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Internal Energy & Evaporation

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, I am came across the following question and can't quite make complete sense of it:

    "Explain, in terms of internal energy, why a liquid cools when it evaporates".

    2. Relevant equations


    3. The attempt at a solution
    Could someone tell me whether this questions is implying random surface evaporation or boiling? Internal energy is the sum of the internal kinetic energy and the internal potential energy. I would say that in order to evaporate a liquid energy must be supplied to overcome the bonds (latent heat of vaporisation). Since this must come from the water, the water itself must lose heat energy and cool down. The thing that is confusing me is that it says in my book that vaporisation occurs at constant temperature so wouldn't the liquid water stay at the same temperature? Or by evaporation does it mean it is not boiling therefore there is no heat being supplied to the system hence the water cools down?

    How would you explain the overall changes to the internal energy of the system? My attempt (assuming we are talking about random evaporation as oppose to boiling where the temperature would therefore not be constant):

    Some of the energy from the water is transferred to surface molecules giving them enough kinetic energy to escape. Since they have evaporated their potential energy will also increase. The potential energy of the water left behind will decrease as will the kinetic energy (and hence temperature).
     
  2. jcsd
  3. Mar 23, 2015 #2
    Is this question posed within the context of a Thermodynamics course?

    Chet
     
  4. Mar 23, 2015 #3
    Yes, why have I put it in the wrong place?
     
  5. Mar 23, 2015 #4
    No, nothing like that. I was just trying to figure out how I would try to address this question.

    Suppose you have a closed vessel that has liquid water in the lower half of the vessel, and vacuum in the upper half of the vessel. The liquid water half is separated from the vacuum half by a barrier. The liquid in the bottom half is at thermodynamic equilibrium at a uniform temperature T0. At time t = 0 the barrier is removed, and the system is allowed to re-equilibrate. Let the total mass of water in the container be m0, and the internal energy per unit mass of water in the initial state be u0. So the initial internal energy of the vessel contents is m0u0. Suppose that, in attaining the final state of the system, the mass of liquid water that evaporates is m, and let the change in internal energy per unit mass when the water goes from liquid water to vapor vapor at the starting temperature be Δu (so that the internal energy per unit mass of the water vapor is u0+Δu). If the temperature of the vessel contents did not change between the initial and final states of the system, what would the internal energy of the remaining liquid water be? What would the internal energy of the water vapor that forms be? What would be the total internal energy of the contents of the container be? How much heat would have to be added to hold the contents of the container at the starting temperature? What if the heat weren't added?

    Chet
     
    Last edited: Mar 23, 2015
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