How much heat is being drawn when water vaporize?

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  • #26
sophiecentaur
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So you've decided to go for the hardest one!
I have one more awkward question for you, though. Is the air stationary or moving? As soon as it moves, the convection will take over as the main heat transfer mechanism, I think.
Boy. This is difficult. The arm waving bit is easy but the quantitative stuff isn't.
 
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Lol, actually I thought this is the easiest option. What did you mean that is easier then?
:P
 
  • #28
sophiecentaur
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I thought that you could have calculated the heat getting into a blackened container from surrounding air easier than needing to work out what's happening right on the surface. After all, wouldn't there be some sort of temperature gradient at the surface? Perhaps not, as the container is insulated.
I must admit, I'm getting into 'problem fatigue' haha.
 
  • #29
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LOL, thank you for the effort. I know its not much, but I click every commercial link here every time I get in...:P

Well, for starters lets go your way. I assume that the blackened body is either water or a material containing waters heat conductivity properties.
Regarding your previous question, eventually of course I want to know what happens with moving air (which evaporates water from an insulated bowl), but first lets say that the air is still.
 
  • #30
sophiecentaur
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Looking at your original problem (without the blackened container to help with the heat supply) I did find a link last night (but went to bed and forgot where I found it. It gave a value for heat transfer at an air interface for still air - about 20W/msquared/degreeC (plus or minus quite a lot)
I'm trying to get my head around how to relate this to the rate at which water can evaporate and the resulting actual temperature difference that it will settle at. I can't get back to this thread for a few days but I will keep thinking.
 
  • #31
haruspex
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enth,

When you say a bowl of water is at such-and-such a temperature, that effectively specifies the average kinetic energy in each molecule. At any instant, though, some will have more, some less, according to a standard probability distribution.
The more energetic ones may have enough energy to escape the bowl and become vapour.
At the same time, the water vapour molecules in the air are bouncing around, gaining and losing energy. Some lose so much that they get caught by the water.

When these two processes are in balance, no net evaporation or condensation occurs.
Make the air a bit drier and fewer are available to recondense, so net evaporation does occur. Make the water a bit warmer and more are available to evaporate.
So at a given water temperature, there's a certain density of vapour above it (the saturation vapour pressure at that temperature) at which the system is in equilibrium.

In all cases, evaporation cools the water and condensation heats it.

The fan in your model is there purely to ensure a steady supply of air at the given temperature and humidity (otherwise the air just above the water gets saturated and cold). Similarly, the surface layer of the water will cool first, inhibiting further evaporation, but as long as it's above 4C convection should kick in and redistribute the heat.

Based on the above, you can do the calculation exactly as SophieC said.
temperature drop in bowl = heat required to vaporise lost water / thermal capacity of remaining water.

Once the water starts to cool, it will be cooler than the air above. This means some heat will flow from the air to the water. As a result, the temperature drop will not be quite as great.
 

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