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It takes 970 BTU's (British Thermal Units) to convert 1 pound of 212 degree F water at atmospheric pressure to 212 degree F steam at atmospheric pressure. How many BTU's are required to evaporate one pound of water at atmospheric pressure? It seems that evaporation of water doesn't require adding any heat, but there is still a change of state from liquid to vapor. What's more, evaporation happens at less than 212 degrees F. Are steam and water vapor the same?
With boiling the hydrogen bonding between all the water molecules are breaking about the same time, allowing individual water molecules to form throughout the continuum. In evaporation the same thing is occurring but on a very limited scale, usually just at the air-water surface.

If one took a gallon of water and boiled it off and a second gallon and let it evaporate, the boiling gallon would take more energy because one also have to increase the temperature thereby adding energy into thermal capacitance. The energy into the hydrogen bonded would be similar for the change of state. Evaporation also makes use of the chemical potential of a concentration gradient associated with an entropy potential.

Evaporation still requires energy. This often absorbed from the rest of the liquid water. The result being the water cooling. Sweat uses this principle with the evaporation absorbing skin heat to make us cooler.
Hi Reallyreally,MDT,

Nice question.

For starters see

It would be nice to get to the bottom of heat loss by evapouration which MDT mentioned .. maybe someone else will post too.

Best wishes,


This looks like a nice sensible account of evaporation...


Water does not have to be at the boiling point to evaporate.  ...  In a given amount of water at a given temperature, some molecules of water will have more energy than others, so some molecules will be able to evaporate, while others remain in the liquid state.  The lower the temperature of the water, the more energy is required for evaporation.  If the water is liquid at a temperature of 0 degrees C, the latent heat of vaporization is 597 cal/g, compared to 540 cal/g at 100 degrees C.  In between, at 50 degrees C, an input of 569 cal/g would be required for evaporation.

Sooo... very roughly it looks like the evaporation of 1 pound of water will cool about 200 pounds of water by 1 degree F.

Pictures of cooling towers and warning ph34r.gif here!

Hope this helps,


Thank you both for the great replies. You were very helpful.
steam is made up of water droplets while water vapour is made up of... well, water vapour, so they are different. huh.gif
Whether boiling or not, the heat needed is the same at the same temperature.

But if you vaporize at a different temperature, you get a difference, because liquid and gaseous water have different heat capacities. Think of two well chosen series of transformations that have identical initial and final states, and you'll see the relation.

Under reasonable conditions, the difference is small as compared to the big heat amount needed to vaporize water. But if you get close to the critical point (374C and 218b) then the vaporization energy gets small, or even zero above this point - in other words, there is no liquid more.
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