecting the drips from an array of air conditioners). https://www.physicsforums.com/images/icons/icon14.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Aug 15, 2012 #5
    Hmm.... I've calculated that the space in the cup will hold 3.2g of water/coffee vapour. But now, is there a way to calculate the rate of heat loss? The speed at which the liquid evaporates?
     
  7. Aug 15, 2012 #6

    NascentOxygen

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    Once it reaches an equilibrium, such as it is, I was thinking there would be no further evaporation. The plastic lid will become covered with a relatively stable layer of condensate. Heat loss from the drink will be from its surface by circulating "air currents" convection, plus radiant losses, conveying heat to the condensate layer on the lid, which transfers it by conduction and radiation through the plastic lid. That's how I picture it, anyway. Feel free to differ.
     
  8. Aug 16, 2012 #7
    Yeah, that sounds right. But what about before equilibirum? I mean, it doesn't reach that state instantly, does it?
     
  9. Aug 16, 2012 #8

    NascentOxygen

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    I have no idea how to analyze the dynamics while the cup equilibrates. Sorry.
     
  10. Aug 16, 2012 #9
    ah, thanks anyway. Know any way to get the attention of those who do?
     
  11. Aug 16, 2012 #10
    How big is that cup?

    The density of air is 1.09 kg/m^3 at 50 degrees celsius, that's only 1.09 mg/(cm)^3.

    partial pressure of water vapour at 50 degrees is 0.125 bar, and a water molecule is lighter than an average air mole cule by a factor of about 18/29, so I get only 0.084 mg for each cubic cm. You'd need 40 litres or more than 10 gallons to get 3.2 grams of water vapour.
     
  12. Aug 16, 2012 #11

    jbriggs444

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    I picture it differently.

    I do not agree that the condensate layer will be stable. It will be dripping down into the hot coffee below. The dominant method of heat transfer will be evaporation/condensation.
     
  13. Aug 16, 2012 #12
    I think that it's stable BECAUSE of the condensation/evaporation cancelling out, isn't it? That's the point of a saturated vapour pressure? But anyway, how long does it take to get to this stage?
     
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