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Evaporative cooling of beverage problem involving thermodynamics and heat concepts

  1. Oct 31, 2008 #1
    A cold beverage can be kep cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperatures of 15 degrees C, the environment has temperature of 32 degrees C and the container is a cylinder with radius of 2.2cm and height of 10cm. Approximate the emissivity as 1 and neglect other energy exchanges. At what rate dm/dt is the container losing water mass.

    Now I have attached the solution from the solutions manual. However my problem with the solution is shouldn't it be considered that heat must be transferred to bring the temperature of the water up to 100 degrees C?
     

    Attached Files:

  2. jcsd
  3. Oct 31, 2008 #2
    Re: Evaporative cooling of beverage problem involving thermodynamics and heat concept

    can you guys access the attachment or is it still pending approval
     
  4. Oct 31, 2008 #3

    LowlyPion

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    Re: Evaporative cooling of beverage problem involving thermodynamics and heat concept

    Generally it's better to put in a link to the picture if you can.
     
  5. Oct 31, 2008 #4
    Re: Evaporative cooling of beverage problem involving thermodynamics and heat concept

    Well I only have the file, it is not available anywhere on the internet. The picture has been approved and now you guys can view the picture.

    Can anyone help me with this problem?
     
  6. Nov 1, 2008 #5

    tiny-tim

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    Welcome to PF!

    Hi laksate ! Welcome to PF! :smile:
    No … evaporation happens without any change in temperature …

    even at "room temperature", some of the molecules are going fast enough to get away! :biggrin:

    For some more detail, se http://en.wikipedia.org/wiki/Thermal_radiation :wink:
     
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