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Latent heat effects thermal equilibrium

  1. Jul 17, 2009 #1
    Can anyone explain that how the ice's temperature is zero degree and water's temperature is 80 degree are mix and their final temperature is zero degree if the both ice and water has the same mass?

    mc(delta T) of water = (mc(delta T) + mL ) of ice
    the result for final temperature is 0 degree but how can it be???

    This is telling me that if I have a perfect insulation container then every drop of water at 80 degree can be zero degree as long as the intial mass is the same, is it true?
  2. jcsd
  3. Jul 17, 2009 #2


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    Gold Member

    Because water has a latent heat. At normal pressure, you need around 80 cal to transform 1 g of 0°C ice into 0°C water, which is the heat needed to cool off 1 g of water by 80°C.
    For further reading : http://en.wikipedia.org/wiki/Enthalpy_of_fusion.
    Last edited: Jul 17, 2009
  4. Jul 17, 2009 #3
    Absolutely wonderful for both the question and solution.

    It's delightful for me, too, to realize that all that 80 celcius worth of kinetic energy in water are absorbed by the ice in order to cause a phase transition from 0 celcius solid to 0 celcius liquid.

    It'd be wonderful to see this in reality, only if i have the tool to precisely control the temperature and mix them with near perfect isolation.
  5. Jul 17, 2009 #4


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    Staff: Mentor

    You can do the experiment in a thermos easily enough.
  6. Jul 18, 2009 #5
    Thanks all for your explanation, they were simple enough for me to understand my problem.
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