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Author Topic: How does one calibrate balances using water?  (Read 13588 times)

CliffordK

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How does one calibrate balances using water?
« Reply #25 on: 02/07/2011 08:25:55 »
Actually,

It would be much easier to measure the mass of water by displacement.

I.E.  measuring negative mass

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).
« Last Edit: 02/07/2011 08:36:23 by CliffordK »

imatfaal

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How does one calibrate balances using water?
« Reply #26 on: 03/07/2011 10:46:57 »
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.

CliffordK

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How does one calibrate balances using water?
« Reply #27 on: 03/07/2011 11:42:20 »
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.

imatfaal

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How does one calibrate balances using water?
« Reply #28 on: 03/07/2011 12:14:01 »
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

Bored chemist

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How does one calibrate balances using water?
« Reply #29 on: 03/07/2011 13:39:13 »
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.

CZARCAR

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How does one calibrate balances using water?
« Reply #30 on: 03/07/2011 22:02:58 »
put it in a sealed plastic bag, weigh it on a digital scale ...

Bored chemist

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How does one calibrate balances using water?
« Reply #31 on: 04/07/2011 18:53:03 »
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.

Atomic-S

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How does one calibrate balances using water?
« Reply #32 on: 16/07/2011 04:59:09 »
Quote
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.

Atomic-S

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How does one calibrate balances using water?
« Reply #33 on: 16/07/2011 05:03:58 »
Quote
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.

Atomic-S

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How does one calibrate balances using water?
« Reply #34 on: 16/07/2011 05:11:03 »
Quote
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.

Atomic-S

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How does one calibrate balances using water?
« Reply #35 on: 16/07/2011 05:26:36 »
Quote
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?
« Last Edit: 16/07/2011 05:33:52 by Atomic-S »

Atomic-S

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How does one calibrate balances using water?
« Reply #36 on: 16/07/2011 05:30:55 »
Quote
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.

Atomic-S

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How does one calibrate balances using water?
« Reply #37 on: 16/07/2011 05:37:32 »
Perhaps this experiment is best carried out in space.

Bored chemist

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How does one calibrate balances using water?
« Reply #38 on: 16/07/2011 19:20:59 »
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.
« Last Edit: 16/07/2011 19:24:05 by Bored chemist »

Atomic-S

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How does one calibrate balances using water?
« Reply #39 on: 28/07/2011 01:50:12 »
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.

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How does one calibrate balances using water?
« Reply #39 on: 28/07/2011 01:50:12 »