Latent heat effects thermal equilibrium

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saykaof
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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?

Where:
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?
 
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saykaof said:
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?
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.
 
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Absolutely wonderful for both the question and solution.

It's delightful for me, too, to realize that all that 80 celsius worth of kinetic energy in water are absorbed by the ice in order to cause a phase transition from 0 celsius solid to 0 celsius 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.
 
Thanks all for your explanation, they were simple enough for me to understand my problem.