# Homework Help: Thermal expansion of water

1. May 19, 2014

### Myr73

A glass bottle is filled with salted water, and a pipette is inserted in the top through the cork. ( Leaving the bottle sealed). A thermometer is also inserted to keep record of the temperature. The glass bottle is then inserted in ice until the water reaches Zero Celsius. Finaly, the temperature and the rise of the water in the pipette was observed.
A graph V(cm^3) vs T(C) was also created.
In this process, did the expansion of the plastic pipette and the glass(regular) affect the data of the rise and fall of the water in the pipette . Explain.

I have that the expansion in the plastic pipette would indeed affect the data, because that is what the reading of the water is done through. And therefore, say the water would have risen 3cm's but because the expansion of the plastic it only rose 2.5cm, then the height would be smaller, also the cross sectional area would be larger, and so the V(cm^3) would be different. Since,Change in V= Change in h X A . I think that is correct.
However when it comes to glass , I can't see how it would affect it because even if the bottle expands and water does not go as high in the bottle, it still rises in the pipette the same height. But I am unsure of this because it seems to me like it should affect it somehow, but how?

2. May 19, 2014

### haruspex

As you say, the volume of the bottle changes (but not increasing, I think).
Write an equation connecting the following:
- the change in volume of the bottle
- the change in volume of the water
- the change in volume of the portion of water that is in the pipette

3. May 21, 2014

### Myr73

Sorry, I forgot to mention that once the temperature reached abit below 0C, it was removed from the cold ice and so the temperature then started to rise. So yes, I believe the glass bottle would contract at first but then it would expand as it heats up. But I still can't see how it would affect the rise of the water in the pipette.

4. May 21, 2014

### haruspex

Let volume of bottle be B, volume of water be W, volume of water within pipette be Wp. What equation relates them?

5. May 21, 2014

### Myr73

Volume Wp-->Change in h= h- h0-->Change in V(wp)= Change in h X A
Change in Volume W=0
Change in Volume B= I don't know, this is only a small part of the work, the expansion of everything but the air and water was first ignored, and then simply asked if the glass expansion would affect the rise in the pipette.

6. May 22, 2014

### haruspex

Forget about the expansion for the moment. At some instant, let volume of bottle be B, volume of water be W, volume of water within pipette be Wp. What very simple equation relates them?

7. May 22, 2014

### Myr73

If you mean volume of water in bottle , then it would be --> W= B + Wp
But if not you meant specifically the volume of of the bottle. then I am not sure

8. May 22, 2014

### haruspex

Isn't the bottle full at all times? How is the volume of the bottle different from the volume of water in it?

9. May 24, 2014

### Myr73

Because when the water from the bottle rises into the pipette, the bottle is no longer full.

10. May 24, 2014

### haruspex

If the bottle is no longer full of water, what else is in the bottle? There's no airspace, right? Why does the water rise in the pipette if the bottle is not full?

11. May 26, 2014

### Myr73

Because the temperature is changing.

12. May 27, 2014

### Myr73

I think the formula for finding the change in height would be close to this,like a thermometer, except it wouldnt be quite the same, because the pipette is plastic,the liquid is water of course, and the liquid is not coming from the pipette itself but the glass bottle..--> Delta L= Vo / A { bwater- bglass}Delta T. Where b water is the coefficient of volume expansion for water and bglass is that for the bottle. However I know it wouldnt be precisely this because again the pipette is plastic. It is trying to find how to incorporate the bglass that I am having trouble with..Any ideas?? Thanks

13. May 27, 2014

### haruspex

How does that make the bottle not full?
Yes, it is just like a thermometer. When the temperature rises, the liquid rises in the narrow part because it has expanded more than the bulb has, and the bulb is full, so the only way for the liquid to go is up. When that happens with a thermometer, does a space appear in the bulb? No, it remains full.

14. May 28, 2014

### Myr73

This is how I understand it to be.Because when the temperature rises, the water leaves the bottle and enters the plastic pipette., witch is only slightly inserted in the bottle, the rest sticks out. When I tested it, the water level went down in the bottle.because some of the water rose into the pippette. But,if it is the same as a thermometer, how do I include the plastic pipette in the formula. The coefficient of volume expansion for plastic is not provided.
Thank you

15. May 28, 2014

### haruspex

If the water level went down in the bottle it was because you had some air trapped in it. The bottle should have been full, no air. When heated, the air expanded and pushed the water out of the bottle. That would have been a much bigger effect than either the expansion of the water or the expansion of the bottle.
Re the expansion of the pipette, its volume is so small that its expansion would be negligible.

16. May 29, 2014

### Myr73

Ok , thanks

17. May 30, 2014

### Myr73

Ok, So I have been using the data that was provided to me by the teacher and the above equation I had mentioned, in order to find the coefficient of volume expansion for the water at different temperatures. We were provided a list to compare it to , with the actual precise coefficients. I expect it to be a little off the list,however my answers seem to be way off.
I had used the following equation , where b is the coefficient--> {DeltaV/Vo Delta T} + b(glass)= b(water)
Any ideas what I have done wrong?

18. May 31, 2014

### haruspex

Please clarify: the data that you are now working with comes from your teacher, not from your own testing, yes? So for these data, the water level does not go down in the bottle?

I don't understand the equation you just posted. You seem to be adding a coefficient to some other term. That's not what you do with coefficients - you multiply them by some other term.

19. May 31, 2014

### Myr73

So I began with the equation --> Delta V= b Vo Delta T --> That is provided in the book--> then I used both coefficients , that is the one of the expansion of glass and the one of the expansion of water. I figured if the glass expands, the water level would not rise as much in the pipette , therefore b= b(water) -b(glass).
--> Delta V= (b (water) - b(glass)) V0 Delta T--> in trying to find b(water), the equation becomes --> b(water)={DeltaV /V0 Dela T} + b(glass).
One I had to find was for 10 C ,therefore I took the
T1=9.5C T2=10.2C V1=318.9732cm^3 V2=318.9799cm^3 b(glass) = 27 X 10^-6C

V0= 0.319L delta V== 0. 0000067L delta T= 0.7 C
My answer is b(water)= 5.70 X 10^-5 (C ) ^-1
the correct b(water) for 10C is 0.088 X 10^-3 (1/k) I believe that is 88 X 10^-6 (C) ^-1

These are the answers I got the closest, the rest were much bigger differences.

20. May 31, 2014

### Myr73

and yes the data is from the prof. I am unsure, if the water level in the bottle went down, I am guessing not, if you are correct, on how it should work.

21. May 31, 2014

### haruspex

I get the same answer from those data.
Quoting the two volumes that way doesn't make a lot of sense. There is no way they could have been measured that accurately. The real data, I presume, are an initial volume of around 0.319L, or a bit under, and a change in volume of 6.7E-6L as seen by the rise in the pipette.
You should estimate the error range for both the change in the volume (how accurately could the height be measured? how accurately is the pipette cross-section known?), and the change in temperature (how accurate is the thermometer reading?).
However, I struggle to explain an error of 50% in the final answer that way.
Where are you getting the true value for water from? http://en.wikipedia.org/wiki/Thermal_expansion and http://www.engineeringtoolbox.com/volumetric-temperature-expansion-d_315.html give 2E-4 at 20C.

22. May 31, 2014

### Myr73

23. May 31, 2014

### haruspex

OK.
Note how the value changes dramatically from 0 to 5 to 10 to 15C. It is quite possible that it is less than 88E-6 between 9.5 and 10C, so your answer need not be far wrong.
You mentioned other results that were 'worse'. Could you post those?

24. Jun 2, 2014

### Myr73

Ok, Thanks... Here they are-->
at 1C --> T1=0.7C T2=1.6C V1= 318.9665 cm^3 V2=318.9598cm^3
--> b(water)==0.00005(C^)-1 or is it ( because Delta V is minus) b=3.66 X 10^-6 C^-1

at 4C --> T1=3.2C T2=4.4C V1=318.9531cm^3 V2=318.9531cm^3
Delta V= 0 Delta T= 1.2C
Delta V=( b(water)- b(glass)) V0 DeltaT--> (since Delta V is 0)--> b(glass)=b(water)= 27 X 10^-6 (C^)-1

at 7C --> T1= 6.8C T2=7.9C V1=318.95645cm^3 V2=318.9598cm^3
Delta V= 0.00335cm^3=0.0000035L Delta T=1.1C --> b(water)=3.697 X 10^-6 ( C)^-1

at 20C--> T1=18.4C T2=21.5C V1=319.1072cm^3 V2=319.1742cm^3
Delta V= 0.067cm^3 =0.000067L Delta T= 3.1C--> b(water)=9.475 X 10^-5 (C) ^-1

25. Jun 2, 2014

### haruspex

I plotted up your results and the official numbers from the link. Looks to me that adding 1E-4 to all your results brings the two sets of numbers into reasonable alignment. But that would imply an expansion coefficient for the glass about 5 times that quoted. You might care to check my analysis.
I don't think I can help any more on this.