Ping: Scientists

[st.cronin]

I performed an experiment today that I don't understand.

Two beakers of equal volumes of water from the same source placed in a contained environment. One water was at 35 degrees celsius. The other water was at 25 degrees celsius.

The environment was gradually de-pressurized, until, as expected, the warmer water began to boil. The pressure was lowered some more, and the cooler water eventually also began to boil, but, what puzzled me was ... the warmer water STOPPED boiling. It was now about 6-7 degrees warmer than the boiling water, but not boiling itself. How is this possible?

[st.cronin]

dola

no heat was added, except possibly ambient heat (I think both waters were warmer than room temperature, though) - the only change was to the pressure.

[terpkristin]

The key here is that it is a closed environment.
Eventually the water gas + air gas came into equilibrium with the remaining water.

[st.cronin]

The key here is that it is a closed environment.
Eventually the water gas + air gas came into equilibrium with the remaining water.

Both waters were in the same environment, the cooler one was boiling. I don't think you have solved the puzzle - or if you have, you have not explained it in a way I understand.

[terpkristin]

I think I have solved it (though I don't think equilibrium is the word I'm looking for) but I'm trying to think of a better way to explain it.
But even with the closed environment, because they started at different temperatures, they will "settle" different.

/tk

[st.cronin]

What do you mean by 'settle'? Is it not true that at a particular temperature and pressure, water will turn to gas? What would cause that process to stop, even at a higher than neccesary temperature?

I admit that I'm over my head here.

[Arctus]

I performed an experiment today that I don't understand.

Two beakers of equal volumes of water from the same source placed in a contained environment. One water was at 35 degrees celsius. The other water was at 25 degrees celsius.

The environment was gradually de-pressurized, until, as expected, the warmer water began to boil. The pressure was lowered some more, and the cooler water eventually also began to boil, but, what puzzled me was ... the warmer water STOPPED boiling. It was now about 6-7 degrees warmer than the boiling water, but not boiling itself. How is this possible?

You must have broken a law of thermodynamics........now get cracking on something useful, like a perpetual motion machine. :D

[SackAttack]

If I'm not mistaken, water will boil at lower temperatures at low pressure, but doesn't it take more energy to bring the water to the boiling point at the lower pressure?

[st.cronin]

If I'm not mistaken, water will boil at lower temperatures at low pressure, but doesn't it take more energy to bring the water to the boiling point at the lower pressure?

Yes to the first part, I don't know to the second. But even if that's true, the first water STARTED boiling, and as the pressure continued to drop, then STOPPED boiling, at about the same time as the cooler water started boiling. It's as though the water was too warm to boil.

[SackAttack]

Yes to the first part, I don't know to the second. But even if that's true, the first water STARTED boiling, and as the pressure continued to drop, then STOPPED boiling, at about the same time as the cooler water started boiling. It's as though the water was too warm to boil.

The reaction still needs energy to sustain itself. If there ain't enough energy, the reaction stops.

That's my guess here. One reaction is starting, but another reached the point where it was no longer self-sustaining.

[Mac Howard]

Does the boiling of the warmer water stop at precisely the same time as the lower temperature water starts? Is there a link between the two or are the two effects independant? If the lower temperature container of water is left out does the boiling stop in the warmer water at the same point?

[KWhit]

I have a theory.

It's possible that the warmer water just got tired of boiling and said "Fuck it, I'm taking a break!"

Oh, and I'm not really a scientist.

[dacman]

It's as though the water was too warm to boil.
Yep. The warmer water reaches a point at which it's superheated. Normally superheating is achieved by adding a large amount of energy suddenly(typically via microwaves), but removing pressure has the same effect.

[terpkristin]

Ok I'm going to try to explain this.
Water boils when its vapor pressure exceeds that of its surroundings. It will boil UNTIL the vapor pressure increases the pressure of the surroundings.

The hotter water boiled first because vapor pressure is related to the temperature of the water. The hotter it is, the more vapor pressure it has.

At some point in the depressurization, combined with the fact that some of the water has turned into a gas, that vapor pressure of the water either equalled or became less than the vapor pressure of its surroundings. It stopped boiling at that point.

Make sense?

/tk

[RPI-Fan]

Yep. The warmer water reaches a point at which it's superheated. Normally superheating is achieved by adding a large amount of energy suddenly(typically via microwaves), but removing pressure has the same effect.

Isn't superheated water typically rather dangerous (from a splattering perspective), and thus the reason you should put a popsicle stick if you try to make very hot water in the microwave?

[dacman]

http://www.phys.unsw.edu.au/~jw/superheating.html

[finkenst]

I have a theory.

It's possible that the warmer water just got tired of boiling and said "Fuck it, I'm taking a break!"

Oh, and I'm not really a scientist.
This, by far, is the best response so far... :)

is it possible there is more gases dissolved in the cooler water and it has reached the point that they are now brekaing the surface pressure of hte cooler water to get into the depressurized environment/

all the best,

tim


disclaimer: i play on networks all day, not much in the chemical world.

[RPI-Fan]

Isn't superheated water typically rather dangerous (from a splattering perspective), and thus the reason you should put a popsicle stick if you try to make very hot water in the microwave?

Woot! :cool:

[terpkristin]

Isn't superheated water typically rather dangerous (from a splattering perspective), and thus the reason you should put a popsicle stick if you try to make very hot water in the microwave?They did that on Mythbusters ;)

/tk

[RPI-Fan]

They did that on Mythbusters ;)

/tk

I learned it from my dad while making oatmeal, about 5 years ago.

:p

[finkenst]

They did that on Mythbusters ;)

/tk
and on good eats one episode.

[Brillig]

Ok, for starters, most of you fail Physics 101. Go back to school, go directly to school. Do not pass Go, do not collect $200.

TK's explanation of vapor pressure is correct, but doesn't address the current situation which I believe (the description is not precise) is that both beakers of water are in the same container, i.e., at the same pressure.

Superheating is a possible answer to the question, however, superheating is a phenomenon which depends on the lack of nucleation sites for gas bubbles to form, a situation which would seem not to apply since the warmer beaker of water started boiling, then stopped.

A more plausible hypothesis is that the "boiling" that was observed was not the vaporization of water, but rather the liberation of dissolved gasses. Since the vapor pressure of gasses dissolved in water is higher than the vapor pressure of the water itself, the gasses will "boil" off well before the water itself begins to vaporize. However, once the gasses are eliminated, the pressure would have to be lowered substantially before the water itself began to boil.

Without more information on the experimental conditions and apparatus, it's hard to say exactly what occurred, but the dissolved gas explanation seems most likely to me at this time.

[finkenst]

Ok, for starters, most of you fail Physics 101. Go back to school, go directly to school. Do not pass Go, do not collect $200.

TK's explanation of vapor pressure is correct, but doesn't address the current situation which I believe (the description is not precise) is that both beakers of water are in the same container, i.e., at the same pressure.

Superheating is a possible answer to the question, however, superheating is a phenomenon which depends on the lack of nucleation sites for gas bubbles to form, a situation which would seem not to apply since the warmer beaker of water started boiling, then stopped.

A more plausible hypothesis is that the "boiling" that was observed was not the vaporization of water, but rather the liberation of dissolved gasses. Since the vapor pressure of gasses dissolved in water is higher than the vapor pressure of the water itself, the gasses will "boil" off well before the water itself begins to vaporize. However, once the gasses are eliminated, the pressure would have to be lowered substantially before the water itself began to boil.

Without more information on the experimental conditions and apparatus, it's hard to say exactly what occurred, but the dissolved gas explanation seems most likely to me at this time.
uh, brillig...

Isn't that what i said?
Post #17 (http://www.operationsports.com/fofc/showpost.php?p=1039465&postcount=17)

[finkenst]

dola.

nevermind, i really can't read...

[Desnudo]

Ok, for starters, most of you fail Physics 101. Go back to school, go directly to school. Do not pass Go, do not collect $200.

TK's explanation of vapor pressure is correct, but doesn't address the current situation which I believe (the description is not precise) is that both beakers of water are in the same container, i.e., at the same pressure.

Superheating is a possible answer to the question, however, superheating is a phenomenon which depends on the lack of nucleation sites for gas bubbles to form, a situation which would seem not to apply since the warmer beaker of water started boiling, then stopped.

A more plausible hypothesis is that the "boiling" that was observed was not the vaporization of water, but rather the liberation of dissolved gasses. Since the vapor pressure of gasses dissolved in water is higher than the vapor pressure of the water itself, the gasses will "boil" off well before the water itself begins to vaporize. However, once the gasses are eliminated, the pressure would have to be lowered substantially before the water itself began to boil.

Without more information on the experimental conditions and apparatus, it's hard to say exactly what occurred, but the dissolved gas explanation seems most likely to me at this time.

Sleep, that's where I'm a viking

[Mac Howard]

Are the bubbles we see when water boils purely dissolved gases or are they also bubbles of water vapour? If the former then why, under normal circumstances, do these bubbles continue right until the water has boiled away entirely? You would have thought the dissolved gases would have been expelled before then. I suspect that at least some of the bubbles are submerged pockets of water vapour - the transition from liquid to gas takes place beneath the surface and forms water vapour bubbles which then rise to the top and break the surface.

If that is the case, are we seeing a change of mode when the temperature reaches a sufficient excess over the boiling temperature at that particular pressure from turbulent to smooth as happens in airflow over a body as speed increases? Perhaps the formation of water vapour bubbles ceases and individual molecules rise.

Another point: the turbulance may be caused because the bubbles of air/water vapour have to overcome the surface tension at the surface of the water - they rise to the surface, hit the resistence of the surface tension bulging the surface and then break through. You'll note that, prior to boiling, bubbles rise but fail to break the surface until boiling begins when the energy has increased sufficiently to allow them to do so. Does the surface tension change with the excess temperature allowing the gas bubbles or molecules to escape freely?

I wonder if the transition from liquid to gas - and the turbulence, which I think is what you mean by "boiling", are two different but connected phenomena and what you're seeing is the cessation of the turbulence but not of the transition.

I should point out that my days as a scientist are a long time in the past and that may well be a load of speculative twaddle :)

[sterlingice]

You must have broken a law of thermodynamics........now get cracking on something useful, like a perpetual motion machine. :DMarge: I'm worried about the kids, Homey. Lisa's becoming very obsessive. This morning I caught her trying to dissect her own raincoat.
Homer: {scoffs} I know. And this perpetual motion machine she made today is a joke! It just keeps going faster and faster.
Marge: And Bart isn't doing very well either. He needs boundaries and structure. There's something about flying a kite at night that's so unwholesome. {looks out window}
Bart: {creepy voice} Hello, Mother dear.
Marge: {closing the curtains} That's it: we have to get them back to school.
Homer: I'm with you, Marge. Lisa! Get in here. {Lisa walks in, chuckling nervously} In this house, we obey the laws of thermodynamics!

(Ok, I've never gotten the "flying the kite at night" joke- anyone know what that's a reference to?)

SI

[st.cronin]

Ok, for starters, most of you fail Physics 101. Go back to school, go directly to school. Do not pass Go, do not collect $200.

TK's explanation of vapor pressure is correct, but doesn't address the current situation which I believe (the description is not precise) is that both beakers of water are in the same container, i.e., at the same pressure.

Superheating is a possible answer to the question, however, superheating is a phenomenon which depends on the lack of nucleation sites for gas bubbles to form, a situation which would seem not to apply since the warmer beaker of water started boiling, then stopped.

A more plausible hypothesis is that the "boiling" that was observed was not the vaporization of water, but rather the liberation of dissolved gasses. Since the vapor pressure of gasses dissolved in water is higher than the vapor pressure of the water itself, the gasses will "boil" off well before the water itself begins to vaporize. However, once the gasses are eliminated, the pressure would have to be lowered substantially before the water itself began to boil.

Without more information on the experimental conditions and apparatus, it's hard to say exactly what occurred, but the dissolved gas explanation seems most likely to me at this time.


I should point out that the water used is ... I forget the wording, it's the opposite of aerated ... it's basically pure h2o. So I don't think 'dissolved gases' can explain this. I performed a second experiment this eve which I will describe in a dola.

[st.cronin]

Ok...

I repeated the experiment several times, same result. I can get water to exist as water above the boiling point for water at a given pressure.

Then I did this (for reasons which I forget - I should probably keep a lab journal.) I took the water which, from being depressurized had boiled and then stopped boiling, and then agitated the water (thinking to reproduce in a different manner an effect I had seen with supercooled water). Nothing happened. Then I took this water, and another sample of water which had NOT been boiled. I carefully made sure the waters were at the same volume and temperature, and put them on the same hot plate. The water which had NOT been boiled rose in temperature faster, and boiled faster. SIGNIFICANTLY faster. Obviously some reaction had taken place in the water by being boiled in the pressurization chamber. I feel as though I am on the brink of solving our oil dependancy situation. I have a theory or two, but I am going to do some reading this weekend before embarrasing myself here by posting them.

[Mr. Wednesday]

Then I took this water, and another sample of water which had NOT been boiled. I carefully made sure the waters were at the same volume and temperature, and put them on the same hot plate. The water which had NOT been boiled rose in temperature faster, and boiled faster. SIGNIFICANTLY faster.Possible explanations for this include impurities somehow being introduced into one of the samples, differences in the containers, and non-uniformities in the heat transfer through the hot plate.

[MrBigglesworth]

If it is not gases being released, it could be that the area right above the water level in the beaker had an increase in pressure, from both heat and water vapor. This localized high pressure kept the water from boiling.

[Glengoyne]

I was going to put my college thermodynamics to use, and start talking about the energy required for the heat of vaporization in your samples. Then I noted that you were changing the pressure, and I can't even begin to remember how pressure enters into the equation.

The Heat of vaporization might explain the sample boiling after placing the containers back on the heat source. Specifically, the required amount of energy required to make the water boil(heat of vaporzization) hadn't been reached, but the process had started. Introducing additional energy, the hotplate, met the heat of vaporization requirement, so the water boiled.

Say it takes 1000 calories(I think the formulas use joules) to bring a volume of water to the boiling point...it then takes an additional 4k or 5k calories to account for the heat of vaporization. Once those calories have been introduced...the water boils.

I think I had better stop talking because I might embarrass myself. I'm contemplating thermodynamics at 1AM, and that just isn't a wise practice.

[MrBigglesworth]

I was going to put my college thermodynamics to use, and start talking about the energy required for the heat of vaporization in your samples. Then I noted that you were changing the pressure, and I can't even begin to remember how pressure enters into the equation.

The Heat of vaporization might explain the sample boiling after placing the containers back on the heat source. Specifically, the required amount of energy required to make the water boil(heat of vaporzization) hadn't been reached, but the process had started. Introducing additional energy, the hotplate, met the heat of vaporization requirement, so the water boiled.

Say it takes 1000 calories(I think the formulas use joules) to bring a volume of water to the boiling point...it then takes an additional 4k or 5k calories to account for the heat of vaporization. Once those calories have been introduced...the water boils.

I think I had better stop talking because I might embarrass myself. I'm contemplating thermodynamics at 1AM, and that just isn't a wise practice.
That's what I thought at first too, but he mentioned that the water was still 6 or 7 degrees warmer than the colder water that was still boiling.

[moriarty]

I was going to put my college thermodynamics to use, and start talking about the energy required for the heat of vaporization in your samples. Then I noted that you were changing the pressure, and I can't even begin to remember how pressure enters into the equation.

Try the ideal gas law. Edit (for the vapor portion, not the liquid).

[rkmsuf]

Try putting Pop Rocks and Coke in your mouth at the same time.

[moriarty]

I performed an experiment today that I don't understand.

Two beakers of equal volumes of water from the same source placed in a contained environment. One water was at 35 degrees celsius. The other water was at 25 degrees celsius.

The environment was gradually de-pressurized, until, as expected, the warmer water began to boil. The pressure was lowered some more, and the cooler water eventually also began to boil, but, what puzzled me was ... the warmer water STOPPED boiling. It was now about 6-7 degrees warmer than the boiling water, but not boiling itself. How is this possible?

I'm having a problem with the entire setup of your experiment. In order to lower the pressure of this "closed" environment, you either need to drop the temperature or remove the volume (typically by vacuuming out air). Since you say you're not adding/removing temperature I assume you're dropping the pressure by vacuuming out the air. At this point, your environment is no longer closed. (You should also keep either temperature or pressure constant, not let both vary).

I suspect the problems you're seeing are due to the fact that the vapor in the environment is not in equilibrium (i.e. even though it's in the same 'sealed' container, the vapor above water A is not the same as the vapor above water B.). So whereas B comes to equilibrium first with the temperature/density of water vapor (and it stops boiling) the density of water vapor above A is different. Adding something to stir the air up (like a fan) to keep the vapor in the container in equilibrium will probably resolve your irregularities.

I also suspect you're pulling air out of the container to reduce the pressure. This can impact the experiment in that the conatiner is no longer a closed environment. Plus if you're pulling vacuum closer to one water sample or the other (and the vapor in the container is not well mixed) you're actually changing the local density of the water vapor near the one water sample.

Rather than a scientific miracle, I think you've discovered the difficulties of a maintaining a controlled experiment. But that's just my $.02 I could be wrong.

[Eaglesfan27]

I was very good at chemistry and physics in college 12 or 13 years ago. Hearing you guys talk makes me regretful of how much physics I've forgotten.

[MrBigglesworth]

I'm having a problem with the entire setup of your experiment. In order to lower the pressure of this "closed" environment, you either need to drop the temperature or remove the volume (typically by vacuuming out air). Since you say you're not adding/removing temperature I assume you're dropping the pressure by vacuuming out the air. At this point, your environment is no longer closed. (You should also keep either temperature or pressure constant, not let both vary).

I suspect the problems you're seeing are due to the fact that the vapor in the environment is not in equilibrium (i.e. even though it's in the same 'sealed' container, the vapor above water A is not the same as the vapor above water B.). So whereas B comes to equilibrium first with the temperature/density of water vapor (and it stops boiling) the density of water vapor above A is different. Adding something to stir the air up (like a fan) to keep the vapor in the container in equilibrium will probably resolve your irregularities.

I also suspect you're pulling air out of the container to reduce the pressure. This can impact the experiment in that the conatiner is no longer a closed environment. Plus if you're pulling vacuum closer to one water sample or the other (and the vapor in the container is not well mixed) you're actually changing the local density of the water vapor near the one water sample.

Rather than a scientific miracle, I think you've discovered the difficulties of a maintaining a controlled experiment. But that's just my $.02 I could be wrong.
Which is what I said, only more detailed :)

[Brillig]

Hmm.

A few pointers.

First, gas diffusion is a fairly fast process, so I don't believe that significant differences in pressure could exist over beaker A and beaker B. If the two beakers are in the same depressurized chamber, I doubt that this could be the explanation for the phenomenon described, although we lack precise measurements for the times involved.

Second, "pure water" is remarkable hard to come by. If you leave a beaker of distilled water out in the open, it will immediately begin to absorb gas from the surrounding atmosphere, albeit slowly. Off the top of my head, the only way I can think of to obtain pure water would be to introduce controlled quantities of hydrogen and oxygen into a combustion chamber and ignite them. I'm guessing you're not doing that.

Third, the vapor pressure of water at 25-35 C is in the range of 20-40 mm Hg. While you haven't described your experimental apparatus at all, (Bad scientist, bad! No biscuit!) this is usually well beyond what one can achieve with a water aspirated vacuum, and usually requires a mechanical vacuum pump. Of course, without pressure readings, it's impossible to say what is going on...

Also, Mr.W is dead-on about the hot-plate. Hot plates not a very effective way to ensure that two objects are receiving the same amount of heat transfer. A hot-oil immersion would probably be a better bet.

Without a great deal more detail on the conditions of the experiment and apparatus used, along with the measurements, it's impossible to say if this is anything other than an artifact of bad methodology. Sorry if that sounds harsh, but so far your experiment has thrown two conclusions (that boiled water has a significantly higher heat capacity than non-boiled water, and that boiled water has a lower water vapor pressure than non-boiled water) that completely defy modern scientific knowledge.

Either you're a genius, or there's something wrong in your experiment methodology or description.

[Desnudo]

Try putting Pop Rocks and Coke in your mouth at the same time.

Are you trying to get people killed? You know internet message boards can serve as criminal evidence.

[rkmsuf]

Are you trying to get people killed? You know internet message boards can serve as criminal evidence.


It's a trap. Real scientists won't fall for this.

[st.cronin]

Hmm.

A few pointers.

First, gas diffusion is a fairly fast process, so I don't believe that significant differences in pressure could exist over beaker A and beaker B. If the two beakers are in the same depressurized chamber, I doubt that this could be the explanation for the phenomenon described, although we lack precise measurements for the times involved.

Second, "pure water" is remarkable hard to come by. If you leave a beaker of distilled water out in the open, it will immediately begin to absorb gas from the surrounding atmosphere, albeit slowly. Off the top of my head, the only way I can think of to obtain pure water would be to introduce controlled quantities of hydrogen and oxygen into a combustion chamber and ignite them. I'm guessing you're not doing that.

Third, the vapor pressure of water at 25-35 C is in the range of 20-40 mm Hg. While you haven't described your experimental apparatus at all, (Bad scientist, bad! No biscuit!) this is usually well beyond what one can achieve with a water aspirated vacuum, and usually requires a mechanical vacuum pump. Of course, without pressure readings, it's impossible to say what is going on...

Also, Mr.W is dead-on about the hot-plate. Hot plates not a very effective way to ensure that two objects are receiving the same amount of heat transfer. A hot-oil immersion would probably be a better bet.

Without a great deal more detail on the conditions of the experiment and apparatus used, along with the measurements, it's impossible to say if this is anything other than an artifact of bad methodology. Sorry if that sounds harsh, but so far your experiment has thrown two conclusions (that boiled water has a significantly higher heat capacity than non-boiled water, and that boiled water has a lower water vapor pressure than non-boiled water) that completely defy modern scientific knowledge.

Either you're a genius, or there's something wrong in your experiment methodology or description.


I was, in fact, using a mechanical vacuum pump. I don't have precise readings because I don't understand what units the gauge was in. I do know that 1) the pressure was higher when the water started boiling and lower when it stopped boiling and 2) the temperature of both waters was affected by the pressure - both waters rose.

Bad methodology doesn't explain the hot plate, in my opinion - it's a calibrated hot plate, designed for exactly what I did.

I agree I am a very bad scientist. I am an amateur stumbling around a lab, without any idea what I'm talking about.

[Desnudo]

Anyone who doesn't know about gas diffusion is probably in the hospital right now.

[moriarty]

Hmm.

A few pointers.

First, gas diffusion is a fairly fast process, so I don't believe that significant differences in pressure could exist over beaker A and beaker B. If the two beakers are in the same depressurized chamber, I doubt that this could be the explanation for the phenomenon described, although we lack precise measurements for the times involved.
.

Gas diffusion isn't that fast at room temperature. Which is important for those of us who like to fart and then leave the room so that someone else can be blamed a few minutes later.

Pressure may equalize quickly at a measureable level, but the temperature / density of water vapor in the air probably would take a while to equalize out in the container depending on how big it is and how much forced circulation of the gas there is.

[MrBigglesworth]

2) the temperature of both waters was affected by the pressure - both waters rose.
Wait a second...if anything shouldn't temperature go down with a drop in pressure (same energy over an increased volume)? Boiling is also an endothermic process, which shouldn't increase the temperature of the water.

And doesn't that maker the warmer water superheated to begin with, since the lower temp water was boiling as well? Being at 35 degrees when it only needs to be 25 degrees to boil is the same as being 110 degrees when it only needs to be 100 degrees under normal pressure.

[Glengoyne]

Try the ideal gas law. Edit (for the vapor portion, not the liquid).Thats the PEE one Tee one and Vee one ratio equations right? That is about the sum total of what I remembered last night. I just couldn't, nor can I now recall, what values belonged in the dividend or divisor. My Physics and chemistry brain cells were apparently among the first ones I've lost. Last weekend I couldn't recall the difference between centripedal accelleration and centrifugal force.

[moriarty]

Thats the PEE one Tee one and Vee one ratio equations right? That is about the sum total of what I remembered last night. I just couldn't, nor can I now recall, what values belonged in the dividend or divisor. My Physics and chemistry brain cells were apparently among the first ones I've lost. Last weekend I couldn't recall the difference between centripedal accelleration and centrifugal force.


There's:
Boyles Law P1V1 = P2V2
The ideal gas law P1V1 = nRT1
and the other law which I always forget the name which bridges the two P1/T1 = P2/T2

[Eaglesfan27]

See, I remembered Boyle's Law as PV=nRT. I knew my memory was failing me :(

[moriarty]

See, I remembered Boyle's Law as PV=nRT. I knew my memory was failing me :(

Well, you might need to double check me. It's been 13 years since i was in a chemistry lab.

[Mr. Wednesday]

The ideal gas equation of state is PV = nRT.

Boyle's Law and Charles' Law are the two subsidiary equations, but I don't remember which one is which.

[Raiders Army]

It was almost 10 years ago that I took Fluids and Thermo. I am no help other than to tell you that I saved a lot of money by switching my car insurance to Geico.

[Brillig]

Gas diffusion isn't that fast at room temperature. Which is important for those of us who like to fart and then leave the room so that someone else can be blamed a few minutes later.

Pressure may equalize quickly at a measureable level, but the temperature / density of water vapor in the air probably would take a while to equalize out in the container depending on how big it is and how much forced circulation of the gas there is.
Bzzt. First, while we don't know how large this experimental apparatus is, the fact that the whole experiment has to be taken down to .04 atmospheres or so of pressure argues that it's quite small. The diffusion in question is almost certainly taking place over a distance of less than .5m, probably around .2m or so. Second, diffusion rate is limited by the collisions of gas molecules. With 1/25 normal atmospheric pressure, the rate of collision is less than that in a normal room by the same factor. So to use your "elegant" analogy, how long does it take the odor of a fart to travel 20cm/25 or 8 millimeters?

Wait a second...if anything shouldn't temperature go down with a drop in pressure (same energy over an increased volume)? Boiling is also an endothermic process, which shouldn't increase the temperature of the water.

And doesn't that maker the warmer water superheated to begin with, since the lower temp water was boiling as well? Being at 35 degrees when it only needs to be 25 degrees to boil is the same as being 110 degrees when it only needs to be 100 degrees under normal pressure.
Well, almost. The temperature of the surrounding gas does go down with a decrease in pressure. That doesn't really effect the water in the beakers though, since the volume is relatively constant. The heat loss in the water from the vaporization is a concern, although cronin states that the temperature of the warmer beaker remained higher. Without additional information on the time-scales involved and the mechanism of temperature measurement, it's unclear whether or not this is an artifact.

Bad methodology doesn't explain the hot plate, in my opinion - it's a calibrated hot plate, designed for exactly what I did.
I have to disagree with this. Hot plates are designed to provide heat transfer to a given temperature, not to give equal heat transfer rates to multiple objects. Even if you could guarantee that the surface of the hot plate was at a uniform temperature, it would be impossible to guarantee the same rate of heat transfer to two different beakers placed upon the hot plates surface without the intervention of an additional medium.

[moriarty]

Bzzt. First, while we don't know how large this experimental apparatus is, the fact that the whole experiment has to be taken down to .04 atmospheres or so of pressure argues that it's quite small. The diffusion in question is almost certainly taking place over a distance of less than .5m, probably around .2m or so. Second, diffusion rate is limited by the collisions of gas molecules. With 1/25 normal atmospheric pressure, the rate of collision is less than that in a normal room by the same factor. So to use your "elegant" analogy, how long does it take the odor of a fart to travel 20cm/25 or 8 millimeters?

Hey captain condescending, you're probably right that it would occur fairly quickly if the pressure is indeed that low and the distances that small. The low pressure will certainly speed things up compared to atmospheric pressure. I'm not smart enough to calculate the diffusion rate of water vapor in air at that pressure, and temperature but it sounds like you can tell me.

In principle the theory is sound though. The humidity of the vapor right over the water interface will be close or equal to the vapor pressure. This will retard the rate of evaporation. Eventually the water vapor will diffuse away allowing more water to evaporate. But the rate of this diffusion is dependent on temperature, pressure AND the amount of time the water has been evaporating (which affects the gradient / drop off of humidity in the air immediately around the water/gas interface).

Well, almost. The temperature of the surrounding gas does go down with a decrease in pressure. That doesn't really effect the water in the beakers though, since the volume is relatively constant. The heat loss in the water from the vaporization is a concern, although cronin states that the temperature of the warmer beaker remained higher. Without additional information on the time-scales involved and the mechanism of temperature measurement, it's unclear whether or not this is an artifact.


He didn't just say it remained higher, he said the temperature of the water went up! MrBigglesworth is right. If St Cronin didn't add any additional heat (other than ambient as he claims) why did the temperature of the water increase?

[MrBigglesworth]

Just a couple of random thoughts, I've been thinking about this on and off all day. I don't have anything concrete, just questions to throw out there:

I kind of like the localized high pressure theory, but St Cronin said that the the water would boil neither in the low pressure atmosphere, nor on the hotplate. Was the beaker tall and thin enough that the vapor pressure stayed high in the beaker that even after taking the time to repressurize and put it on the hotplate? I wouldn't think that would be possible, just throwing it out there.

Secondly, he mentioned that the temp in the beakers went up. Gases dissolve better in lower temperature and higher pressure. Was the combination of dropping pressure and increasing temperature enough to have the gases 'boil' out of the solution at a substantial rate to call the water 'boiling'?

I'm starting to like the dissolved gases hypothesis better.

[MrBigglesworth]

If St Cronin didn't add any additional heat (other than ambient as he claims) why did the temperature of the water increase?
Are there any gasses where coming out of solution would be an exothermic process?

[Mr. Wednesday]

I have to disagree with this. Hot plates are designed to provide heat transfer to a given temperature, not to give equal heat transfer rates to multiple objects. Even if you could guarantee that the surface of the hot plate was at a uniform temperature, it would be impossible to guarantee the same rate of heat transfer to two different beakers placed upon the hot plates surface without the intervention of an additional medium. This reminds me of another potential problem here -- the heat transfer rate is going to depend on the area of contact between the two surfaces. Irregularities in either of them could result in significant areas where no contact is made -- something akin to the idea behind using thermal grease between your CPU and your heat sink in your desktop computer.

[Brillig]

So apparently OPEC is still in business...