About Water - Boiling and Freezing

[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

 

[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.

 

[WhyIsItSo]

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.

 

[fargoth]

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

 

[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.

 

[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.

 

[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.

 

[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.

 

[3trQN]

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?

 

[WhyIsItSo]

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.

 

[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 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

 

[3trQN]

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

 

[WhyIsItSo]

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

<br /> t_{cooling}={Nk\over 2\varepsilon\sigma A}\; \left [\frac{1}{T_{final} ^3}-\frac{1}{T_{hot} ^3}\right ]<br />
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.

 

[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).
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.

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...

 

[WhyIsItSo]

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).

 

[WhyIsItSo]

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.

 

[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!

 

[fargoth]

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.

 

[WhyIsItSo]

the force which keeps water in it's liquid phase is the inter-molecular electric attraction, not the outside pressure.

I follow you, but...

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.

 

[Bystander]

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.

 

[WhyIsItSo]

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.

Approach - Radiative Cooling
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!

 

[WhyIsItSo]

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?

 

[Bystander]

Doing fine.

 

[WhyIsItSo]

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

 

[fargoth]

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...

 

[WhyIsItSo]

I Need Help Please!

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 \\<br /> = 308^4K^4 - 273^4K^4 \\<br /> =3.4445*10^{9}K^4

Is that valid math?

 

[Bystander]

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?

 

[WhyIsItSo]

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?

 

[WhyIsItSo]

Hmmm. Seems LaTeX is still broken.

 

[fargoth]

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...

 

[Bystander]

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.

 

[WhyIsItSo]

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.

 

[fargoth]

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...

 

[3trQN]

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.

 

[fargoth]

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.

 

[WhyIsItSo]

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..

Not re-read, learn it for the first time. Still, it did take me several days to absorb enough to "see" the problem. I thank Bystander for his scathing criticism; he sent me down the right track. You had started there, of course, but I had no frame of reference to get from where I was (no real knowledge) to your contention. I had to brush up (a lot) on my algebra, and Thermodynamics is something I barely touched back in High School (some 20-odd years ago).

I know it's taken a while for me to get here, but in my defense I've also been reading/learning Mechanics, a few issues of Quantum Physics, and learning to build a Linux box with Sun Application Server (I'm trying to build a Distributed Architecture system). Note I know almost nothing about Linux.

At my ripe old age of ?? I'm doing a BS in Software Engineering. I'm also studying towards Java certification.

I have a lot going on :)

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

Indeed, but I think I'm going to have to learn those integrals and limits Calculus equations. "Simple" algebra won't cut it, unless maybe I find a neat formula somewhere. But then, as you've noticed, and as my screen name suggests, I like to know "why" things work the way they do. As the saying goes, "Give me a fish, I eat for a day. Teach me to fish, I eat for life".

Watch this space!

 

[Bystander]

Sorta getting back to mainstream physics --- few points, vocabulary notes for people to tidy things up a bit:

"nucleation" is NOT a fancy word for "boiling;"

vapor pressure can be approximated over short temperature ranges as lnP = A(1/T) + B, not

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...,

whatever this is;

"free expansion" of an ideal gas results in a zero temperature change, and air is nearly ideal at ambient conditions --- pumping down the bell jar does not result in any sensible cooling effect;

"boiling" is a "superheat" phenomena, in which liquid is NOT at uniform temperature, and liquid below the surface is at a pressure that is higher by "rho^.g^.h," and reaches a temperature sufficient that the vapor pressure is equal to or greater than the higher pressure at depth, and vapor bubbles form from "nucleation sites, then rise."​

I'll applaud your enthusiasm for learning, and caution you to examine detail more closely rather than grabbing single ideas and running off into the woods with them.

 

[WhyIsItSo]

Sorta getting back to mainstream physics --- few points, vocabulary notes for people to tidy things up a bit:

"nucleation" is NOT a fancy word for "boiling;"

vapor pressure can be approximated over short temperature ranges as lnP = A(1/T) + B, not whatever this is;​

Could you point me to a resource where I can read what that means?

"free expansion" of an ideal gas results in a zero temperature change, and air is nearly ideal at ambient conditions --- pumping down the bell jar does not result in any sensible cooling effect;

I thought this was the basic principle of how air conditioning worked. Why your propane bottle gets very cold if you open the valve and let it expel its contents.

If not due to the lowering of pressure (hence expansion of gas), why is it cooling?

"boiling" is a "superheat" phenomena, in which liquid is NOT at uniform temperature, and liquid below the surface is at a pressure that is higher by "rho^.g^.h," and reaches a temperature sufficient that the vapor pressure is equal to or greater than the higher pressure at depth, and vapor bubbles form from "nucleation sites, then rise."

According to http://www.phys.unsw.edu.au/~jw/superheating.html" , superheating is very distinct from boiling, and in fact occurs when conditions are not favorable to nucleation. While I grant you the article may not be said to state "nucleation is boiling", it does seem clear that "boiling" is merely a lay term for the visually observed process of pockets of vapor (caused by nucleation) forming and "popping" at the surface.

Now, I acknowledge you know much more than I do about Physics, but there is nothing wrong with my ability to follow logic, and your explanation seems to lack a point relevant to your statement "boiling is a superheat phenomena" (I'll explain myself shortly). Either I have missed something (hence do not in fact udnerstand what you said), or some information is missing.

Explanation
My reasoning is as follows. While I am not familiar with the formula you gave, I intuit you refer to increased pressure due to the increasing weight of water the deeper you go. Ok. Hence, the water can reach a higher temperature before nucleating. Ok.

Where is the connection to the state of superheating?

So, assuming 1 atmosphere in my kitchen, the water at the bottom of the pot might reach 213F before having enough energy to nucleate; are you saying this water is superheated? If so, the state called Superheated must be independent of pressure. Therefore if I set up an experiment at 2atmospheres, in which case water should nucleate at 424F, once it gets around 213F or so it is superheated?

Extended to say a volcanic event 5 miles deep on the ocean floor where sea water may be superheated. Is that the type of cirumstance to which you refer?

Finally, I suspect all of the above is irrelevant. With enthalpy of vaporization in mind, I would postulate that provided the water has indeed started nucleating, that all energy put into the water (via the stove element) is being expelled by the process of nucleation, and the liquid water remains at 212F (adjusted for pressure) throughout. Further more, the said adjustment for pressure is:
1. negligible in the systems we have discussed in this thread, and
2. compensated for by the nucleation going on. If hotter liquid rises into a lower pressure area, it either forms a vapor bubble or more likely, since it is easier to increase the size of a vapor bubble than it is to create one in the first place, contribute its higher energy to nearby vapor bubbles.

Hence, a balance between temperature and pressure, with regard to the Phase change of liquid to gas, is maintained, again provided nucleation is occurring.

And that is the point of the article I reffered to, that superheating occurs when the liquid is hot enough to nucleate, but lacks any "seed" to get started, such as imperfections in the surface of the container, or air/vapor bubbles in the container.

In conclusion, it seems this whole issue boils down to a definition of "boiling" (pun intended).

I had always thought the term was simply a common-use label given to the visual event of (usually) water forming bubbles of vapor which naturally rise to the surface (being about 1/16,000 the weight/volume of liquid water). When I discovered "nucleation", it certainly seemed to fit as the physics explanation of the phenomenon we call boiling.

As I have explained, I do not as yet see how your statements contradict that concept.

 

[fargoth]

Why your propane bottle gets very cold if you open the valve and let it expel its contents?

is there liquid propane in the buttle?
i suspect its pretty much the same as your experiment...

 

[WhyIsItSo]

is there liquid propane in the buttle?
i suspect its pretty much the same as your experiment...

Yes there is. Next time you take your BBQ bottle to have it filled, shake it around and listen to the liquid inside slosh.

I have always assumed that was due to there being enough pressure to force (compress) the gas into a liquid state.

But then, I've been shot down in flames so many times, I'm starting to wonder why my feet stay on the ground

 

[Bystander]

Could you point me to a resource where I can read what that means?

http://www.google.com/search?hl=en&q=vapor+pressure+equations&btnG=Google+Search
Start at the top, and quit when you've picked up the idea; there're as many vapor pressure equations as there are people who've measured vapor pressures; all of 'em are forms, or elaborations of what I gave you.

I thought this was the basic principle of how air conditioning worked. Why your propane bottle gets very cold if you open the valve and let it expel its contents.

Open a propane bottle at room T, and liquid propane boils, cooling the bottle --- same as water in a vacuum --- it's got a much higher vapor pressure at room T.

If not due to the lowering of pressure (hence expansion of gas), why is it cooling?

Joule-Thompson expansion through a throttling valve (not free expansion); you're looking a (dT/dP)_H, positive below J-T inversion T (much greater than room T for most gases --- means cooling on expansion), zero at the J-T inversion T, and negative above J-T inversion (H, He at room T). Air conditioners depend on both the phase change of the refrigerant and the throttled expansion of the gas.

According to http://www.phys.unsw.edu.au/~jw/superheating.html" , superheating is very distinct from boiling, and in fact occurs when conditions are not favorable to nucleation. While I grant you the article may not be said to state "nucleation is boiling", it does seem clear that "boiling" is merely a lay term for the visually observed process of pockets of vapor (caused by nucleation) forming and "popping" at the surface.

"Superheating" is used in many contexts --- enough so, that calling it "distinct from boiling" isn't really possible. "Boiling" is boiling, lay, scientific, or the witches in Macbeth. Water at depth is hotter than water at the surface in "quasi-equilibrium" with vapor over a boiling vessel --- it's superheated; as it rises (lower density than overlying water), it reaches a level where pressure has dropped enough that it explosively vaporizes.

(snip)Finally, I suspect all of the above is irrelevant. With enthalpy of vaporization in mind, I would postulate that provided the water has indeed started nucleating,

Don't get bullheaded with misuse of the word --- "nucleation" is the process in which a phase change is initiated by pronounced, small scale discontinuities in chemical potential, be those "nucleation sites" dust motes, cosmic ray tracks, acoustic compressions and rarefactions, seed crystals, adsorbed air, surface irregularities in containers, or whatever else.

that all energy put into the water (via the stove element) is being expelled by the process of BOILING, and the liquid water remains at 212F (adjusted for pressure) throughout. Further more, the said adjustment for pressure is:
1. negligible in the systems we have discussed in this thread, and
2. compensated for by the BOILINGgoing on. If hotter liquid rises into a lower pressure area, it either forms a vapor bubble or more likely, since it is easier to increase the size of a vapor bubble than it is to create one in the first place, contribute its higher energy to nearby vapor bubbles.

Hence, a balance between temperature and pressure, with regard to the Phase change of liquid to gas, is maintained, again provided BOILING is occurring.

And that is the point of the article I reffered to, that superheating occurs when the liquid is hot enough to BOIL, but lacks any "seed" to "nucleate" the process, such as imperfections in the surface of the container, or air/vapor bubbles in the container.

In conclusion, it seems this whole issue boils down to a definition of "boiling" (pun intended).

I had always thought the term was simply a common-use label given to the visual event of (usually) water forming bubbles of vapor which naturally rise to the surface (being about 1/16,000 the weight/volume of liquid water). When I discovered "nucleation", it certainly seemed to fit as the physics explanation of the phenomenon we call boiling.

As I have explained, I do not as yet see how your statements contradict that concept.

 

[WhyIsItSo]

Bystander,

It seems I have a head full of misconceptions.

Since High School, I have understood that a Centrifuge is named after a force that does not exist, but try to explain that to someone who doesn't "get it".

I like to count myself as open to new ideas, but I'm getting a headache! Since coming to this forum I've learned some things that have turned some of my concepts on their head, such as:

Though we draw free body diagrams happily showing gravity as a downward force, it is not a force at all, rather a bedning of space time (or some such).

Depressurization of a gas or liquid filled container does not cool it because of the loss of pressure, per se, but because of boiling (which at first blush appears to be a contradiction).

I am working on adjusting what I think I know, but it is amazing how many concepts are built upon just those two "facts" alone. I have a lot of rethinking to do...

...and two uni assignments to finish by midnight tonight :)

TO BE CONTINUED...

 

[WhyIsItSo]

Definitions

I think I need to clarify definitions for myself before proceeding. Too many concepts are
getting confused due to my (apparent) fuzzy-ness in this regard.

Superheating​

In physics, superheating (sometimes referred to as boiling retardation, boiling delay, or defervescence) is the phenomenon in which a liquid is heated to a temperature higher than its standard boiling point, without actually boiling. This can be caused by rapidly heating a homogeneous substance while leaving it undisturbed (so as to avoid the introduction of bubbles at nucleation sites).

Because a superheated fluid is the result of artificial circumstances, it is metastable, and is disrupted as soon as the circumstances abate, leading to the liquid boiling very suddenly and violently (a steam explosion). Superheating is sometimes a concern with microwave ovens, some of which can quickly heat distilled water without physical disturbance. A person agitating a container full of superheated water by attempting to remove it from a microwave could easily be scalded. (Wikipedia, 2006c)

The phrase “without actually boiling” disassociates boiling and superheating.

Boiling​

There is a longer explanation, but the short version is “the temperature at which the vapor pressure of the liquid equals the pressure of the surroundings.” (Wikipedia, 2006a)

Nucleation​

“Nucleation is the onset of a phase transition in a small but stable region. The phase transition can be the formation of a bubble or of a crystal from a liquid. Creation of liquid dropplets in saturated vapour is also characterised by nucleation” (Wikipedia, 2006b)

For example:

“Pure water freezes at −42°C rather than at its melting temperature of 0°C if no crystal nuclei, such as dust particles, are present to form an ice nucleus.” (ibid).

Conclusions​

So, Superheating is not Boiling. It is a state where a liquid is hot enough to boil, but is not boiling. Likewise, Supercooling is not freezing, but is the condition where a liquid is cold enough to freeze, but doesn't.

In both cases, the Super- state is due to conditions unfavourable to nucleation.

As for nucleation, I see now it was invalid to use at as “boiling”. Vaporizing nucleation is a state that can occur anywhere from the boiling point through to and including the upper limit of superheating.

Finally, you questioned what I was talking about regarding that “B” value. That was my incomplete attempt at deriving a boiling point constant. To finish and tidy up that concept…

Since the boiling point of a liquid is a function of Temp/Pressure, we can write

Let B = Boiling Point, T = Temperature, P = Pressure, then
B=\frac{xT}{yP}

B=n\, T/P
where x, y are some constants and
n=\frac{x}{y}
triple point < T < critical temperature
triple point < P < critical pressure

The International Union of Pure and Applied Chemistry (IUPAC) recommends pressure use the unit bar which equals 100kPa. At this pressure, water boils at 372.76K. So, the boiling point constant B for water is

B=\frac{372.76K}{100kPa}

B=0.37276\, K/kPa

Useful? I don’t know yet. It may prove a convenient term to insert into the equations I next have to learn how to do. We’ll see.

References​

Wikipedia. (2006a). Boiling point. Retrieved August 22, 2006, from Wikipedia Web site: http://en.wikipedia.org/wiki/Boiling_point

Wikipedia. (2006b). Nucleation. Retrieved August 22, 2006, from Wikipedia Web site: http://en.wikipedia.org/wiki/Nucleation

Wikipedia. (2006c). Superheating. Retrieved August 22, 2006, from Wikipedia Web site: http://en.wikipedia.org/wiki/Superheating

 

[Bystander]

(snip)
(snip)

Conclusions​

So, Superheating is not Boiling. It is a state where a liquid is hot enough to boil, but is not boiling. Likewise, Supercooling is not freezing, but is the condition where a liquid is cold enough to freeze, but doesn't.

In both cases, the Super- state is due to conditions unfavourable to nucleation.

One more time,

"Superheating" is used in many contexts --- enough so, that calling it "distinct from boiling" isn't really possible. "Boiling" is boiling, lay, scientific, or the witches in Macbeth. Water at depth is hotter than water at the surface in "quasi-equilibrium" with vapor over a boiling vessel --- it's superheated; as it rises (lower density than overlying water), it reaches a level where pressure has dropped enough that it explosively vaporizes.

There're plenty of vapor pressure data in the literature for which PIs measured boiler temperatures. Those data are useless. Boilers superheat. That's the big trick to measuring vapor pressures, avoiding the effects of superheating.

(snip)Finally, you questioned what I was talking about regarding that “B” value. That was my incomplete attempt at deriving a boiling point constant. To finish and tidy up that concept…

Since the boiling point of a liquid is a function of Temp/Pressure, we can write

Let B = Boiling Point, T = Temperature, P = Pressure, then
B=\frac{xT}{yP}

B=n\, T/P
where x, y are some constants and
n=\frac{x}{y}
triple point < T < critical temperature
triple point < P < critical pressure

The International Union of Pure and Applied Chemistry (IUPAC) recommends pressure use the unit bar which equals 100kPa. At this pressure, water boils at 372.76K. So, the boiling point constant B for water is

B=\frac{372.76K}{100kPa}

B=0.37276\, K/kPa

Useful? I don’t know yet. It may prove a convenient term to insert into the equations I next have to learn how to do. We’ll see.(snip)

You're Frank Sinatra? Gonna "do it your way?"

You've been given a generic vapor pressure equation; you've been given a Google list of sources on vapor pressure equations; is there a point to inventing a "square wheel?" Is there a point to inventing nonsense terms, "boiling point constant?"

If you are interested in understanding the demonstration (freezing water in a vacuum chamber), ask questions, do not assert nonsense. Say what you know, say what you think might be happening, ask whether you're in the ballpark, ask where you go from there.

 

[WhyIsItSo]

One more time,

"Superheating" is used in many contexts --- enough so, that calling it "distinct from boiling" isn't really possible. "Boiling" is boiling, lay, scientific, or the witches in Macbeth. Water at depth is hotter than water at the surface in "quasi-equilibrium" with vapor over a boiling vessel --- it's superheated; as it rises (lower density than overlying water), it reaches a level where pressure has dropped enough that it explosively vaporizes.

And right back atcha!

Superheating is the state where a liquid has reached the temperature/pressure ratio to boil, but is not boiling due to lack of anything to seed nucleation.

"it reaches a level where pressure has dropped enough that it explosively vaporizes"

I don't recall the upper limit of superheating, but I remember reading there is one. For a pressure P, there is a temperature T at which the liquid will start boiling if there is some seed for nucleation. If it does not boil, and you increase T, or reduce P, then it is superheated. If you now introduce some seed to trigger nucleation, it will violently boil and expend energy in the form of heat (not to mention possibly expelling liquid from the container) until it lowers T such that the T/P ratio is once again at boiling point. If you do not introduce anything to trigger nucleation, but continue to increase T and/or decrease P, you will reach the limit of superheating which is the point at which there is enough energy in the water to overcome P, at which point it explosively vaporizes.

I think you miss the fact that it is called boiling point. Get the point? It is not a range, unlike superheating.

I think that is the issue you missed about why I calculated a boiling point constant for water. Maybe it is in fact useless, but the exercise was fruitful. If nothing else, it illustrated the fact you are apparently missing; the relationship between T and P for boiling point is fixed and specific within the bounds of triple point and critical heat/pressure.

Recalling enthalpy of vaporization, provided again that a seed exists to trigger nucleation for the phase change, water at 1 atm will be at boiling point (phase change) at 212F. Adding more and more heat will not superheat the water, just increase the rate of vaporization. The liquid water will remain at 212F until the phase change is complete - you've boiled the pot dry!

You're Frank Sinatra? Gonna "do it your way?"

You've been given a generic vapor pressure equation; you've been given a Google list of sources on vapor pressure equations; is there a point to inventing a "square wheel?" Is there a point to inventing nonsense terms, "boiling point constant?"

If you are interested in understanding the demonstration (freezing water in a vacuum chamber), ask questions, do not assert nonsense. Say what you know, say what you think might be happening, ask whether you're in the ballpark, ask where you go from there.

Thanks to you, I have read about these issues. On the matter of superheating, I've read multiple sources to check and double check my understanding. I am not learning impared nor stupid. If you still do not agree with what I have said, take your own advice; I've provided concise arguments, in logical form, to support my contention. In your rebuttal, address each of my arguments and clearly state what you believe is wrong with them. I have cited my sources, you have not. Have you considered the possibility that this subject is somewhat vague in your own mind? Perhaps your understanding is based on long unused knowledge, and some errors have crept into your thining.

If not, then logically explain where you think I'm wrong. Your over-generalized argument about higher temperatures rising to lower pressures could possibly be a sign of superheating, but it can just as likely occur during normal boiling. Be more specific.

And Bystander, being derogatory towards me is not doing anything for your own credibility. Neither does it bring anything constructive to this thread. I've shown myself to be willing to be corrected, but I'm not accepting information merely because you say so. If you truly want to correct a misconception of mine, then be concise, be accurate, and be logical.

Your insults won't drive me away, but ill-considered arguments might.

 

[Bystander]

(snip)Superheating is the state where a liquid has reached the temperature/pressure ratio to boil, but is not boiling due to lack of anything to seed nucleation.

You've found one of many definitions of "superheat;" don't marry it. You're going to find lots of words in the sciences that have different uses in different contexts --- IUPAC and IUPAP have enough work to keep them busy for a loonnngggg time.

(snip)I don't recall the upper limit of superheating, but I remember reading there is one.

Theoreticians have been kicking the question around for many moons, and getting nowhere. Experimentalists have played with it off and on for just as long, or longer, and push observed superheats up from time to time. Critical T is an obvious, intuitive, upper limit, but no one's established anything.

(snip)I think you miss the fact that it is called boiling point. (snip)

--- and, it ain't a "point." "Boiling point," or "normal boiling point" is taken by convention to be the temperature at which the vapor pressure of a liquid equals the standard atmosphere (whatever the latest convention happens to have fixed upon). At that temperature and pressure, the liquid and vapor are in equilibrium; that is, there is no net evaporation of liquid, or condensation of vapor; there is no "boiling" at the b.p. (n.b.p.). "Boiling" is a dynamic process which transfers heat via vapor from a boiler to machinery, reflux condensers, the atmosphere, wherever; the energy transfer proceeds through the movement of energy along temperature and pressure gradients. The "degree of superheat" (one definition of several) in the boiler is a function of boiler geometry, boiler size, amount of liquid being boiled, expansivities of liquid and vapor, thermal conductivities of liquid and vapor, viscosities of liquid and vapor, interfacial tension between liquid and vapor, rate at which heat is supplied to the boiler, and assorted other parameters.

I think that is the issue you missed about why I calculated a boiling point constant for water. Maybe it is in fact useless, but the exercise was fruitful. If nothing else, it illustrated the fact you are apparently missing; the relationship between T and P for boiling point is fixed and specific within the bounds of triple point and critical heat/pressure.

If you haven't read any of the vapor pressure information I Googled for you, just say so.

Recalling enthalpy of vaporization, provided again that a seed exists to trigger nucleation for the phase change, water at 1 atm will be at boiling point (phase change) at 212F. Adding more and more heat will not superheat the water, just increase the rate of vaporization. The liquid water will remain at 212F until the phase change is complete - you've boiled the pot dry!

The number of physicists who think a boiling liquid can be used as a constant temperature bath never ceases to amaze me --- explains why they make lousy chemical engineers --- sleep through the old-fashioned smash-mouth thermo --- hit the junior year and Kittel, and they start "boiling" "spherical chickens" with zero surface tension. Try Perry, Chemical Engineers' Handbook --- university libraries, or interlibrary loan --- you don't wanta spend that much money. Adamson, Physical Chemistry of Surfaces might help --- strikes me the Kelvin equation for small droplets, or bubbles, (nucleated sites) is treated in some detail.

Thanks to you, I have read about these issues. (snip)

Keep reading.

And Bystander, being derogatory towards me is not doing anything for your own credibility. Neither does it bring anything constructive to this thread. I've shown myself to be willing to be corrected, but I'm not accepting information merely because you say so. If you truly want to correct a misconception of mine, then be concise, be accurate, and be logical.

Your insults won't drive me away, but ill-considered arguments might.

You get the respect you demonstrate for the field --- type your way in here with a philosophy degree, and Wiki, and assert explanations of the universe? You are to be congratulated for not presuming to derive the universe, but you ain't going to be taken seriously.

"Rome was not built in a day" --- you're not going to turn yourself into a physicist with an afternoon in Wiki. Get off the philosophical high-horse, state what you know and understand, state what you've read and the conflicts you see between that and what you know or read elsewhere, ask questions to resolve those conflicts or clarify concepts. Wiki's decent for overviews, picking up key words to Google, and generally getting your "foot in the door" far as acquainting yourself with a topic --- it's useless as an authoritative source.

Bullheaded assertion of nonsense is welcome in the philosophy forums (just below "Other Sciences" on the PF main page) --- it is not welcome in the science, math, and engineering forums.

 

[vmars]

Hey this is a great thread.
Thanks all !
Now all that remains, for me, is to repeat the experiment.

Please, if somone knows:
What equipment, EXACLTY, would I need,
and where could I purchase it?
Thanks in advance!
...Vern