Download What is the Physical Nature of “Latent Heating?

What is the Physical Nature of “Latent Heating?
Latent heating is one of the most important micro-scale processes that occurs in the atmosphere:
Heating during condensation adds buoyancy to updrafts, which strongly drives the turbulent
motions within clouds. Cooling during evaporation can induce downdrafts, build cold-pools, etc.
However, try to look up a physical description of latent heating in any textbook on the Atmospheric Sciences. Typically, you will not find a single correct description of how air warms up or
cools down under phase change processes. (Although there is one exception that I know of.) In
any case, a proper understanding of the nature of latent heating can help us in our understanding
of how cloud processes work.
Temperature:
Before considering how condensation affects heating/cooling, let’s first consider the molecular
nature of temperature: Kinetic Theory tells us that the temperature of a liquid or gas is determined
by the average kinetic energy (KE) of the molecules in the liquid or gas:
T ∝ KE ∝ v
2
N [# molecules cm-3]
N [# molecules cm-3]
Of course, the average KE is proportional to the average of the square of the velocity of the molecules.
Now, in a gas or liquid, molecules do not move with the same speed. In fact, a gas or liquid posT = cold velocities.
sess a distribution of molecular
T = hot
KEcrit
KE
KEcrit
KE
[For the case of a gas, the molecular motions are random, so the distribution of velocities is
described by a Boltzmann distribution.] In the above figure, note that the KE of the molecules is
distributed with a pronounced peak at some particular KE (many molecules have this particular
KE). Also, notice that if the gas or liquid is subjected to a heating, and therefore becomes hot, the
distribution of KE “flattens” out and there are more molecules with higher KE than in the case
when the liquid or gas was cold. [The difference between the hot and cold distributions is greatly
exaggerated for illustrative purposes.]
Now, think about solid ice or liquid. If the ice is heated, molecules will be oscillating more rapidly and more will gain enough energy to break the hydrogen bonds at the surface; hence more
sublimation (evaporation) will take place. In the above figure, more molecules have KE greater
than KEcrit in the warmer ice, so more molecules will be able to break free of the hydrogen bonds.
Depositional/Condensational Heating:
With the above description in mind, let us consider what happens during the process of condensation. [In reality, when condensation takes place, it is really net condensation. Whenever a liquid/
ice and its vapor are in contact, both evaporation and condensation are taking place simultaneously. But more about that later.] Lets consider what happens at the surface of the liquid:
Vapor
Molecule
H
Oxygen
δ+
Attraction by
Electrostatic Force
δ-
Electrostatic attaraction
of Hydrogen (in vapor H2O)
by Oxygen near surface of liquid
leads to acceleration of vapor
molecules towards surface.
Surface of
Liquid/ice
As vapor molecules approach the surface of the liquid, they are attracted by electrostatic forces.
The negatively charged (oxygen) side of the H2O molecules in the liquid attract the positively
charged hydrogen side of the vapor molecules. This attraction leads to an increase in the kinetic
energy (KE) of the vapor molecules as they approach the surface of the liquid.
Potential Energy
hydrogen bond
strength
vapor
molecule
Intermolecular
Distance
∆EH
Think about the case where you have a magnet placed close to a metal object. If we hold the magnet in place, the metal object will move towards the magnet, hence gaining kinetic energy. In fact,
the vapor molecule will gain ∆EH of energy through the attractive forces as it “falls” into a lower
potential energy state. This situation is roughly analogous to a ball rolling down a frictionless
incline. The ball will gain kinetic energy through gravitational forces instead of through electrostatic forces. Now, when the ball reaches the bottom of the hill (in this analogy) it doesn’t stop. It
has gained KE so it starts to move back uphill on the other side. The same sort of thing happens to
the vapor molecule. It has gained KE from the electrostatic forces and this KE allows the mole-
cule to continue to reduce the intermolecular distance (i.e. it “rises up” the other side of the potential energy curve). However, as the distance is reduced, the attracted molecule interacts with the
electric fields from other molecules around it in the ice. In the ice almost all of the H2O molecules
are joined together by hydrogen bonds. (In a solid, the molecules are hydrogen bonded together,
however they don’t sit still. They oscillate back-and-forth in a manner similar to that of balls connected by springs...in other words, a harmonic oscillator.) Because of this, the KE gained by the
accelerated H2O molecule is transferred to the other molecules throughout the ice. (Of course, we
could say exactly the same things for liquid.) This increase in the KE of the molecules in the ice
manifests itself as a rise in the ice temperature.
Air Temperature
Of course, we know that phase changes causes enhancements of air temperatures & we have said
that deposition (condensation) produces an increase in the temperature of the ice (liquid). Consider the fact that the molecules that make up the air are continually colliding with the ice crystal
(or water drop). During these collisions, which are largely elastic, energy is transferred between
the molecules in the ice (liquid) and molecules in the air. Now the molecules in the ice (liquid)
have more KE (and a higher temperature) than the air because deposition (condensation) is taking
place. Hence, during collisions, net KE is transferred from the molecules in the ice (liquid) to
molecules in the air (of course, this is what we call conduction). This leads to a warming of the
air.
Dispelling a Myth
It is now time to do away with an explanation of the nature of latent heating that I have seen not
only in print (e.g. Rogers and Yau’s text) but also heard from the mouths of learned colleagues.
This explanation goes something like the following:
Since molecules in the vapor have higher kinetic energy than molecules in the liquid, energy is
required to evaporate the liquid and a cooling must take place.
Of course, this is incorrect. In the first place, we know that the magnitude of latent heating is 51 kJ
mol-1 for ice. Now, a mole of gas, at 273 K, should have KEgas = NakT = 2 kJ mol-1, where Na is
Avogadro’s Number. So, the magnitude of the energy is not even close to that associated with
latent heating.
In the second place, the above myth would have to violate the conservation of energy principle.
Consider the total KE of the vapor and liquid system:
K E total =
∑ K E i, liquid + ∑ K E j, vapor
i
j
where the sum is over all of the molecules in the liquid (i) and vapor (j). If we were to simply
transfer a vapor molecule of some KE from liquid to the vapor state (without any electrostatic
forces and bond-breaking), the total KE would remain the same! Thus, the temperature of the liquid-vapor system could not change because there has been no net increase or decrease in the systems KE.
Potential Energy
Temperature Dependence of Latent Heating:
We must explain one further aspect of latent heating & this is an important one. Latent heating has
a temperature dependence. The latent heating for vaporization and sublimation both increase as
the temperature decreases (see Fig. 3-12 in Pruppacher and Klett and Table 2.1 in Rogers and
Yau). First of all, let us consider the case of ice since almost all of the H-bonds (hydrogen bonds)
have formed in ice as compared to liquid. Consider the potential well diagram for H-bonded water
below:
Intermolecular
Distance
∆EH,4
∆EH,3
∆EH,2
∆EH,1
It turns out that, as the temperature of our ice crystal is raised, the depth of the potential well
(bond strength) begins to decrease. This is due to the fact that the frequency of the oscillations
increases as energy is added to the system. (From quantum physics we know that energy goes up
as hν where ν is the frequency). Because the frequency goes up, the depth of the potential well
decreases (i.e. the bond strength decreases). This is exaggerated in the above figure which shows a
schematic of the various energy levels as temperature is increased. Hence, as the temperature is
raised, it becomes easier to remove an H2O molecule from the ice surface. Hence, the magnitude
of the latent heating decreases!
Please realized that for individual molecules, the spacing of these energy levels is discrete, as
quantum physics tells us they must be. However, because there are large numbers of molecules
participating in the phase change process, and since our typical atmospheric temperatures are relatively high, the quantum nature of this process becomes “smoothed out” on average. Hence, we
see a continuous variation of the latent heating with temperature.
Liquid vs. Ice
If you compare the variation of the latent heat of sublimation (Ls) with the latent heat of vaporization (Lv) given in Table 2.1 of Rogers and Yau, you will notice that as T decreases from 0 to -40,
Lv increases much more rapidly than does Ls. The increase in Ls was explained above. However,
now we need to explain why Lv would increase faster than Ls. The reason for this behavior has to
do with the number of bonds that are broken in liquid water & how this number varies with temperature.
If we examine Table 3.1 in Pruppacher and Klett, we see that the percentage of broken bonds in
liquid water increases as T increases. This is to be expected for reasons that we discussed in class:
Higher temperatures lead to more energetic molecules & so the chance that bonds will be broken
in the liquid increases. Herein lies the reason for the more rapid increase of Lv with decreasing T:
It is due to the fact that fewer bonds are broken as T decreases. Hence, in liquid water, Lv will
increase with decreasing T because:
(1) Bond energy itself is increasing (as Table 3.1 of Pruppacher & Klett shows)
(2) Fewer bonds remain broken in the liquid.
Think of it this way, let’s say I have some liquid water at -10C. If I evaporate all of this liquid
water, while keeping the temperature constant, I’d need to supply a certain amount of energy.
Now, let’s say I cool this water to -20C. Now, I’d need to supply a greater amount of energy if I
wanted to evaporate the water. Why? Well in the first place, the bond energies have increased. But,
in addition to that, I have fewer broken bonds at -20C than at -10C. This means that I will have to
supply an even greater amount of energy.
Please note one extra fact: As T approaches -45C, Lv approaches Ls! At this temperature, the two
latent heats become the same. (Why should this be the case?)