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Question: Evaporation - beyond boiling point

  1. Nov 15, 2006 #1
    Hi, I've been wonderring what happens to a substance once boiling point is reached, in particular the rate of evaporation.

    Am I right in thinking that further energy will cause the rate of evaporation to increase, while temperature of substance remains constant?

    Is there a point where if enough energy is given, instanteneous evaporation occurs?

    Cheers
     
  2. jcsd
  3. Nov 15, 2006 #2
    There will be phase coexistence. Further input of energy will not raise the temperature.

    Think of a nuke in water.
     
  4. Nov 16, 2006 #3
    The boiling (evaporation) of the liquid is actual an equilibrium between the liquid phase and the gas phase. The boiling point is the tempurate at which the equilibrium shifts to gas phase. The more energy you can put in the faster the equilibrium shifts. The rate of evaperation is governed by the the energy input. The faster you put energy into a system the faster the material will boil. If you can get the entire energy required boil a give mass of liquid into the liquid all at once....boom it flash evaporates!

    Try this link: http://cnx.org/content/m12596/latest/
     
  5. Nov 16, 2006 #4
    So, if the energy required to break all intermolecular bonds in a given volume of substance is equal to or is less than the Power(Energy per second) put into substance (assuming 100% efficiency), then total evaporation of the substence will occur in no longer than a second?

    Is this the requirement? A Nuke would vapourise just about anything, let alone water, I'm more interested in what the limits are. If heat energy is just kinetic energy of particles, what limits are imposed on how quickly the momentum of the particles can be transfered to further particles? In other words, how quickly can you heat something? Is there a limit regardless of how much energy you pummel at a substance/object per second?
     
  6. Nov 17, 2006 #5
    The is a limit to how fast heat moves through a substance if it is transferred by kinetic energy. For heat transfer by radiative means then the limit would be governed by the speed of light and the ability of material to absorb the photon and turn it into kinetic energy. There is a definite rate at which an atom/molecule absorbs a photon and then rearranges to accommodate the energy. Even if the photon is infrared (for example in a molecule) it takes time for the energy to transfer to the actual movement of the atoms involved. I actually took a kinetics class and our final project was to calculate the rate at which this sort of thing happened.
     
  7. Feb 3, 2009 #6
    Hi All,

    Not quite on the same question, but I have a question along the same line. If water was being boiled in an open container and all things remained constant (atmospheric pressure, humidity, heat source), would the rate of evaporation stay linear over time once the boiling point was reached, even though the volume of liquid is decreasing or would the rate of evaporation increase because of the amount of volume the same amount of energy is being applied to has decreased?

    Thanks in advance for your help
     
  8. Feb 3, 2009 #7

    Mapes

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    Hi Ragged, welcome to PF. The evaporation rate would be constant because the input power and exposed area are constant.
     
  9. Feb 3, 2009 #8

    LURCH

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    Carrtying the same idea a bit further; what about a cone-shaped container? There, the surface area would continually decrease, but the ratio of surface area to volume would remain constant. As the water level lowers, would the evaporation rate remain constant?
     
  10. Feb 3, 2009 #9

    Mapes

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    At a constant temperature less than the boiling point, the evaporation rate would decrease because the surface area has decreased.

    At constant power input at the boiling temperature, the evaporation rate (mol/s) would be constant because the rate of energy acquired by the gas phase through vaporization must be constant to balance the power input. The evaporation rate per area increases as the surface area decreases.

    At constant power input at a temperature less than boiling... need to think about it some more.

    EDIT: Posted too fast the first time, please disregard my earlier answer.
     
    Last edited: Feb 3, 2009
  11. Oct 31, 2009 #10
    To further this Discussion. What equation would I use to qualify the evaporation rate of boiling water given a varying amount of energy input as well as a varying surface area?
     
  12. Oct 31, 2009 #11

    russ_watters

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    Once it is boiling, nothing else matters besides latent heat of vaporization and energy input. Divide energy by heat of vaporization and you get mass flow of vapor.
     
  13. Oct 31, 2009 #12
    What you saying is that once you reach boiling point, whether under a vacuum or open to atmosphere, the only variables that play a role are heat energy input divided by heat of vaporization. I thought the amount of surface area also played a major role?
     
  14. Nov 1, 2009 #13
    Note that the boiling point is reached at the temperature at which the vapor pressure equals the atmospheric pressure (or at whatever pressure the liquid is kept at). What then happens is that inside the liguid small vapor bubbles can form. Any such bubble will be at the atmospheric pressure, so they can only form if the vapor pressure is equal to the atmospheric pressure (or larger).

    If you are below the boiling point, the liquid can still evaporate, but that happens at the surface of the liquid. The molecules can escape the liquid from the surface and they can also be absorbed into the liquid. Equilibrium is reached when the partial pressure of the gas of the molecules equals the vapor pressure (in case of water you would then be at 100% relative humidity).
     
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