# About Water - Boiling and Freezing

• WhyIsItSo
In summary, the conversation discusses the scientific terms and processes involved in boiling and freezing water, particularly in relation to pressure and atmospheric changes. The experiment of freezing boiling water is also described, with the conclusion that both pressure and energy loss contribute to the cooling of the water.
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

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.

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fargoth said:
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).
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.

the formula you must have though about was $$NT=PV$$
but the N decreases too when you use the pump.
Could you expound on what those symbols represent?

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

the loss of heat from the water surface is done mostly by radiation, because there's 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

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.

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.

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.

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.

Is this the case?:
When the vacuum 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?

3trQN said:
Im afraid my maths is too poor to deal with the numbers though. What are you calculating?

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.

When the vacuum 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.
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.

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 I am 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

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

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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.

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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 - that's 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...

3trQN said:
Now i ask, why does a molecule in liquid phase suddenly get enough kinetic energy to escape the liquid?
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).

fargoth said:
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). the water had a surface area of about 15cm^2.
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 - that's why the "deeper" water didn't freeze yet.
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.
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.
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?
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.
My calculations do in fact include the ambient. I'll post separately to show the full math...
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...
I'm very curious to see what you come up with.

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!

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WhyIsItSo said:
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).

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.

fargoth said:
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.
...Would you not agree that boiling the water greatly increases that probability? And if we boil it by reducing pressure rather than inputting more energy...

Just a thought.

Congratulations! You've ignored heat capacity, enthalpies of fusion and vaporization, misapplied Stefan-Boltzman, and generally made a mockery of established physics.

Start at the beginning: state your assumptions one at a time for people to correct; state the approach you're going to take toward calculations based on the corrected assumptions, again for corrections; then proceed with the calculations one step at a time for corrections.

Bystander said:
Congratulations! You've ignored heat capacity, enthalpies of fusion and vaporization, misapplied Stefan-Boltzman, and generally made a mockery of established physics.
OUCH! That's going to smart for a while :) Still, I'm willing to learn, and this issue has bugged me for most of my life, so...

Start at the beginning: state your assumptions one at a time for people to correct; state the approach you're going to take toward calculations based on the corrected assumptions, again for corrections; then proceed with the calculations one step at a time for corrections.
Scenario
An experiment to demonstrate that water can be transformed from a boiling state to a frozen state in what appears to the naked eye to be an instantaneous transformation. The water is at ambient temperature and is made to nucleate (boil) by the drastic and rapid reduction in pressure within the "chamber" such that the boiling point of the water is kept below its lowering temperature.

It is by observation that the water boils for a short (about 1 minute) period of time, then apparently instantly freezes.

The Issue
By what mechanism(s) does the water freeze.

Assumptions
1. My assumption has always been that the chamber is greatly cooled as a result in the extreme drop of pressure.

1a. Therefore the water is cooled to freezing by this environment.

2. Fargoth asserted that the (near-) vacuum indicated that only Radiative Cooling could explain my assumption, and that this would not be sufficient for the time alloted.

2a. Fargoth assumes cooling through boiling is the mechanism by which the water reaches freezing point.

As this is my contender for the primary mechanism for cooling the water, I endeavored to show mathematically that this was sufficient. Apparently I've made a mockery of established physics. That doesn't surprise me since I've said several times that it didn't feel right to me. I don't "know" why it is wrong, but something in the back of my mind is telling me the math I used is nonsense. So, correct away!

Heat Capacity​
From http://www.ausetute.com.au/heatcapa.html I have for water:

$$C_g = 4.18 JC^{-1}g^{-1}$$

where $$C_g$$ is Specific Heat Energy.

and the formula for calculating heat energy gained or lost...

$$q = m * C_g * (T_f - T_i)$$

where
q = heat energy (joules)
m = mass (grams)
$$T_f$$ = final temerpature (Celsius)
$$T_i$$ = initial temperature (Celsius)

Since I am using 20cc of water, which weighs 20grams, and I want to drop the water temp from 22C to 0C...

$$q = 20g * 4.18JC^{-1}g^{-1} * \left (0C - 22C\right ) = -1839.2J$$

So the water loses $$1.8392*10^3 Joules$$ in this process. Now I need to calculate how long that will take. OK so far?

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Doing fine.

Phase Change​
Actually, I'm not finished with energy yet.

To reach water temp of 0C we must lose 1839.2J

From http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html#c1

...it takes 334J/g to transform water at 0C to ice at 0C. Since I have 20g of water, that's an additional 6680J !

In round numbers, I actually need to find out how long it takes to lose 8520J

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

but when you integrate don't drop the $$T_{ambient}$$ (which you did... it actually becomes quite hard to solve this integral when you keep it, guess that's why they dropped it.

if you want to be more realistic, you need to add the fact that the water below the surface transmit their heat due to heat conductivity, and it's only the surface water which lose the heat through radiation.

oh, and as for you argument that when you have atmosphere you lose less heat to radiation - it isn't true, it doesn't matter if the radiation is absorbed in the atmosphere, the only thing that matters is that it transferred the energy from the water.
theres just a second mechanism that carries the heat when there's atmosphere...

now, i can't seem to estimate the rate in which the water evaporate in the boiling process... the boiling process happens due to imperfections (seeds), and it depends on too many things... evaporation through surface is easy enough, but boiling...

In Stefan's constant:

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

What does the "K" represent? My assumption is that it is Kelvin.

If that is so, then I don't see how to complete the math in the Stefan-Boltzmann law:

$$P = \varepsilon\sigma A \left (T^4 - T_C ^4\right )$$

P = power, which means some unit of energy per time, such as "watt" or "J/s". Since Stefan's constant introduces "watt", I'm going to need to cancel out all other units.

Also, I can see that the temperatures in this formula must be expressed in Kelvin since they must be positive values; the difference between these values raised to the power of 4 is a key factor, and two values of say -20C and 20C would cancel out to zero once the math inside the parentheses was done.

If I let $$am^2$$ represent A, and expand Stefan's constant than rewrite the formula:

$$P=\frac{\varepsilon\ 5.6703*10^{-8}watt\,am^2\left (T^4 - T_C ^4 \right )}{m^2K^4}$$

$$\varepsilon$$ has no units. I can cancel out $$m^2$$, so now I see that I must somehow be left with the unit $$K^4$$ in the denominator. I don't know how to manipulate this to achieve that. If my radiator is 308K and surroundings are 273K, then do I write and manipulate as:

$$\left (308K\right )^4 - \left (273K\right )^4 \\ = 308^4K^4 - 273^4K^4 \\ =3.4445*10^{9}K^4$$

Is that valid math?

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Gettin' ahead of yourself. Before going through the calculation, STATE the assumptions you're making about radiative heat transfer --- you are making a couple, and they're getting you into trouble.

Give you a hint --- when's the last time you set your ice cream in the sun to freeze it?

Bystander said:
Gettin' ahead of yourself. Before going through the calculation, STATE the assumptions you're making about radiative heat transfer --- you are making a couple, and they're getting you into trouble.

Give you a hint --- when's the last time you set your ice cream in the sun to freeze it?
I understand. I think however that remembering I need to lose 42.6J/s to reach the required temperature by radiative cooling only within 200 seconds, I have already shown that this is insufficient.

In an ideal environment where there is no input of radiation, the best initial rate of power output, even at 0K, is barely over 1J/s. This is an absolute best case (actually, and unrealistic case) and is still way below what would be needed for that water to freeze via radiative cooling only.

I believe I can already conclude that radiative cooling is NOT the primary mechanism that freezes the water.

I am suspecting now that fargoth is on the right track. Enthalpy of vaporization is looking like a much better explanation.

$$H=U+PV$$ gives the formula for enathalpy. I'm clear on how to proceed with this, however.

The experiment induces a large $$\Delta P$$, reducing it significantly and rapidly. I would imagine if we plot the P/t curve, it would be an asymptote.

Assumptions
1. Since I am only interested in showing if enthalpy of vaporization is sufficient explanation for the loss of energy that results in freezing the water, I am also satified with gross calculations, therefore I will treat delta P as a constant for the duration of the experiment.

2. There may be phenomena I am not aware of that would have unexpected results due to significant and instantaneous loss of pressure. Since this is not actually the case (as in assumption 1) any such results will be overlooked.

Procedure
Show that enthalpy of vaporization can explain the loss of 8520J in approximately 200 seconds or less.

Observation
It was clear from the experiment that the water did in fact boil, and did in fact freeze after a short period of time. I do not remember the exact time it took, but an upper limit of 200 seconds has been set. So the water went from equilibrium to boiling to frozen within this period.

To prepare for testing this theory, I'll need to get some realistic numbers about vacuum pumps; I need to have some reasonable number for the pressure inside the experiment. This will be needed for determinging the work done in the form $$W=P\Delta V$$

How does this look so far?

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Hmmm. Seems LaTeX is still broken.

as you know, i already started going in this direction, but couldn't finish, because as i said, i can't estimate how many bubble are being created in the boiling process and what is their size)

i can give you some tips though:
the energy lost from the boiling process is the number of molecules that vaporized times the energy of the inter-molecular attraction.

the number of molecules with sufficiant energy to break free is an integral over this:
$$Ce^{-\frac{E}{kT}}$$
from E=W to infinity (where W is the energy of the bond).
and C is determind by integrating over all velocities and saying its equal to the total number of molecules.
normaly we could just say we're in equilibrium when $$\frac{n_{gas}}{n_{liguid}}=e^{-\frac{W}{kT}}$$
because some molecules will get back and "stick" to the water, while others will escape, but we constantly get the gas away, so we're never in equilibrium and no molecules are getting to the water state.now, we have to estimate how many of the molecules that can escape actually form the bubbles or are close enough to the surface to get away...

without bubbles, we can count only these who are in a distance in the length of a molecule from the surface, and say the rate of escape is their velocity devided by that distance distance.
we'll have to integrate over the velocities which are higher then the escape velocity.

anyway, it's pretty easy, but that would just be the cooling effect of sweating, or blowing over a hot cup of tea... in boiling there are the bubbles which let much more of these high energy molecules escape...

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Want an upper limit on time? What's the vapor pressure of water at the triple point? What's Langmuir (F/A = 1/A(d(mv)/dt)) say about evaporation rate? Laboratory vacuum pumps handle tenths of liters per sec to maybe 10 liters for "large" units. For "reconstructing" the demonstration in the OP, stick with 1-3 dm3/s.

OMG, it just hit me.

Thanks to Bystander, I went and read about Phase changes. I was explaining the concept of "boiling" to my wife, and in that process it hit me like a brick on the head.

When you boil a pot of water, it is at 212F more or less. If you set the stove to provide just enough heat to keep it boiling, it is at 212F. If you turn the heat way up THE ENERGY GOES TO NUCLEATING and the temperature of the water IS STILL 212F!

The water will be boiling up all over the place and making my wife really angry with me, but during the phase change the temperature of the water is still 212.

Since by observation the water is part liquid, part gas, it is by definition in this Phase change state.

This does not answer how much energy is lost by radiation, convection, transference, etc., but I think it is rather apparent empirically that the energy going to nucleation is the gross component, and since the resultant vapor is released from the environment (the pot of water/the crucible in my experiment) this energy is thus lost to the "environment".

By observation, ice melts rather slowly in an ambient environment. It is being heated by various source; ambient temperature of the environment gas, radiation, etc. Since the experiment is performed in an environment approaching a vaccum, it follows that external heating processes are minimized. All I can think of is radiation from light sources. A vacuum is therefore a good insulator (gee, I guess that's why people make vacuum flasks, yeah?). Not perfect, but I am taking the partial vacuum of the experiment as sufficient so as to make temperature loss or gain from external factors to be negligible.

Continuing, I had assumed that the environment within the glass container would get very cold due to the loss of pressure. It does, no doubt, but then since we remove most of this atmosphere, this factor becomes negligible.

Water is not compressable, therefore the loss of pressure does not cool it in the same way it does for a gass.

This is going to fun without LaTeX...

Let B = boiling point for water. 212F = 373K (close enough). I believe standard atmospheric pressure is about 1.29Kg/m^3.

B = 373 / 1.29 = 289. Without exploring what happens as P -> 0, we can see that for water to be boiling, B = 289. If B < 289, it is liquid (or solid). If B > 289, it is gas.

So, the experiment starts with water at about 295K (22C).
295 / 1.29 = 229 < 289 t.f. is not boiling.
However, merely halve the pressure...
295 / 0.645 = 457 > 289 t.f. the water is now boiling - vigorously.

Enthalpy is given as E = U + PV. It so happens that the enthalpy of 1 mole of water (liquid) at 1 atmosphere at 298K is -285.83. What happens when we halve the pressure? In most cases, the volume will double (and this is an example of "work"), but liquid water is not compressable! Over time, the enthalpy will change, but what happens immediately? Since V does not change, U (internal energy) must!

That suggests U must increase, yet we know from observation that the reverse is true. Where is this energy going?

It goes into the work of driving the phase change of water from liquid to gas. Relative to the energy needed to change the temperature by one degree, this is a massive amount of energy.

When water changes from liguid to gas, it increases volume approximately 1,600 times. This does not relate directly to the quantity V above, but there is a correlation - I just need to figure out what it is.

Here is what I think truly matters.

For water, it takes about 1 calorie to cool 1 gram of liquid 1 degree. It takes about 539 calories to convert 1 gram of liquid to gas. Therefore, for every gram of gas released, we have dissipated enough energy to cool 539 grams of liquid by 1 degree.

Let me back up a moment.

Assumption:
1. 1 gram (approx) of liquid is expended by nucleation in time t. This uses 539 calories.
2. Leaving 19 grams in liquid form, to cool this from 22C to 0C requires 418 calories be lost.
3. To phase change 19 grams from liquid to solid requires 1,514 calories be lost.
4. Loss of another 3 grams via nucleation expends 1,617 calories.
5. Mass of liquid would have been reduced to 16 grams by this process.
6. Phase change liquid to solid of 16 grams requires 1,275 calories.
7. Therefore somewhere between the loss of 3 and 4 grams of liquid due to nucleation / phase change liquid to gas, sufficient energy has left the "system" to result in solid form (ice).

So, now I need to work out the math to test this. If my reasoning so far is valid, then I need only discover the rate of this conversion, and thence the time it takes.

well, I've been telling you the same things only in different words, i guess you had to re-read it elsewhere and take the time for it to take its roots..

anyway, I am happy you got it

maybe you'd even come up with some simple way to get the rate as a function of temperature and pressure...

WhyIsItSo said:
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).

Ahh great reply, thanks for pointing that out.

3trQN said:
Ahh great reply, thanks for pointing that out.

no no, it isn't true... i mean, the nucleation condiotions do depend on pressure, but not the ability of water to leave the water surface...

some molecules have more energy then avarage, and if they are near the surface they can escape - pressure has nothing to do with it, the force that keeps the molecules together in liquid phase is their attraction, not pressure, and you don't change this force.
low pressure would just make it easier for high energy molecules to form bubbles, and allow them to escape without being close to the surface.

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