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How much heat is being drawn when water vaporize?

  1. Feb 29, 2012 #1
    Hello,

    So, I'm not much of a physicist but I thought that at least this question will be simple enough to answer with some internet browsing... though it seems apparently that I can't tell my left from my right... :P

    I have a bowl of water at room temperature (25 C). I blow at it with a fan who transports wind at say 25C with relative humidity of 40%.
    And now I'm trying to figure out what amount of energy have dissapeared from the liquid if say 10 grams were evaporated...



    I read in wikipedia all about latent heat and heat of vaporization, but I still cant really figure much out.
    Thank you very much, I know its fairly rudimentary question.
    :)
     
  2. jcsd
  3. Feb 29, 2012 #2

    Redbelly98

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    Welcome to Physics Forums.

    Heat of vaporization might be given in terms of Joules per gram, and you have 10 grams of water that evaporated. So it's a straightforward multiplication to get the energy change.
     
  4. Feb 29, 2012 #3
    total heat = mass x specific heat of vaporization. just plug it in. now if you want to account for the heat that's used by the fan, that gets tricky but lets assume its negligible so its still an easy equation.
     
  5. Mar 1, 2012 #4

    sophiecentaur

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    it's referred to as latent heat, I believe, if anyone wants to look it up.

    Actually, the devil is in the detail in this situation. It is true that the amount of heat needed to vaporise 10g of water is easily calculated but the rate of cooling and the final temperature reached will depend on humidity and air speed (and the total mass of water you started with).

    There is a quaint piece of equipment called a Wet and Dry Bulb Thermometer which can be used to measure air humidity. It is just what the name suggests and the wet bulb is wrapped in wet cloth. You whirl it around in the air and note the difference in temperature between the wet bulb and dry bulb. In highly saturated air the temperature difference will be small but, in very dry air, the temperature difference will be significant as the wet bulb cools down. There will be a table, printed on the back, to let you work out the humidity from the two readings.
     
  6. Mar 1, 2012 #5
    Thank you for your answers, but here is the thing that got me stuck-
    Vaporization occurs when the surface molecules of the water recieve energy from the air, allowing them to escape.
    Latent heat of water vapor, as far as I understood (and please correct if I'm wrong) is the entire energy that water molecule posseses and with which it leaves the liquid system.
    Meaning, Latent Heat=Energy taken from water+Energy taken from air. So how can I calculate only the energy which left the water?








    edit: How much energy does a gram of 25C water (or any material really) carry?
    Should the count be from freezing point (0C) or maybe from absolute zero?
    Or am I completly off-base here?
     
    Last edited: Mar 1, 2012
  7. Mar 2, 2012 #6

    sophiecentaur

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    The water will lose heat or gain heat from the air, depending on the temperature difference (could be either way).

    The reason that heat is lost is that it is the faster water molecules (the ones with way above mean KE) that will escape. This means that the mean energy (temperature) reduces.

    With a covered (insulated) container, the final situation will be that air and water end up at the same temperature and the vapour pressure in the air is the same as the vapour pressure on the water surface - the humidity will reach a steady value and as many molecules will land on the surface as leave the surface.

    If there is a current of new air over the surface of the water, many of those faster water molecules which have made it into the air will be carried away, along with their 'above-mean' KE so the water temperature will continue to drop until, as before, the vapour pressures on either side of the boundary are the same.

    For most thermodynamic problems like this, you are only interested in changes of energy and not absolute energy because absolute energy is not really relevant. (Who cares about the energy needed to bring your gram of water fro 0K to room temperature?)
     
  8. Mar 2, 2012 #7

    Redbelly98

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    The temperature difference is zero:
     
  9. Mar 2, 2012 #8

    sophiecentaur

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    But only at the start. The water will cool down. The energy needed to evaporate the water will come from the Air. The air will have thermal energy (KE of randomly moving fast molecules, which have an RMS speed of about 500m/s) and a very small amount of KE die to the total mass moving slowly (say 2m/s - and I think that only relates to the molecules near the surface). The only significant source of energy will be the thermal energy of the Air.

    So the total energy involved will be mass X latent heat of vaporisation and it will all have to get there by cooling the air. If you want to work out the rate of cooling then that's a different problem - a bit harder.
     
  10. Mar 2, 2012 #9

    Redbelly98

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    Okay, at first I was thinking the 25C air would maintain the water at the same temperature. But it makes sense that the water temperature would drop to some lower value, and the air would actually maintain that lower temperature for the water.
     
  11. Mar 3, 2012 #10

    sophiecentaur

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    I always like to tackle things like this from the Energy point of view. The energy for this, in the long run, must come from the air. Whatever the source of the energy, the latent heat of vaporisation is the same value.
    The time factor is a much more difficult thing to discuss. It would relate to air speed and would be very non-linear as the surface would form wavelets, which would change the surface area plus the air flow. The rate of evaporation would be related to the final temperature of the water.
     
  12. Mar 4, 2012 #11
    Hmm...
    Thank you, I think that I almost understand you. So if I want to know what the temperature of the (25C) water would be after 10g were evaporated, I should multiply the amount of water (0.01 kg) by the latent heat of evaporation (2257 kj/kg for water) which would give me the result of 22.57 kj, and... umm...
    Not sure what next. :P
     
  13. Mar 5, 2012 #12

    sophiecentaur

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    Then: Heat supplied will be
    Total mass X shc X temp drop
    Equate the two and solve for unknown.
     
  14. Mar 8, 2012 #13
    So...
    22.57kj= 22570 j

    22570(j)=10(g)*4.186(j)*X(C)?

    22570/41.86=X
    x=539?


    I obviously screwed the conversion somewhere...

    PS-4.186 (j/g) is the specific heat of water...


    PS2-
    Uhh... Actually on second thought, I think that this is a misunderstanding problem and not a conversion one.
    We have the value of the ENTIRE heat a water molecule possesses in order to become vapour (22.57).
    We also have the values for energy required to raise this molecule by 1C (41.86).

    But the formula you cited (Heat Added(Q)=Specific Heat(C)*Mass(m)*Change in temperature (Delta T))- Has 2 variables. Change in temperature and Heat ADDED (so that the molecule would have its ENTIRE energy).

    Therefore...Uhhh... well I'm not sure about the formula I should make.

    I hope that I was understood.
    Sorry for holding the rest of the class, I've always been like this. XD
     
  15. Mar 8, 2012 #14

    sophiecentaur

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    I'm not sure exactly what your thought experiment consists of; it may not be realistic. The 10g of water can't 'contain' enough thermal energy for it to evaporate on its own. I think that's why you got a dodgy answer. It implies that the water needs to 'lose 540 degrees Celcius - the most it can lose is 300 and it'll be at Zeero K!!! haha
    There must be some other source of heat (the rest of the water, the air, the Sun or a heater) to make the 10g evaporate.
     
  16. Mar 18, 2012 #15
    Ummm...
    But are you sure that the entire use of the formulas was correct?

    Don't we have two unknowns here?
    'Change in temperature' and 'Heat added'?
    How can we equate Q(Heat Added) to the overall heat of the system (the vapor molecules)? One is only a part of the other.


    Thank you.
     
  17. Mar 18, 2012 #16

    sophiecentaur

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    I don't know that the simple formula for latent heat of vaporisation holds over all temperatures (particularly when sublimation comes into it. Also, the SCH of water changes with temperature.
    But the general principle holds and the experimental details are so vague that I reckon my conclusion is valid.
     
  18. Mar 21, 2012 #17
    Ok, so if my scenario is unrealistic, what is a realistic scenario then(when the water isn't hot so it's not the source of the energy)?
    Why is it that when i'm in a room that shows around 25C on the thermometer and I blow a fan over a glass dish (or a towel) with little water- the water dissapears?
     
  19. Mar 21, 2012 #18

    sophiecentaur

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    Heat gets there from elsewhere because the room temperature drops.
     
  20. Mar 22, 2012 #19
    I see...
    Well, I suppose that the heat comes from the air. So, are you saying that in a room temperature if I evaporate half a cup of water by blowing air, the temperature of the remainding water would not change?
     
  21. Mar 22, 2012 #20

    sophiecentaur

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    In an infinite room, the temperature of the water will drop until the latent heat times the mass lost per second is balanced by the rate at which heat goes from room to water.
     
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