Why weight change when I put effervescence tablet ?

[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.