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When the sun is out after a rain, I can see what appears to be steam rising off a wooden bridge nearby. I'm pretty sure this is water turning into a gas.

However, I thought water had to reach 100 degrees C to be able to turn into a gas.

Is there an edge case, for small amounts of water perhaps, that allows it to evaporate?

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5 Answers

up vote 15 down vote accepted

Evaporation is a different process to boiling. The first is a surface effect that happens at any time, while the latter is a bulk transformation that only happens when the conditions are correct. Technically the water is not turning into a gas, but random movement of the surface molecules allows some of them enough energy to escape from the surface into the air. The rate at which they leave the surface depends on a number of factors - for instance the temperature of both air and water, the humidity of the air, and the size of the surface exposed. When the bridge is 'steaming': the wood is marginally warmer than the air (due to the sun shine), the air is very humid (it has just been raining) and the water is spread out to expose a very large surface area. In fact, since the air is cooler and almost saturated with water, the molecules of water are almost immediately condensing into micro-droplets in the air - which is why you can see them.

BTW - steam is completely transparent. If you can see it then it is water vapour. Consider a kettle boiling - the white plume only occurs a short distance above the spout. Below that it is steam, above it has cooled into vapour.

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I would add that even ice evaporates with the same process, in this case called sublimation. That is how we have no frost freezers and refrigerators. –  anna v May 27 '11 at 5:57
    
""Below that it is steam, above it has cooled into vapour."" Is this really the meaning of steam vs. vapour? –  Georg May 27 '11 at 8:32
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Dear @Peter and @Georg. Unfortunately, it seems that Peter got vapor and steam mixed up in his answer (v1), see e.g., wikipedia en.wikipedia.org/wiki/Water_vapor and en.wikipedia.org/wiki/Steam –  Qmechanic May 27 '11 at 9:57
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I don't the the rate of evaporation depends upon the temperature of the overlying air, i.e. it is a function of the thermodynamic processes withing the liquid (or solid, like in ann's example). Water vapor is also going the other way, from the air onto the surface, and this is affected by the thermo conditions of the air, i.e. humidity doesn't prevent evaporation, it competes against it. –  Omega Centauri May 27 '11 at 16:45
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@Qmechanic - I just had a look at the Wiki articles. All I can say is that those definitions are exactly the opposite of what I learnt at school. I'll have to be more careful of terminology in future :-) –  Peter May 30 '11 at 23:03
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For every temperature, there is some amount of water vapor that can exist as gas mixed in with the air. This is called the saturation pressure of water at that temperature. The relative humidity is the amount of water vapor, expressed as a percentage of the saturation pressure. As you increase the temperature, the saturation pressure increases.

Steam is water in its gaseous phase.

You can't see water vapor, you can't see steam, but you can see mist, which is water droplets suspended in the air.

When you boil water on the stove, you get steam. This then cools when it comes into contact with the air, increasing the relative humidity above 100%, so the water vapor condenses into mist.

If the relative humidity is bigger than 100%, water vapor will condense from the air, becoming dew and/or mist. If the relative humidity is less than 100%, water will evaporate into the air, becoming water vapor.

If the wooden bridge is warmer than the surrounding air, and the relative humidity is around 100%, then water will evaporate off of the wooden bridge, turning into water vapor (the relative humidity is lower right next to the bridge, because the bridge is warmer). When the air containing this water vapor rises and cools, water condenses out of it, turning into the mist that you see.

Here is a graph of the saturation pressure (from this website). Note that at 100°C, the pressure is $\approx10^5$ Pa $=1000\,$hPa, which is roughly atmospheric pressure. This means that at 100°C, you can have pure water vapor at atmospheric pressure. This is why water boils at 100°C at sea level—a bubble of steam can form below the surface of the water. At higher altitudes, the boiling point can be substantially lower.

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(This is a comment to your question, but I can only post it as an answer)

Aside from kinetic considerations, what people often misunderstand is the thermodynamic description of the process. If you have a pure liquid (say water) then it can only boil (transform into vapour in the bulk) and not evaporate. At an external pressure of 1 atm, water boils at 373 K approx. The phase rule tells you that boiling temperature can only depend on external pressure. When you have a mixture, such as the two-phase problem of water and air (regard for simplicity the air is a pure substance as well) then you have to consider the composition of both phases. There will be some water vapour in the gas phase that will be given by the vapour pressure, and some air mixed in the liquid (given by Henry's law). This is the reason why in equilibrium, at any temperature below 373K you have humidity in the air and some air in the water. It is precisely this air what you observe as forming bubbles in a container when you start heating, quite before the liquid itself boils. (As an aside, the amount of CO2 in the oceans is a major component of the balance of atmospheric CO2 at a given temperature.)

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What about answering questions You know about? –  Georg May 28 '11 at 11:12
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George - Can be more clear in the areas in which he is incorrect and then perhaps offer an improved version? –  Bill Slugg May 28 '11 at 15:03
    
My answer is based on pretty standard thermodynamics without uncovering any detail. Where did I go wrong? Or maybe I was too sloppy? –  perplexity May 28 '11 at 16:59
    
Almost all is incorrect, or as eg Henrys law off topic here. The crown of obvious nonsense is water boiling at 273 K ROFL. –  Georg May 28 '11 at 17:00
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Sure, that was an obvious mistake. I have edited the answer. Henry's law (gas solubility in the liquid) is the reciprocal phenomena of evaporation: you have a mixture of both components (water and air, say nitrogen) in both phases (gas and liquid) and you need to consider this phase equilibrium in order to understand why there must be water vapour in the air in thermal equilibrium with the liquid water, although you are quite below the boiling point. That was all I was trying to explain, your great master;-) –  perplexity May 28 '11 at 17:14
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Below "boiling point" (not always 100C), water can exist in both gas and liquid phase, and has a temperature-dependent vapour pressure, which represents a point of equilibrium between liquid water wanting to evaporate and water vapour wanting to condense. When liquid water meets dry air, it is not in equilibrium; water molecules evaporate off the surface until the amount of water in the air creates enough vapour pressure to achieve equilibrium.

When water is heated to a temperature of 100C, the vapour pressure equals that of sea-level air pressure. Since the air pressure can no longer overcome the vapour pressure of the water, the water boils.

At higher elevations, air pressure is lower; as water is heated, its vapour pressure overcomes ambient air pressure at a lower temperature i.e. the boiling point is lower.

Vice-versa for higher pressures.

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Steam rising from a warm bridge is vaprization of water. Boiling water is vaporization of water. Getting cooled down by a breeze after a sweaty workout is vaporization of water. All result in the same phase change with the same latent heat of vaporization of 540 cal./gram, which is a very powerful cooling effect.

Boiling water is a subset of vaporization of water, wherein the heating of the water is fast enough that vaporization is forced to occur very rapidly AND there is enough water such that the vaporization occurs under water.

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""Boiling water is a subset of vaporization of water, wherein the heating of the water is fast enough that vaporization is forced to occur very rapidly AND there is enough water such that the vaporization occurs under water. "" This definition can be improved, very much so. :=( –  Georg May 27 '11 at 20:15
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@Georg: If it can be improved, then do so. –  wnoise May 27 '11 at 20:55
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protected by Qmechanic Mar 17 '13 at 22:10

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