About Water - Boiling and Freezing

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The discussion explores the scientific principles behind boiling and freezing water, particularly how atmospheric pressure affects these processes. It highlights that boiling involves nucleation, where bubbles form as pressure decreases, making it easier for water to boil at lower temperatures. The freezing point of water is also influenced by pressure, but significant changes require extreme conditions, making everyday variations negligible. An experiment is described where water is rapidly cooled and freezes after boiling due to reduced pressure, illustrating the interplay between temperature and pressure. The conversation delves into the mechanisms of heat loss during boiling and the role of evaporation versus radiative cooling in this process.
  • #31
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.
 
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  • #32
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.
 
  • #33
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 :biggrin:

maybe you'd even come up with some simple way to get the rate as a function of temperature and pressure...
 
  • #34
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.
 
  • #35
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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  • #36
fargoth said:
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.:bugeye: "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!
 
  • #37
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
WhyIsItSo said:
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.
 
  • #38
Bystander said:
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.
 
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  • #39
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...
 
  • #40
fargoth said:
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:eek:
 
  • #41
WhyIsItSo said:
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.
 
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  • #42
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...
 
  • #43
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
 
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  • #44
WhyIsItSo said:
(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,
Bystander said:
"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.
 
  • #45
Bystander said:
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.
 
  • #46
WhyIsItSo said:
(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.
 
  • #47
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
 

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