Naked Science Forum
Non Life Sciences => Technology => Topic started by: Atomic-S on 18/06/2011 21:32:50
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How would one determine the precise volume of a sample of highly purified water at its minimum density, and then copy its mass precisely to be used as a calibration weight?
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How would one determine the precise volume of a sample of highly purified water at its minimum density, and then copy its mass precisely to be used as a calibration weight?
Very carefully.
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Normally water used as a masspiece is used in multi ton lots, mostly for testing cranes, weighbridges and other applications where you need a masspiece in the range of 20 to 200 tons. It is easier to move water in a tanker and pump it up to where the big bag is than it is to have multiple lowbed trailers and trucks to move large masses of iron there to do the work. It is easy to get a water meter and calibrate it to within 1% at a certain flow rate and pressure, and this enables you to fill a big bag with accuracy of around 1% or better, which is what is needed to do these tests and certifications. For smaller masses under 100kg you will use solid masspieces in a protective case for this purpose. For under 500g they are generally in a velvet lined box, and are handled with kid gloves. 1g and under is handles with tweezers.
These masspieces all need regular checking and certification against a National standard masspiece, at a certified supplier, that is itself certified against the International Standard kilogram.
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When they wanted to measure the density of mercury to more places of decimals than most people would care about they made a fused quartz box with optically flat sides and measured it's volume by measuring the lengths of the sides. Then they weighed it, filled it with mercury and weighed it again to get the mass of the known volume of mercury.
You could do something similar with water and some well defined volume (like a volumetric flask).
BTW, do you know how they calibrate volumetric flasks?
Incidentally Sean, What Atomic S is proposing is exactly what they did to make the kilogram in the first place. If they could do it then (and get it right within about 30ppm) then you should be able to do it today.
The question is how much work would it be, and would it be easier to buy a calibration mass.
Atomic S
I just bet you meant to say maximum density; the minimum is practically zero.
Why do you want to do this anyway?
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There is work on replacing the kilogram with a standard that does not depend on a reference to a block of metal held in a vault, but rather one that is based on the number of atoms in a block of Silicon ( as you can get that in multi kilogram lots with impurity levels approaching a single atom per kilogram). This will bring these standards into line as being determined by properties of matter, rather than arbitrary objects.
This allows the reproduction of primary standards to be made with a precision equal to the original, as opposed to losing accuracy with each time that you use one to make a sub standard, you can make every one who needs that accuracy a standard that will be intrinsically accurate. You can do this now with atomic clocks, where they are now small enough to almost literally fit inside a wristwatch, though they do need an external power supply still.
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Technically there are two kilogram standards now.
One is a cubic deciliter of water at the maximum density at 4°C (actually 3.984 °C)
And the other one is the platinum–iridium standard.
Introducing a third standard would just add more confusion.
One of the problems with length (as well as volume) is that it is dependent on thermal expansion. With volume you have thermal expansion of the container as well as expansion of the liquid inside of the container.
So, the meter (and thus the cubic deciliter) is defined as: length of the path traveled by light in vacuum in 1⁄299,792,458 of a second which is a distance that is difficult to measure, but perhaps better than defining it based on the distance from the equator to the North Pole.
Anyway, for your task, find a volumetric flask that is calibrated for 1L of water at 25°C (condensation will be a problem at 4°C). Tare your balance with the flask, and fill with water at 25°C, and you'll be all set. You would probably be better off by purchasing a good set of brass weight standards made anywhere EXCEPT China.
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The platinum-iridium standard is a bit flakey. http://www.npr.org/templates/story/story.php?storyId=112003322
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The platinum-iridium standard is a bit flakey. http://www.npr.org/templates/story/story.php?storyId=112003322
Hmmm
Has anybody ever checked whether the standard kilogram is radioactive? Fortunately it was made before the Hiroshima Bomb. What about background radiation affecting it?
It would seem easy enough to go back to the original definition of a kilogram as the mass of 1 liter of water = 1 cubic decimeter of liquid water at the temperature of its maximum density (just below 4°C), and at 1 ATM.
Then the scientists can go back and argue until the cows come home as far as how many watts that is actually equal to.
So many of our measurements are centered around water anyway.
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"Technically there are two kilogram standards now.
One is a cubic deciliter of water at the maximum density at 4°C (actually 3.984 °C)
And the other one is the platinum–iridium standard."
No, there is only one.
It's the bit of metal.
They knew when they made it that, though they had done their best, there was every chance that the bit of metal was not actually the right mass so they made the decision that this object was, by definition, correct.
It's easier to compare the mass of two objects than to make exactly cubic decimetre (N.B decimetre, not "decilitre" as you said).
The definition of the litre is the volume of a kilogramme of water at maximum density and it is not the same as the dm^3. As I said, it's out by about 30 parts per million.
"Introducing a third standard would just add more confusion."
Nope,
for practically everybody it wouldn't make any difference at all.
For the people wishing to establish the mass of things accurately it would be a great benefit.
At the moment they have to refer their measurements back to a lump of metal in a vault just outside Paris. Not many people actually have access to it so they have to refer to a copy.
The copies are not perfect.
Every now an then they take the copies back and compare them to the original.
They usually find that all the masses have changed a bit.
So, in order to state their results accurately, the people weighing things have to say which version (i.e. what date) of the prototype kilogram they are referring to.
While changing the definition might cause some confusion, it won't be nearly as bad as having a "standard" that not only changes but changes in an unpredictable way.
"One of the problems with length (as well as volume) is that it is dependent on thermal expansion."
Nope, it doesn't, because you either measure the rate of expansion or you keep the temperature constant.
They didn't choose fused quartz for the container I mentioned earlier because they thought it looked pretty. It has a very low thermal expansion coefficient (of the order of 0.52 part per million per K) and it's easy to keep the temperature of the equipment stable to 0.01K so the errors due to thermal expansion are about a part in 200,000,000 which is normally considered small. (Actually, because quartz is transparent, you can measure the expansion to a silly degree of accuracy, rather than just 2 digits like I quoted, so the error margins are even smaller).
"length of the path traveled by light in vacuum in 1⁄299,792,458 of a second which is a distance that is difficult to measure,"
No, it's actually quite easy to measure. That's why they chose it.
"Anyway, for your task, find a volumetric flask that is calibrated for 1L of water at 25°C (condensation will be a problem at 4°C). Tare your balance with the flask, and fill with water at 25°C, and you'll be all set."
All set to get an answer that's wrong by about 0.1% because you forgot to allow for the density of the air.
"You would probably be better off by purchasing a good set of brass weight standards made anywhere EXCEPT China."
Nope, buy them anywhere, but don't try to get them cheap, and buy stainless steel ones while you are at it- but be sure to ask the supplier for the density of the steel.
"Has anybody ever checked whether the standard kilogram is radioactive?"
It's mainly platinum so of course it's radioactive. Roughly two atoms in ten thousand of it are radioactive with a decay half life of about a million million years.
"It would seem easy enough to go back to the original definition of a kilogram as the mass of 1 liter of water = 1 cubic decimeter of liquid water at the temperature of its maximum density (just below 4°C), and at 1 ATM."
Not when you are trying to make it good to 12 significant figures. Your "box" full of water needs to be of accurately known size. If one side is out by a part in a million million then you have failed.
"Then the scientists can go back and argue until the cows come home as far as how many watts that is actually equal to."
Nope, they won't. Not even if they use a Watt balance to do some of the measurements. The Watt just isn't the right unit for mass- though the Watt second is one contender.
Did you not think they would have looked into this quite carefully?
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Air will not contribute to the weight of a liter of water (unless you are meaning dissolved gases).
I'm sure I've seen discussions on whether an open box full of air would weigh the same as a closed box full of air vs a flattened piece of cardboard.
A meniscus, of course, is another issue, and has to be considered as part of the calculation.
I was surprised to learn that the boiling point of pure water at 1 ATM isn't considered to be 100°C now [:-\]
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It might seem a bit silly to spend so much energy defining a new standard for mass, but mass is such an important construct in all science that, to me at least, basing the standard on a single lump of metal seems slightly unsatisfactory.
For example (not that I'm saying this is the case, and I know it's silly), what if the fluctuations in weight between the standard and its copies were a consequence of preferential variations in the weights of certain elements? Are we sure the compositions of the copies are "exactly" the same as the original? How do we even know what the exact chemical composition of the original is?
No prob - let's cut a chunk off and analyse it........doh!
That's the problem. Nobody can reproduce the current standard, and science is not too happy about things that cannot be reproduced.
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"Air will not contribute to the weight of a liter of water (unless you are meaning dissolved gases)."
Would you care to explain that to these people?
http://www.npl.co.uk/upload/pdf/buoycornote.pdf
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"Air will not contribute to the weight of a liter of water (unless you are meaning dissolved gases)."
Would you care to explain that to these people?
http://www.npl.co.uk/upload/pdf/buoycornote.pdf
Ok...
So if you are weighing a Helium Balloon.
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Not only is the pressure important (1 ATM), but the density of the gas it is being weighed in (Hydrogen, Air, Argon, water, etc).
So, what you actually have is a displacement of a volume of air which has a mass, just like if you weighed the object in water in which case it would displace a volume of water.
So, technically, it would be best to do all masses in a perfect vacuum which would remove the density aspect of the measurement. Except, it is much easier to measure stuff in air, and some things, for example humans, don't react well to being measured in a perfect vacuum.
But.
Once you define your standard such as one cubic decimeter of pure water at its minimum density at 1 ATM. Then the actual density of the medium of measurement is irrelevant except when comparing that to other items with different densities, and thus the "standard" density of the air and the fudge factors. Of course, the definition of 1 ATM also depends on the kg.
So...
If you calculate 1 dm3 of water as having:
H: 1.0079 g/mol
O: 15.9994 g/mol
H2O: 2*1.0079 g/mol + 15.9994 g/mol = 18.0152 g/mole
So, 1 kg of water = 1000 g / 18.0152 g/mol = 55.5087 moles
And you get 5.5087 moles * 6.02214078×1023 molecules / mole
And you get: 3.3428x1025 molecules of H2O, or you can break it down to its constituent parts of Hydrogen and Oxygen atoms (and thus avoiding ions H+, OH-, H3O+, etc).
So...
The mass of 3.3428x1025 molecules of H2O (or the equivalent of hydrogen and water) would be the same independent of pressure, temperature, density of the fluid it is being measured in, or whatever.
And... 3.3428x1025 molecules of H2O at its minimum density at 1 ATM would occupy 1 dm3 regardless of the density of the fluid it is being measured in. All this being in an ideal world, of course, as I have just calculated backwards from mass rather than volume.
Now... Avogadro's Number is also based on the mass of Carbon 12 (12C), and thus is also based on the definition of the Kilo, so one does end up going in circles a bit.
You still have isotopes... 1H, 2H, 3H, 16O, 17O, 18O, and etc, but you could certainly make a universal definition of water, for example 1H216O, or your favorite isotopic mixture.
Earth may expand, shrink, or change shape over time, so the meter must be defined independent of the actual size of Earth.
But, personally, I think there is a problem with allowing the Celsius scale to drift from the 0°C to 100°C melting to boiling point of water, and mass to drift away from 1g water = 1 cc water, not that there is anything intrinsic about water other than it being very common, and that it has been used in the past as a calibration standard.
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The denser the object you use as a standard, the less accurately you need to know the air density to correct for it to a given level of precision of weighing.
That's part of the reason they used Pt/Ir in the first place and it's a reason why water is a poor choice.
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Atomic S
I just bet you meant to say maximum density; the minimum is practically zero.
You are quite correct; pardon my lack of attention. Yes, the maximum density.
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Normally water used as a masspiece is used in multi ton lots, mostly for testing cranes, weighbridges and other applications where you need a masspiece in the range of 20 to 200 tons. It is easier to move water in a tanker and pump it up to where the big bag is than it is to have multiple lowbed trailers and trucks to move large masses of iron there to do the work. It is easy to get a water meter and calibrate it to within 1% at a certain flow rate and pressure, and this enables you to fill a big bag with accuracy of around 1% or better, which is what is needed to do these tests and certifications.
I did not know this. Then again, I don't have much experience with heavy construction or industry.
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Why do you want to do this anyway?
Because of the wavering value of the existing standard, which suggests there should be an experient that could be performed anywhere to establish it.
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but rather one that is based on the number of atoms in a block of Silicon ( as you can get that in multi kilogram lots with impurity levels approaching a single atom per kilogram).
Wow! This stuff would appear to have potential.
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atomic clocks, where they are now small enough to almost literally fit inside a wristwatch,
I want one of those for Christmas.
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Anyway, for your task, find a volumetric flask that is calibrated for 1L of water at 25°C (condensation will be a problem at 4°C). Tare your balance with the flask, and fill with water at 25°C, and you'll be all set.
That will probably work for my local market, but will not be accurate enough for scientific work.
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I was surprised to learn that the boiling point of pure water at 1 ATM isn't considered to be 100°C now
I didn't know that either; water isn't what it once was.
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For example (not that I'm saying this is the case, and I know it's silly), what if the fluctuations in weight between the standard and its copies were a consequence of preferential variations in the weights of certain elements? Are we sure the compositions of the copies are "exactly" the same as the original? How do we even know what the exact chemical composition of the original is?
What difference would the chemical compositions make? If the mass of the standard is 1 kilogram, and if the copy is established by it in a vacuum, then the mass of the copy should also be 1 kilogram no matter what its chemical composition. The question, of course, is, will the masses of the original and the copy remain equal over time? They should unless certain processes alter them. The kinds of processes that can alter them are evaporation, absorbtion from the environment, and radioactive decay. If none of those processes are taking place, then the masses should remain equal whenever they are brought together in the same reference frame.
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"Air will not contribute to the weight of a liter of water (unless you are meaning dissolved gases)."
Would you care to explain that to these people?
http://www.npl.co.uk/upload/pdf/buoycornote.pdf
I am not concerned about the weight of the water but about its mass. The point is to establish an exact mass of 1 Kg. That is unaffected by what it is immersed in.
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I might also add that, as for using the standard mass as a calibration weight, yes, at some point one has to take into consideration the environment in which the final weighing will be done. Suffice it to say, that if one has at hand an accurate calibration mass of 1 kg, and places it on a balance located in air for the purpose of calibrating the balance, the balance thus calibrated will correctly indicate 1 Kg every time an object of 1 Kg is place on it, subject, however, to the complication that the weighed object would have to thave the same density as the calibration mass. If it does not, then one must apply a correction, based on the difference in densities between the two objects and the current density of the air. None of this, however, is definitive of establishing the original standard.
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How would one determine the precise volume of a sample of highly purified water at its minimum density, and then copy its mass precisely to be used as a calibration weight?
Very carefully.
As has become clear in this discussion. Thermal expansion, dissolved air, and menisci being among the complications. I would envision carrying out the entire operation in an air-free chamber. An optically precise fused quartz box would be meticulously sanitized, brought to the temperature of water at maximum density, and its dimensions would then be determined interferometrically with great accuracy. (Alternatively, they could be determined at a different temperature and then corected mathematically if the expansions coefficient of the quartz were known accurately enough). The box, including lid, would be tared on a very sensitive comparison balance inside the chamber. A vessel containing water that had been highly purified in the absence of air so as to eliminate dissolved gases, would be place in the chamber and brought to the temperature of maximum density in equilibrium with its vapor. The lid of the box would be removed, and the box would be immersed in the water. Immersion rather than simple filling would be used in order to keep the pressures on the inside and outside of the box equalized, so as to eliminate distortion from internal water weight. The lid would be also immersed but not fully closed on the box. The system would be allowed to come to thermal equilibrium at the temperature of maximum density The lid would then be slowly closed on the box. At this point we have to hope that no significant number of water molecules become trapped around the rim of the lid where it seats. I don't know what the chances of that would be. Assuming that it is not a problem and that we get a perfect seal, the box would then be removed from the water, the water vessel removed, and all residual water vapor pumped out. That would dry out the outside of the box, so that the only water attached to the box would be what was inside. (We continue to assume that the perfect seal of the lid would prevent any water from the inside from evaporating. The full box would then be moved to the balance, and its new weight would be used through the balance to construct a weight just matching the extra water, all being carried out in a vacuum of course. A very delicate operation.
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Actually,
It would be much easier to measure the mass of water by displacement.
I.E. measuring negative mass [;D]
Take a one cubic decimeter chunk of glass or metal. A sphere? Of course calibrating your volume as necessary.
Weigh it.
Drop it into water (with adequate controls).
Weigh again.
The difference is exactly 1 kilo (assuming one uses the definition 1dm3 of water at maximum density = 1kg).
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Sorry Cliff - you have lost me there. If you have a bowl of water on a scale and add a cubic decimeter object; then the weight shown on your scale will go up by what ever the objects weight is.
If you add the object and displace a similar amount of water (that is no longer part of weight calc) then you will measure the difference between the weight of a litre of water and your object (and you still need a good mass reference). You could measure the water displaced - but that has potential for great inaccuracy.
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Hang the cubic decimeter by a fishing line.
Weigh it.
Drop it in a tank of water, and check the weight again (suspending it by the fishing line)
The difference in weight will be 1 kg.
It just seems like it would be easier to build the perfect decimeter cube than a cubic decimeter tank.
Of course, as mentioned, the big thing is to know exactly how much water is in the tank... or how much water is being displaced, so you would use the same procedure with a 1.5dm cube, or 1.5l tank.
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Thanks Cliff - I reread your earlier post in the light of the above explanation and they both made sense now. Of course the above still suffers from the potential inaccuracies of mass of displaced air and the density of your water again
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Surface tension effects on the fishing line would makle life interesting.
You might find this
http://www.ptb.de/en/aktuelles/archiv/presseinfos/pi2011/pitext/pi110127.html
interesting.
Also, if you are going to use water, could you let me know what water you plant to use?
Most water contains a little heavy water. If you do the experiment with pure H2O (I.e. no DHO) you will get an answer that makes all previous determinations wrong.
On the other hand, how much DHO should you use?
Your best bet would probably be to go with this
http://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water
Incidentally, the cube shaped box full of water wouldn't work. The vapour pressure of the water would lift the lid on the box as soon as it was under vacuum.
You could clamp the lid, but the clamping force would distort the box and the distortion would be different from that of the empty box (because water is practically incompressible).
You could use another interferometric set of measurements to find the size of the clamped box full of water (and surrounded by vacuum) but to do that you would need to know the refractive index of the water (at some poorly defined pressure) to 9 or 10 significant figures.
Have fun measuring that.
As I asked before, do you guys not think that the people who do this for a living have already thought about this sort of thing?
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put it in a sealed plastic bag, weigh it on a digital scale ...
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put it in a sealed plastic bag, weigh it on a digital scale ...
And guess the volume.
With a bit of luck you should get that within a factor of two, which may be good enough for some measurements.
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Take a one cubic decimeter chunk of glass or metal. A sphere? Of course calibrating your volume as necessary.
Weigh it.
Drop it into water (with adequate controls).
Weigh again.
The difference is exactly 1 kilo (assuming one uses the definition 1dm3 of water at maximum density = 1kg).
This could be a good idea. It would eliminate some problems associated with water in a box.
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Thanks Cliff - I reread your earlier post in the light of the above explanation and they both made sense now. Of course the above still suffers from the potential inaccuracies of mass of displaced air and the density of your water again
Do the experiment in an evacuated environment. The only gas will be water vapor near the freezing point. Or, the object could first be weight in high vacuum and then transferred to the chamber having water in equilibrium with its vapor. Once the object is immersed, it is immaterial what kind of vapor pressure exists above the liquid as far as the object itself is concerned; however the fishing line and other apparatus would be affected. One can deal with that problem, or decide to carry out the entire experiment in water vapor and then apply a correction where required, which may be simpler but could cause accuracy problems because the density of the vapor might not be all that precisely known.
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Also, if you are going to use water, could you let me know what water you plant to use?
Most water contains a little heavy water. If you do the experiment with pure H2O (I.e. no DHO) you will get an answer that makes all previous determinations wrong.
On the other hand, how much DHO should you use?
Your best bet would probably be to go with this
http://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water
Undoubtedly.
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Incidentally, the cube shaped box full of water wouldn't work. The vapour pressure of the water would lift the lid on the box as soon as it was under vacuum.
You could clamp the lid, but the clamping force would distort the box and the distortion would be different from that of the empty box (because water is practically incompressible).
That depends on whether the vapor pressure would exceed the weight of the lid. It may not. However, if the lid has weight, then it could introduce distortion. The whole box is subject to distortion, for that matter, by reason of its weight & how it may be held at any moment. Assuming for the moment that the lid would be heavy enough to prevent vapor at around 4 C from pushing it off, the major distortion issue centers around the configuration of the box when it is outside the water, and then when it is inside the water and subject to its bouyancy. This raises a number of issues, of course. The panels of which the box is made should be optically flat (at least). How does can one verify their flatness when they are subject to gravity, which will tend to introduce distortion and that distortion will vary depending upon how they are suspended. Once in water, we might say that they are effectively weightless and therefore undistorted. This suggests manufacturing the panels underwater, but that of course poses other problems. I think the only way you can deal with this is carefully measure the spring constant of the quartz, and do the interferometric measurements in a highly repeatable manner, so that it is known precisely what suspension configuration they pertain to. Then, the changes introduced by placing this system under water can be calculated by a knowledge of the behavior of the quartz combined with the change in suspensional circumstances introduced by immersion. When the system is withdrawn full, another set of distortions come into play due to the now uncompensated weight of the interior water. One might be able to minimize these problems by borrowing technology from modern telescopes in which distortions of the mirrors are dealt with by a fancy, and in some cases dynamic, suspension system. Any way you look at it, these issues make the experiment cumbersome and difficult. But then again, when was good science ever cheap and easy?
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You could use another interferometric set of measurements to find the size of the clamped box full of water (and surrounded by vacuum) but to do that you would need to know the refractive index of the water (at some poorly defined pressure) to 9 or 10 significant figures.
Have fun measuring that
Only if the light actually has to go through the water. There may be some way of measuring by using only light reflected from the outside surfaces of the box.
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Perhaps this experiment is best carried out in space.
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Perhaps this experiment is best carried out in space.
Indeed, that way the weight of the water is know and independent of the volume taken because it's exactly zero.
On the other hand, that might make weighing it a bit tricky.
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Interesting complication; however if the volume of water could be precisely acquired in space, it could then be brought back to Earth to complete the experiment.