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How does clay pot cool water

  1. Apr 17, 2015 #1
    From things i've read it states that water oozes through pores of clay pot to outer surface and absorbs the heat form outer air.The heat absorbed is latent heat for water and specific heat for air outside which lowers the air temperature and as evaporation on outer surface takes place water inside gets cooled.
    My doubts are
    1) If the water gains latent heat and evaporates then there should be no temperature change within the water inside.Then how come water inside pot cools down
  2. jcsd
  3. Apr 17, 2015 #2


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    From what you are describing, water in an uglazed clay pot used to cool air, by absorbing heat from the air, the water in the pot gets warmer, not cooler.
  4. Apr 17, 2015 #3
    The heat to evaporate the water exiting the pores is supplied mainly by the clay pot (although part of it also comes from the surrounding air). The evaporation causes the outer surface of the clay pot to cool, just like evaporation of sweat from your body cools your skin on a hot day. The cooler surface temperature of the clay pot causes heat to be conducted from the water inside the pot to the cool outer surface, thus cooling the liquid inside.

  5. Apr 18, 2015 #4
    @ Chestermiller
    If a part of latent of vaporization of water is taken from pot then the resulting air would have a higher enthalpy value than the initial condition since some part of heat is externally added from pot.Then how come the air has an constant enthalpy value as stated in http://en.wikipedia.org/wiki/Evaporative_cooler
  6. Apr 18, 2015 #5
    The is no "extra heat" available for heating anything. The evaporating water acts as a heat sink, not as a heat source. So, if anything, the air next to the pot will be cooler than the surrounding air.

  7. Apr 18, 2015 #6


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    I think that statement adds confusion. The water on the surface of the pot will lose heat because the water molecules that are blown away from the surface are those with higher Kinetic Energy than average. They take net energy from the cooler +water+ beer and lower the temperature.

    The limit to the amount of cooling is set when the temperature gets low enough for fewer molecules to be lost. A fast current of air will, of course, increase the cooling effect as the faster molecules can never return to the pot surface.
  8. Apr 18, 2015 #7


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    That's an idealization that assumes the system is in a perfect steady-state. In a perfect steady-state, the enthalpy of the air and water are equal and remain equal because while the evaporating water is adding enthalpy to the air, the air is also heating the water, at exactly the same rate, so the enthalpy remains constant in both.
  9. Apr 18, 2015 #8


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    Are we not losing sight of the fact that this sort of cooling system actually works?? Some of the comments in the thread seem to be indicating that it doesn't, which is nonsense.
    The reason is that it isn't an equilibrium situation because the external air is constantly being changed. It's pretty obvious that, if you kept the unglazed pot in a plastic bag, the cooling wouldn't work. What is needed is an explanation of why it works when it works.
  10. Apr 18, 2015 #9
    The enthalpy that the vapor adds to the air is at the same temperature as the evaporating liquid (i.e., cold). At steady state, the water vapor is diffusing away from the surface of the pot and into the surroundings (which have infinite capacity to accept it). The transport of water vapor away from the pot is also aided by convection in the air. So the rate of evaporation is constant, and the rate of cooling is constant.

    Russ, in your above post, I think you meant to use the word "constant" in all locations in the second sentence where you used the word "equal."

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