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Evaporative Cooling

  1. Jun 6, 2006 #1
    Right now, we're doing thermal physics.

    For one of our assignments, the teacher asked us to make our own experiment and write on a lab on it. For this lab, we had to investigate a question she gave us:

    If water is left in a container open to the air, its temperature will drop below the room temperature. What 2 factors determine the amount by which the temperature is depressed?

    For one of the factors, I tried added increasing amounts of salt into various beakers with water with the same initial temperatures - hoping that the salt will interfere with the final temperature when I left it overnight. However, all the temperatures were still the same when I recorded them in the morning.

    So now I'm stumped, what other factors can I test on that I can do at home - all I have is a thermometer and a beaker...

    Thanks :)
     
  2. jcsd
  3. Jun 6, 2006 #2

    Tide

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    Have you thought about varying the relative humidity? What else might affect evaporation rates?
     
  4. Jun 6, 2006 #3
    Humidity is a factor that will affect it but how could I vary it in my house without any devices :confused: I can probably do something with the temperature of the surrounding environment ... but there isn't much else that would affect the amount by which the temperature will go down.

    Also, I was wondering why my little experiment with the salt didn't work. I know, for example, that the salt would interfere with the freezing point of the water (it interfering with the lattice of the water molecules) but then I'm having trouble applying this to the evaporation process.
     
  5. Jun 7, 2006 #4

    Tide

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    The effect of salinity on evaporation is a little tricky. My understanding is that higher salinity will decrease the evaporation rate. However, the addition of salt to water will increase the temperature of the water (slightly) thereby increasing the evaporation rate. If you measured the evaporation rate shortly after adding salt to the water your results could have gone either way. Try it again but give the water time to "cool down" before making your evaporation rate measurements.

    As for relative humidity, there are several possibilities. You could work in a small room, such as a bathroom, with the door closed and suspend wet towels for a period, run the shower or something similar to increase the relative humidity.

    Also, have you considered the effect of air convection? A simple fan would be suitable to test this effect.
     
  6. Jun 8, 2006 #5

    lightgrav

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    most people would increase the relative humidity by putting a lid on the jar.
    You might try placing an open jar in the refrigerator ... how much colder than the 'fridge does its water get?
     
  7. Jun 9, 2006 #6

    Tide

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    With such a small volume (jar with lid) you quickly saturate the air so you'll have difficulty measuring the evaporation rate because it exactly balances the condensation rate.
     
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