Why is it so difficult to cool below the dewpoint?

One of the first rules of thumb that novice weather forecasters learn (or, at least, one of the first rules of thumb that I learned) was that when forecasting the overnight low temperature, it should never be more than a degree or two cooler than the dewpoint temperature.  This rule of thumb applies very broadly--it's very hard to cool temperatures down below the dewpoint temperature.  Why is this?

Let's go back to what the dewpoint temperature actually represents.  In a basic sense, it's the temperature that air with a given water vapor content would have to cool to before it becomes saturated.  What does it mean for air to become "saturated"?

Saturation is often described in terms of the amount of water vapor that the air can "hold" at a given temperature and pressure.  While this serves as a fairly good analogy, it's not exactly the way things work.  Air can "hold" as much water vapor as it needs to--it's the rates of evaporation and condensation that define saturation.

In a given parcel of air, there's always evaporation and condensation going on concurrently.  When a parcel of air is not yet saturated (it's temperature is greater than the dewpoint temperature), the rate of evaporation of water into the parcel is greater than the rate of condensation of water out of the parcel.  As such, there is the possibility of a net gain of water vapor into the parcel, assuming that there is water present to evaporate into the parcel.  If there's no real liquid water around to evaporate into the parcel of air, then obviously it won't be constantly gaining water vapor.  But if the parcel is unsaturated, then the potential exists for more water to be evaporated into the parcel than would condense out.

As the parcel comes close to this "saturation" level (for example, by the temperature cooling to the dewpoint temperature), the rate of condensation out of the parcel increases.  Think of it this way--the more water molecules we're adding to the parcel, the greater the probability than molecules will condense out of the vapor phase to form liquid.  When we reach the saturation level, the rate of evaporation into the parcel exactly equals the rate of condensation out of the parcel.  This is why we think of saturation as a "limit" describing how much water the parcel can "hold"--once it reaches saturation, there is no net gain of water vapor into (or out of) the parcel. The evaporation into the parcel and the condensation out of the parcel are equal.

Finally, if we were to somehow get beyond this saturation level ("supersaturation"), by, for example, somehow cooling the parcel below its dewpoint, then the rate of condensation out of the parcel would become greater than the rate of evaporation into the parcel.  There is a net loss of water vapor from the parcel until enough water vapor has been lost so that the rates of evaporation and condensation are once again equal (we lose water until we get back to saturation).

I tried to sum up these three situations in the diagram below.

So what does this have to do with cooling below the dewpoint temperature?  A lot. It turns out that when water condenses from water vapor into liquid water, it releases a large amount of heat.  In fact, experimentally we know that water releases 2.5x10^6 Joules of heat energy for every kilogram that condenses from vapor to liquid.  This is called the Latent Heat of Vaporization (or Condensation).  Think about it this way--water vapor molecules are in a gaseous phase--they move around much more quickly than water molecules in their liquid phase.  So, to transition from fast-moving gas molecules to slower-moving liquid molecules, some energy has to be lost.  This energy is lost in the form of radiating heat.

So let's do a simple experiment here.  Let's say that it's a cool spring evening and we've cooled down overnight until our temperature has reached the dewpoint temperature which happens to be 10 degrees Celsius (50 degrees Fahrenheit).  Since we've cooled to our dewpoint temperature, we know that the air is saturated.  So, any further cooling is going to make the air supersaturated, and condensation out will dominate over evaporation into the air until enough water vapor has been lost to bring the air back to saturation.

We can use a thermodynamic chart to find just what mass of water vapor air at that temperature has in it at saturation.  We'll assume that our pressure is 1000mb (a rather typical surface pressure).  I apologize for the poor quality of these charts--it's hard to find a high-resolution thermodynamic diagram online.  I've exaggerated the drawing a bit so it's hard to match up the saturation mixing ratio lines (which tell us the mass of water vapor in the air).  You'll have to trust my numbers here...

Stuve thermodynamic diagram showing the initial pressure, temperature and mixing ratio of water vapor.  The blue vertical lines are temperature contours, the horizontal blue lines are pressure contours (the red dot is on the 1000mb line) and the brown, dashed, tilted lines are mixing ratio contours.  The red dot lies between the 5.0 and 10.0 mixing ratio contours at a value of about 8 g/kg.

We can see that at 1000 mb and 10 degrees Celsius, the "saturation mixing ratio" is about 8 grams per kilogram.  That means, if the air is saturated at 10 degrees Celsius and 1000 mb, then there are 10 grams of water vapor per every kilogram of air.

Now, let's say that we try to cool the air by 2 degrees to 8 degrees Celsius (about 46 degrees Fahrenheit).  If we cool the air, the saturation mixing ratio is going to change.  Let's return to the chart...

Same as previous diagram, but now showing the new position of the cooled air.  The pressure line is the same (1000mb), but the temperature has dropped from 10 Celsius to 8 Celsius and consequently the saturation mixing ratio has dropped from 8 g/kg to 7 g/kg.

We see that at 8 degrees Celsius, the saturation mixing ratio has dropped to 7 grams per kilogram.  Remember, if we cool the parcel below its dewpoint, it becomes supersaturated, which means condensation dominates.  That condensation will continue to dominate until it has removed enough water vapor to bring the air back into saturation again.  In this case, the amount of water that will have to be condensed out to return to saturation is 8 - 7 = 1 gram of water (per kilogram of air).  So if we try to cool the air two degrees Celsius below its dewpoint, we'll condense out 1 gram of water vapor (per kilogram of air).

But, as I mentioned before, when water vapor changes from gas to liquid phase it releases latent heat.  We can calculate how many Joules of heat evaporating one gram of water will release (1 gram = .001 kilograms) by using that latent heat of vaporization (Lv) that I mentioned earlier.

So just evaporating that one gram of water releases some 2,500 Joules of heat.  That heat energy is going to go toward warming the air around the condensing water.  We can use a form of the first law of thermodynamics to calculate just how much the air will warm.

We'll assume here that the second term on the right hand side involving changes in pressure (delta-P) equals zero.  All that's saying is that we're assuming all of the released heat goes to warming the air and none of it does work on the air by changing its pressure and volume.  Remember on our thermodynamic diagram we stayed at the same pressure when we tried to cool the air--so the change in pressure (delta-P) was zero.

The Cp in this equation is the specific heat capacity of air.  For perfectly dry air this works out to about 1004 Joules per kilogram of air per degree Kelvin (Celsius).  Since this isn't perfectly dry air (it has 10 g/kg of water vapor in it, after all...), that number is actually adjusted a bit--it works out to be around 1012 Joules per kilogram of air per degree Kelvin (Celsius).  We know the amount of heat added (delta-Q) on the left hand side--that's the 2,500 Joules we're getting from the condensation of the water vapor.  We also know that since our mixing ratios were defined as grams per kilogram of air, the mass of air we're concerned with is just 1kg.  Knowing all this, we can solve for the change in temperature (delta-T):

Our change in temperature works out to be about 2.5 degrees Celsius of warming.  But remember--we only cooled the parcel by 2 degrees from its dewpoint!  So, the evaporation of the excess water vapor generated by cooling the air 2 degrees actually releases enough heat to warm the air up by 2.5 degrees--that's a net warming of the air!

It should hopefully be clear now why it is so difficult to cool the air below its dewpoint.  As soon as you cool below the dewpoint, water vapor condenses out faster than it is evaporated in.  This excess condensation releases latent heat into the air, which in turn warms the air enough to compensate for the cooling you tried to do in the first place.  It's pretty much a hopeless endeavour--you try to cool, but the condensation that results cancels out all cooling.  This is why the temperature rarely drops more than a degree or two below the dewpoint.

This is a handy tool to have when trying to make a forecast for the low temperature overnight.  Its more or less an absolute minimum possible temperature that could happen (assuming no fronts come through or there are other radical changes to the air mass).  Of course, if you do reach the dewpoint, all that condensation of water vapor as the air keeps trying to cool results in lots of little water droplets--and you get fog.  In fact, on mornings if you wake up and there's fog outside, you know that the air cooled to the dewpoint overnight--and is still at the dewpoint right then.

Looking even more into what this means, note how just a small amount of water vapor--just one gram in a whole kilogram of air--released enough energy to warm the air by 2.5 degrees Celsius.  This is why warm, moist air is so important when looking at the formation of thunderstorms.  The heat capacity of water is pretty incredible, and it provides that extra fuel that makes storms so violent.