# About Water - Boiling and Freezing

1. Aug 16, 2006

### WhyIsItSo

Water – Boiling and Freezing
What started as a question has become my own answer. While waiting the 15 minutes or so for this site to send me my new authorization, I found my answers to…

Many years ago I watched a science show hosted by a professor called Julius Sumner-Miller, who would demonstrate some experiment then ask his standard question, “Why is it so?”. One day I watched him freeze boiling water. Fascinating stuff.

For some reason I recently started wondering what “boiling” means in scientific terms. Also, the experiment works on the premise of altering the boiling point of water through manipulating atmospheric pressure, but I wondered if the freezing point is also affected this way. Here is what I found out.

Boiling
In an article concerned with mildly superheating water (context is the dangers with heating water in a microwave oven) (Wolfe, 2005), Wolfe talks about a process called nucleation; essentially the event of a bubble of steam forming. It requires a certain amount of energy to overcome the pressure of the surrounding water to allow a bubble to form, and this is why increasing pressure makes it harder for nucleation to occur. So when we talk about the “boiling point” of water for a given pressure, nucleation is the process we are discussing.

Freezing and Pressure
What about the freezing point of water; is it affected by pressure? Turns out it is, as I suppose common sense would dictate, but then remember we are talking about a substance that breaks a few “rules”, such as expanding as its temperature continues to fall below freezing. I found a formula stated as:

dTm/dP = Tm*dV/dHm (Calder)

Calder supplies the numbers for water, and concludes this formula shows that to lower the freezing point of water by 1 degree C, one would have to apply 135 atmospheres. So, for most day-to-day applications, one could assume that variations in atmospheric pressure are not making significant changes to the freezing point of water.

Why Then Did the Experiment Work
The above were just the knots I simply had to undo for myself. If you are curious, the following describes the experiment.

The original experiment I saw used a shallow dish of water placed inside a cylinder of glass, domed on top and sitting on a steel plate with a vacuum pump attached in the center. The experiment began with the water at room temperature (say 35 C) and the pressure at ambient (say 1 Atmosphere). When the vacuum pump was turned on, the pressure inside the container fell rapidly causing two things to happen; the boiling point of the water soon fell below 35 C (and continued to fall), and the temperature inside the container fell abruptly (result of lowering pressure), thus began cooling the water. The experiment was a race between the falling boiling point, and the falling temperature of the water. The observation I made was that the water was boiling (nucleating) for a few seconds, then abruptly formed into ice; and I mean in the blink of an eye. The why is simply that by evacuating the container, the boiling point was reduced below 0 C very quickly. It took a little longer for the water to cool to 0 C, so we watched it “boil” for a short time, then it hit 0 C and bam, just like that it was frozen.

I should probably note that the freezing point was slightly increased, so the actual temperature at freezing may have been closer to 1 C than 0 C. Rather insignificant compared to the relatively large change pressure has on boiling point.

References
Calder, V. (n.d.). Freezing Water. Retrieved August 16, 2006, from Argonne National Labarotory Web site: http://www.newton.dep.anl.gov/webpages/askasci/chem00/chem00543.htm

Wolfe, J. (2005). Superheating and microwave ovens. Retrieved August 16, 2006, from University of New South Wales Web site: http://www.phys.unsw.edu.au/~jw/superheating.html

2. Aug 16, 2006

### fargoth

i think it's the boiling that takes away the heat of the water, not the drop in pressure
-well, it is the drop in pressure that makes the water boil... but the cooling effect is due to the energy lost to boiling (the molecules with higher energy get away).

the formula you must have though about was $$NT=PV$$
but the N decreases too when you use the pump.

oh, and welcome to PF.
i'm glad you could have worked most of it by yourself.

Last edited: Aug 16, 2006
3. Aug 16, 2006

### WhyIsItSo

Now you have raised an interesting point.

I would counter that the temperature inside the container did indeed drop a great deal, and I would be comfortable asserting it could easily have fallen below freezing point. The "dish" of water was very shallow; it allowed just enough depth to overcome the surface tension of the water and allow it to "pool". My point is that there was a great deal of surface area relative to the volume; cooling by energy exchange with the "atmosphere" would have been efficient. I would say then the fall in temperature contributed to cooling the water.

Nevertheless, your point is one I hadn't considered, and it appears valid. Now you have me curious all over again (and just when I thought I'd finally put this one to bed).

Therefore we need some formulae to express the energy loss by these two methods to determine which was the most significant contributor.

I have good analytical abilities, but no training, so I'm limited to what seems to make sense (not terribly reliable, I know).

Here is what I'm thinking that makes me favor my original assumption regarding heat loss. Remembering that we rapidly achieve minimal pressure in this environment (not a perfect vacuum, of course, but very low pressure indeed), therefore it requires very little energy for the water to nucleate. Wouldn't it follow then that this process, being driven by so little energy, would be capable of only minimal energy loss?

On the other hand, take a high pressure container (the gass bottle from your BBQ) and let the pressure out rapidly. That tank will get very cold, and it is not unusual to see ice form on its surface even on a hot day.

The same thing occurs if you start at ambient pressure and evacuate the air; there is a very significant drop in temperature. It would seem to me that this very large differential in temperature (water and environment) would be doing most of the work in thermal exchange.

Could you expound on what those symbols represent?

Thank you kindly.

4. Aug 16, 2006

### fargoth

the loss of heat from the water surface is done mostly by radiation, because theres almost no air in the container, so even though the surface area is big, the heat conduction is pretty small...

i'll have to do some math before i'll continue though.

i'll be back

5. Aug 16, 2006

### WhyIsItSo

fargoth,

I'm researching heat loss by radiation. Most of the formulae I'm finding are based on Kelvin temperatures. I'm working on a start temperature of 308.15K (35C) down to 273.15K (0C).

I've got numbers to find values for as yet. For example, the emissivity of water (a number between 1 and 0). I'll present my findings when I'm done.

6. Aug 16, 2006

### WhyIsItSo

Volume of water... Let's see, this is a 30-35 year old memory, but I suspect it was probably about 20CC.

Surface area was perhaps 16cm^2.

7. Aug 16, 2006

### WhyIsItSo

Ok, here is what I have. I've made some approximations; necessarily so since I do not know the exact conditions of the experiment I saw on TV so long ago.

I also must apologize I do not know how to input a formula, nor could I find a guide for doing so here. If, as an aside, you could point me in the right direction. Or is it Latex? I think I have a page of codes for that somewhere. Anyway...

I'm using the Stefan-Boltzmann Law from http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

I'm probably going to get torn to pieces over this because I'm mostly omitting units... it's just to ugly to try to recreate here without better formatting skills *gulp*

Radiative Cooling time is given as:

Nk 1 1
----- [ -------- - --------]
2eoA Tfinal^3 Thot^3

Where
N = number of particles = 1 x 10^23 (approx)
k = Boltzmann Constant = 1.38066x10^-23
e = emissivity = 0.95 (approx for water)
0 = Stefan's constant = 5.6703x10^-8
A = surface area m^2 = 0.0016
Tfinal = final temp (K) = 273.15
Thot = start temp (K) = 308.15

Well, I plugged all that in and come up with 120 seconds. Now, I started at 35C for the water. That's probably way too high. Taking 22C instead (295.15K) yields 38 seconds.

After all that work, of course, there are probably some who are laughing their heads off at me (the moron) who has no idea what he's talking about. Very true, I don't. I just did some research and this law seemed to fit what I was looking for. Even to my ignorant eye, there seems to be something missing from all this. Even so, the numbers I got out of this are about what I expected, so maybe I'm not completely ignorant :)

I shall leave it to my betters to set me straight.

8. Aug 16, 2006

### WhyIsItSo

Ok, that formula didn't come out right; the formatting is messed up. Here is a different way of expressing it:

Nk / 2eoA (1 / Tfinal^3 - 1 / Thot^3)

Let's hope you can make sense out of that.

9. Aug 16, 2006

### 3trQN

Is this the case?:
When the vaccum pump reduces the pressure in the bell-jar, it shifts the equillibrium of the liquid/vapour phse equillibrium and more liquid evaporates (that is vaporises at a temperature below its boiling point) to re-establish equillibrium.

You can think of this as an inversion of a ball on a table. A ball on a table has a potential energy in the gravitational field. It will allways move to a lower gravitational potential energy unless its opposed by the table, there is an equillibrium of forces between the tabled and the ball.

Now remove the table and the ball falls. Its like that for the liquid phase molecules.

There is an equillibrium of the vapour and liquid at the vapour pressure, the rapid evaporation occurs when this equillibrium is disturbed and lots of heat is lost from the liquid phase into the gas phase ( as the kinetic energy of the gas phase molecules escaping the liquid).

This results in a rapid cooling of the liquid, which would, i think, be enough to freeze it.

Thermodynamically, you could say that the energy removed from a system in equillibrium like that is redistributed about the system, and as such, part of it is cooled. Like taking a half of a cup of hot water out of a full one, it has less entropy.

Im afraid my maths is too poor to deal with the numbers though. What are you calculating?

10. Aug 16, 2006

### WhyIsItSo

fargoth challenged my assumption about the reason the water cools. He states that it is due to evaporation. I assumed it was due to radiative cooling.

All that math I just threw in here was trying to support my assertion that the cold temperature inside the container is sufficient to rapidly freeze the shallow crucible of water.

fargoth is pursuing numbers to show heat loss through the boiling of the water.

Almost; lowering the pressure lowers the boiling point. The water is still vaporizing at its boiling point, that temperature just happens to be lowered by the vacuum.

11. Aug 16, 2006

### fargoth

hi, i had some friends over, so i haven't done anything yet...
i'll do it after a good night sleep... i'll post my conclusions as soon as im done...

i haven't checked your furmula, but if you don't really understand how they got this formula, i suggest reading some thermodynamics books... some times you must make some values constant to get it, and make some assumptions that don't really fit into every situation... so i advise against googling formulas up.

anyway, about writing... it's latex, put you text inside tex field -[te*]x^2+\frac{1}{x}=\sqrt{x}+e^{x+3}[/te*] where te* is tex
so you'd get $$x^2+\frac{1}{x}=\sqrt{x}+e^{x+3}$$

you can just click on the formula to see how it was done...

goodnight

12. Aug 16, 2006

### 3trQN

Now i ask, why does a molecule in liquid phase suddenly get enough kinetic energy to escape the liquid?

Last edited: Aug 16, 2006
13. Aug 16, 2006

### WhyIsItSo

Ok, let me try my latex skills on that formula...

$$t_{cooling}={Nk\over 2\varepsilon\sigma A}\; \left [\frac{1}{T_{final} ^3}-\frac{1}{T_{hot} ^3}\right ]$$
where:
$$t_{cooling}=$$ time to cool (in seconds)

$$N\approx1*10^{23}$$ Number of particles

$$k=1.38066*10^{-23}$$ Boltzmann's Constant

$$\varepsilon\approx0.95$$ Emissivity of water (distilled)

$$\sigma=5.6703*10^{-8}$$ Stefan's Constant

$$A=0.0016m^2$$ Surface area

$$T_{final}=273.15$$ Final Temp Kelvin

$$T_{hot}=308.15$$ Start Temp Kelvin

Now that looks a little better.

Last edited: Aug 16, 2006
14. Aug 17, 2006

### fargoth

good morning, though i haven't finished my part of the work, i had to look at yours.

I did a little experiment, to see if it makes any sense... i took a measuring cup, filled it with about 5cc, and put it in the fridge (about 269K).
it took the water about 8 minutes to freeze, and not all of it froze.

so, the assumption that the water's temperature is homogenous isn't correct, because it has less then ideal heat conductivity - thats why the "deeper" water didn't freeze yet.

and ignoring the ambient temperature is true for the sun for example, but you can't do it if the ambient and "hot" temperatures are close, as it is in this case.
in our case, there is radiation from the outside that gets absorbed in the water, its not only the water which radiates.

in my experiment there was atmosphere, which means that in your experiment these water would cool slower, the lack of atmosphere slows the cooling process because the heat conduction is done only by radiation and not with the help of the surrounding gas.

so, even though i haven't done my calculations for the loss of temperature due to boiling, i can say that it wasn't the enviorement's temperature that caused the water in your experiment to freeze, as they froze much faster then the water in my experiment.

i'll continue to develop the formula for heat loss due to boiling just out of curiousity, but i think the conclusions can be made right now...

15. Aug 17, 2006

### WhyIsItSo

Ask the question the other way around. How does a molecule in a liquid with some amount of energy suddenly find that energy is enough to escape the liquid.

Answer: By changing the condition that prevented that molecule from escaping before. In other words, pressure is what was preventing the liquid from nucleating before, but we drastically reduce the pressure when we run the experiment, so that same energy is now enough for the liquid to nucleate (boil).

16. Aug 17, 2006

### WhyIsItSo

First I'll acknowledge that Boltzmann's Law does tend to yield slightly shorter cooling times than actual observation would measure; but not drastically so.

That said, I embarked on this study of Radiative Cooling for the very reason that you rightly pointed out that the "experiment" creates a near vacuum, therefore radiative was the only realistic scenario for cooling; notwithstanding whatever you come up with for the evaporative cooling.

Also, your experiment has some significant deviations from the original. For one thing, the container (measuring cup) would have a much greater capacity to hold heat than the container in the original; a very shallow metal dish about 1mm thick. Your container provided insulation.
You use the presence of gasses in your freezer to argue they were assisting cooling; I should think a vacuum would be ideal for the transmission of photons, however, so would not Radiative Cooling be more efficient then?
My calculations do in fact include the ambient. I'll post separately to show the full math...
I'm very curious to see what you come up with.

17. Aug 17, 2006

### WhyIsItSo

To address the issue of Ambient Heat you raised. The following math is taken from:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html

The Stefan-Boltzmann law.

$$P=\frac{dE}{dt}=\varepsilon\sigma A\left(T_{hot} ^4 - T_{ambient} ^4\right)$$

where

$$\sigma=5.6703*10^{-8}\frac{watt}{m^2K^4}$$
$$\varepsilon=0.95$$ emissivity

I did make some quick calculations for P (power emitted) and E (energy) and got

$$P\approx 3.1299442*10^{-1} watts$$
$$E\approx 6.816641*10^{-6}$$

The important issue here is that $$T_{ambient} ^4$$ would be dropped from the equation for extremely hot objects such as the Sun; I did not do this.

Continuing the math... Here is where you might take me to task, because I use equipartition of energy to say...

$$E = N\frac{3}{2}kT$$

where $$N \approx 1.2612447 * 10^{23}$$ number of particles

Using the chain rule for differentiation

$$\frac{dE}{dt} = \frac{dE}{dT}\frac{dT}{dt} = \frac{3}{2}Nk\frac{dT}{dt} = \varepsilon\sigma AT_hot ^4$$

Rearranging gives us

$$dt = \frac{3Nk}{2\varepsilon\sigma AT_hot ^4}dT$$

And integrating gives the cooling time

$$t_{cooling} = \frac{-3Nk}{2\varepsilon\sigma A} \int_{T_{hot}} ^{T_{final}} \frac{1}{T^4}dT = \frac{Nk}{2\varepsilon\sigma A} \left [\frac{1}{T_{final} ^3} - \frac{1}{T_{hot ^3}}\right ]$$

substituting

$$= \frac{1.2612447*10^{23} * 1.38066*10^{-23}}{2 * 0.95 * 5.6703*10^{-8} * 0.0016} \left [\frac{1}{273.15^3} - \frac{1}{295.15^3}\right ]$$
note I substituted 22C for the "hot" temperature.

$$= \frac{1.7413501}{1.7237*10^{-10}} \left [4.9067*10^{-8} - 3.8893*10^{-8}\right ]$$

$$= 1.0102*10^{10} * 1.0174*10^{-8}$$

$$= 102.77775$$

So, I arrive at my assertion that the experiment could easily freeze the water in about 103 seconds. Of course, I've no idea what I'm talking about, but damn it all looks good!

Last edited: Aug 17, 2006
18. Aug 17, 2006

### fargoth

well, actually, you have to make a distinction between boiling and evaporating.
while nucleation is easier under lesser pressure, water molecules leave and enter the surface spontaneously, if you have less water molecules entering then exiting the water liquid phase, the water evaporate.

the force which keeps water in it's liquid phase is the inter-molecular electric attraction, not the outside pressure.
now and then probability dictates that a molecule with enough energy would get to the surface and escape... if you have little water vapor (i.e. the air is very dry) the water would evaporate quicly regardless of the pressure.

19. Aug 17, 2006