Why weight change when I put effervescence tablet ?

[Simon Bridge]

1. How the bubble go there is important - just magicing it in place means you have added mass to the system (the air in the bubble), displaced some water - which compresses the air already there.
You could push the bubble down from the surface somehow ... the bubble displaces water equal to the volume of the bubble (but the bubble probably compresses too).
You could have a bottle of pressurized gas... you get the idea? It adds complications where you want to remove them.

What I'd suggest is doing away with the tablet and the water, adding a valve and a bottle of compressed CO2.
pressurize the inner container from the bottle ... see if you get weight loss.

You could repeat by carbonating some water, also from the gas bottle.

Remember to include the gas bottle in the weighing.

2. Pretty much - pressure changes cannot travel instantly.
But they will be pretty fast for the small size of your system.

 

[Gh778]

2. Pretty much - pressure changes cannot travel instantly.
But they will be pretty fast for the small size of your system.

Watch new image please (theoretical system)

With height of gas = H = 0.1 m
Surface at contact = S = 0.0036 m²
Speed of sound = V = 340 m/s (decrease when CO_{2} is added but increase with pressure)
Gravity = g = 9.81 m/s²
Difference of pressure = P = 100000 Pa
Duration of bubbles = T = 120 s

Modification of force due to the speed of pressure = \frac{PHS}{TVg}

AN = 0.09 g

Pressure at top of water come from :

- each new bubble increase pressure
- bubble increase in size => water move up (very short time)

 

[Buckleymanor]

Where in that link is it demonstrated that the overall weight-reading on a scale decreases?
(Or changes at all?) That's what you need to demonstrate.
All I see in the linnk is that the density of the air inside the stopper changes, changing the buoyancy force on the stopper - not the entire bottle.

In Gh778's setup - the air inside the container increases density because additional gas is being added - even though the bubbles decrease the mean density of the liquid, and there is reduced density from the disintegration of the tablet. The overall volume of the setup does no change and neither does the mass (it has been asserted).


Consider: If a bottle of compressed helium is in an evacuated chamber, and the whole is sitting, in air at STP, on a sensitive balance. The bottle is opened (by an automated mechanism, remote control, something) allowing the helium to escape and fill the room. All else remaining equal: what do you think would happen to the reading on the balance?

You can even imagine the mass of the chamber+bottle is lighter than the air it displaces so that, filled with helium, it would normally float in the air... and we tether it to the balance.

Maybe it is not as obviouse as imagined.
The water inside the bottle represents the air surrounding the container in Gh778's experiment.
The stopper in the link represents Gh778's container.The change in density of the air inside the stopper represents the change in density inside the container.If its sealed properly the container will expand slightly just like the bottle in the video did before it exploded.
It's not a complete sealed system to the atmosphere.
It will either expand, leak, or if too much gas is produced explode.
The change in the buoyancy force of the stopper as the air expands inside it is the same effect the container will have within the atmosphere when the tablet is inserted, bubbles and expansion happen and it becomes more buoyant.

If you were to place a scale below the stopper inside the bottle when you compressed the bottle and as a consiquence the air inside the stopper would also compress, the diver "stopper" would sink.
This would register on the scale if you did the opposite the diver would rise and there would be no reading on the scale.

 

[Gh778]

I measured the additional pressure inside the container (like last drawing show), it's only 38000 Pa with one tablet and a volume big enough for have 0.1 m in height and 36 cm², so the result is 0.034 g. Maybe the lost of weight come from this. This could explain the difference of weight is always the same with different volumes, because If HS (volume) increase, P (pressure) decrease in the same proportion (if temperature is the same).

With height of gas = H = 0.1 m
Surface at contact = S = 0.0036 m²
Speed of sound = V = 340 m/s
Gravity = g = 9.81 m/s²
Difference of pressure = P = 38000 Pa
Duration of bubbles = T = 120 s

 

[Gh778]

I think there is an error in my study with delay due to the pressure. Pressure is not coming at 833 Pa in each second, like steps, it's not an air cylinder very fast. So:

1/ How could I calculate the good differential pressure with a tablet ?
2/ I can replace tablet by air cylinder very fast, this change the Archimedes law ?
3/ With a system at constant speed V=10 m/s, 100 m/s or more in theory, if Archimedes law is changed this would say I can move up less weight on 10 meters for examples and move down the "real" weight. But this would say energy recover can be higher than needed. All energy give by the air cylinder can be recover from pressure of temperature, or energy of accelerating gas goes to momentum of movement ?
4/ Is it possible to reduce weight on a closed system, even gravity is present ?
5/ Why it's not possible to use this system for elevating a rocket ? It's possible to replace gas by water, the only problem is the elevation of temperature.
6/ If there is an up acceleration, the air at bottom add pressure at bottom and cancel a part of move up. But:
6.1/ an object can be on levitation with this ?
6.2/ if the object is heavy compare to mass of air, the up acceleration is only a small part of acceleration of gas ?

 

[Simon Bridge]

Maybe it is not as obviouse as imagined.
The water inside the bottle represents the air surrounding the container in Gh778's experiment.
The stopper in the link represents Gh778's container. The change in density of the air inside the stopper represents the change in density inside the container.

In the archemedes diver, the air stopper (dropper?) changes density because surrounding water from the bottle is pushed into (or out of) it through the hole in the bottom.

Are you supposing that, following your analogy, air from outside the container is forced into (or out of) the container? You realize that the container is supposed to be airtight right?

Finally, the overall density of the dropper + contents, in the diver, changes.
In Gh778's experiment, the overall density of the container + contents does not change.

 

[Simon Bridge]

I think there is an error in my study with delay due to the pressure. Pressure is not coming at 833 Pa in each second, like steps, it's not an air cylinder very fast.

I don't know what you are trying to say. Sorry.
I'm going to try to answer your questions though:

1/ How could I calculate the good differential pressure with a tablet ?

No you can't - you have to measure it.

2/ I can replace tablet by air cylinder very fast, this change the Archimedes law ?

Archemedes law is a law of nature - it is the same for all systems.
Using an a cylinder of compressed gas (I suggest using CO2) and a valve you can pressurze the container quite quickly - it that what you mean?

3/ With a system at constant speed V=10 m/s, 100 m/s or more in theory, if Archimedes law is changed this would say I can move up less weight on 10 meters for examples and move down the "real" weight. But this would say energy recover can be higher than needed. All energy give by the air cylinder can be recover from pressure of temperature, or energy of accelerating gas goes to momentum of movement ?

The laws of physics are the same at a constant speed. Total energy will be conserved as always ... as is momentum. Not all the energy expended will be recoverable though, since some of the processes in your experiment are irreversible.

4/ Is it possible to reduce weight on a closed system, even gravity is present ?

One can reduce the mass in a closed system by turning into energy by E=mc² ... this happens, for instance, in nuclear decay. Apart from that, no, mass is conserved in a closed system ... it is what "closed system" means.
Mass changing normally means that your system is not closed.

5/ Why it's not possible to use this system for elevating a rocket ?

No reason - it is possible to use this system as a rocket. Making high-energy gas at high pressure via a chemical reaction is how most rockets work. BUT - you have to put a hole in the container to get a rocket action.

6/ If there is an up acceleration, the air at bottom add pressure at bottom and cancel a part of move up. But:
6.1/ an object can be on levitation with this ?
6.2/ if the object is heavy compare to mass of air, the up acceleration is only a small part of acceleration of gas ?

The weight-loss of your containers is not evidence of some sort of anti-gravity or levitation method, no.

 

[Gh778]

No you can't - you have to measure it.

I measured the pressure at start and after 2 minutes, the difference is 38000 Pa, but how get the differential pressure and take account of speed of sound ?

One can reduce the mass in a closed system by turning into energy by E=mc2 ... this happens, for instance, in nuclear decay. Apart from that, no, mass is conserved in a closed system ... it is what "closed system" means.
Mass changing normally means that your system is not closed.

not mass but weight, mass is always the same, but weight can decrease with only gravity like external force, nothing else ?

No reason - it is possible to use this system as a rocket. Making high-energy gas at high pressure via a chemical reaction is how most rockets work. BUT - you have to put a hole in the container to get a rocket action.

A part of the system is on the floor. I would like to say, all the system, not a part.

 

[Simon Bridge]

I measured the pressure at start and after 2 minutes, the difference is 38000 Pa, but how get the differential pressure and take account of speed of sound ?

Is http://physics.stackexchange.com/questions/80878/how-speeds-pressure-modify-the-archimedes-law ?

You mean you want to find how the pressure varies within the container?
For that you use several pressure gauges located at different places.
But if you want the equilibrium speed of sound in your gas mixture, you can estimate it.
http://en.wikipedia.org/wiki/Ideal_gas#Speed_of_sound
... you should take Buckleymanor's ideas about buoyancy with some caution.

not mass but weight, mass is always the same, but weight can decrease with only gravity like external force, nothing else ?

Stroctly speaking, weight is the force of gravity on an object. Therefore only a change in the local gravitational field or the change in mass of the object will change the weight.

However, the weight read by a balance is not like that. A balance shows you the net force needed to support the object. If gravity is the only force to balance out, then the scale will read the weight. In your case, this is probably not the case.

I gave you a list of four possible ways the weight-reading could change earlier.

A part of the system is on the floor. I would like to say, all the system, not a part.

Um... OK: it's on the floor... now what?

 

[Gh778]

A friend has let me a balance with a precision of 0.001g and now all is fine the weight is the same. Even this change a little during bubbles move up, it's not more the accuracy.

But now, in a standard bottle, I would like to understand why weight is not greater when bubbles move up, due to the speed of sound in the container. If there is a differential of pressure, the top surface receive pressure with a delay why this don't change the weight ?

 

[Buckleymanor]

In the archemedes diver, the air stopper (dropper?) changes density because surrounding water from the bottle is pushed into (or out of) it through the hole in the bottom.

Are you supposing that, following your analogy, air from outside the container is forced into (or out of) the container? You realize that the container is supposed to be airtight right?

Finally, the overall density of the dropper + contents, in the diver, changes.
In Gh778's experiment, the overall density of the container + contents does not change.

If there was no hole but greater pressure applied to the bottle the same effect would happen.

No air enters or escapes from the container yes it should be airtight.

Yes the overall density of the dropper +contents,in the diver, changes.

In Gh778's experiment, the overall density of the container +contents does not change.
Well it should not but the container is not made from an inelastic substance so if you introduce a tablet that has a chemical reaction and bubbles are produced because it is airtight? what happens to the container.
If it expands the overall density changes and the weight goes down.
If it leaks the weight goes down.
If it explodes the weight goes down.
If the scales are faulty then non of the above matter

 

[Simon Bridge]

If there was no hole but greater pressure applied to the bottle the same effect would happen.

Only if the dropper could change shape.
Note: When you play with one of these, the change in the volume of the dropper is easily seen.

...the container is not made from an inelastic substance so if you introduce a tablet that has a chemical reaction and bubbles are produced because it is airtight? what happens to the container.

The container expands... this has already been covered, several times, earlier in the thread.
Please reread page 1.

tldr: Gh778 reports no visible change in the overall volume of the container.

So - have you calculated how big the change in size needs to be in order to produce the 30-60mg change in the weight reading?

I did one earlier but it was contested and I haven't revisited it since.

Initial weight reading is: $W=mg-\rho Vg \qquad \text{...(1)}$
Final weight reading is: $W-\Delta W = mg-\rho (V+\Delta V)g\qquad \text{...(2)}$
$\quad$...where $\small \rho\approx 1.204$ g/l is the density of the air at room temperature and sea level.

Subtract (2) from (1): $$\Delta W = mg-\rho Vg -\big(mg-\rho (V+\Delta V)g\big)\\
= \rho \Delta V g\\
\Rightarrow (\Delta x)^3 = \frac{\Delta W}{\rho g}
$$​

... would give the linear dimension change for isomorphic expansion.
Let's see how big this is:

$\Delta W = (0.01g)\text{ grams}\\ \rho= 1.204\text{ g/l} = 0.001204\text{ g/cm}^3$
$$\Delta x = \left(\frac{0.01}{0.0001204}\right)^{1/3} = 4.4\text{ cm}$$​

... that's for a 10mg weight-loss ... 30mg is 6.3cm and 60mg is 7.9cm ... the kinds of figures reported.
... I think he'd notice don't you?

 

[Simon Bridge]

A friend has let me a balance with a precision of 0.001g and now all is fine the weight is the same.

I'm sorry, you need to be clear in your meaning.

Are you saying that there is zero weight-loss when using the new balance?

 

[Gh778]

I had done several tests and I can say the weight decrease a little more than the accuracy (0.001g) of a good balance. But it can be a problem of a gasket or anything else.

The rotation of Earth is not in equation ? for 1kg, the rotation decrease pressure of 0.03 N (centripetal forces) at 6400 km, if water move up (with lower density), the centripetal force must decrease a little more the weight, no ?

 

[jbriggs444]

I had done several tests and I can say the weight decrease a little more than the accuracy (0.001g) of a good balance. But it can be a problem of a gasket or anything else.

The rotation of Earth is not in equation ? for 1kg, the rotation decrease pressure of 0.03 N (centripetal forces) at 6400 km, if water move up (with lower density), the centripetal force must decrease a little more the weight, no ?

Are you suggesting that because the water becomes less dense and moves up slightly, it is then farther from the center of the Earth and subject to greater centripetal acceleration?

Can you calculate the expected magnitude of such an effect?

Can you design an experiment to determine whether (for instance) a one meter change in elevation will result in a change in apparent weight that is detectable with your apparatus?

 

[Simon Bridge]

I had done several tests and I can say the weight decrease a little more than the accuracy (0.001g) of a good balance. But it can be a problem of a gasket or anything else.

If two weight readings differ by little more than the accuracy of the scale, then that is no difference at all and your results are an artifact of the measuring process.

Presumably if you used scales capable of 10x more accuracy you'd get 1/10 again the weight loss.
It is not reasonable to think that the process somehow knows the accuracy of your scales and adjusts itself to suit right? So it must be something to do with the scales themselves that we have yet to identify.

http://www.slideshare.net/diverzippy/uncertainty-and-equipment-error

I suspect small wobbles throwing the calibration off.