Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
There is a whole in the pressure tube but I will try to fix it with tape.
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
I'm not sure how to get the pressure tube to have any affect but the water just falls back down like normal. I thought it would stay in the tube because of increased air in the bottle causing increased pressure but I guess not.
This is usually not a problem. Did you cap the end according to instructions?
** What happens when you remove the pressure-release cap? **
Yes, the air will escape because it is under pressure inside the bottle so it will be forced out.
** What happened when you blew a little air into the bottle? **
I still do not see a air column in the pressure indicating tube. Nothing happens in the pressure indicating tube and the water goes up in the vertical tube so I assume the pressure increased.
It changed length because the pressure increased and more air was forced up the tube. Yes, it moved back to its original position.
The water traveled up the vertical tube.
Because the air increased the number of molecules in the bottle increased which means increased collisions with the container which means increased pressure.If the pressure goes up the water should rise.
** Your estimate of the pressure difference due to a 1% change in pressure, the corresponding change in air column height, and the required change in air temperature: **
if the gas pressure changes by 1% it changes by 1000 N/m^2.
If the pressure increases by 1% then the temperature increases by the same amount according to PV=nRT when P/T = constant. P2/P1 = T2/T1 so the temperature must change by 1% as well.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3K
1%
1%
All of the values are proportional to each other so it is also a 1% change when P2 and T2 are proportional.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
9K
.9K
This is obtained because assuming the top of the bottle is about 30 cm high so 1 cm is about 3.33% so the temperature must raise the same amount 3*3.33 = about 9K
If the temperature was 900 K then a 1% change would correspond to 9 K. If atmospheric pressure corresponded to a water column 30 cm high then about 9 K would be right.
However a 1 cm different does not correspond to anything close to 1% of atmospheric pressure.
How much extra pressure does an extra cm of fluid depth add? What proportion is this of atmospheric pressure? What is the temperature of the system, and how much temperature difference would this proportion correspond to?
** water column position (cm) vs. thermometer temperature (Celsius) **
I cannot get the data because the hole is about 5cm from the top of the bottle and the air is escaping even with packaging tape around it.
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
No data
** Water column heights after pouring warm water over the bottle: **
The water slowly rises in both tubes but much more in the vertical tube it changes about 1 cm average. I think the pressure tube level shouldn't rise as much as it did but the leak may cause less pressure in the end of the tube.
** Response of the system to indirect thermal energy from your hands: **
Yes, it seems that the hands could have warmed the air which warmed the bottle which warmed the air inside and it is noticeable. The water raised about 6cm.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
24.5 0
24.5 0.1
24.6 0.5
24.5 0.3
24.6 0.6
24.6 0.7
24.5 0.3
24.4 0.3
24.25 0.1
24.3 0.2
This data is the second set because the first set seemed to be two high with as much as 3 cm changes which were probably due to gravity.
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
The water moves through the tube towards the end quicker than when the tube was vertical and farther as well about 32 cm.
** Pressure change due to movement of water in horizonal tube, volume change due to 10 cm change in water position, percent change in air volume, change in temperature, difference if air started at 600 K: **
The pressure decreased proportional to the amount of water leaving the bottle.
7.07*10^-7 m^3
depends on how much water is in the bottle, the same percent
the same percent
it would double
The pressure required to push the water across the horizontal was rather constant and increasing but as the water leaves the bottle the pressure decreases inside so more pressure must be applied to the outside.All precents should be equal if all other variables are constant.
You need extra pressure to support the increased depth of water in the tube. The volume of water in the tube is very small compared to the volume in the bottle; only when the tube is near horizontal is this volume significant.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
Because the container deformed as well. Yes
** Pressure change to raise water 6 cm, necessary temperature change in vicinity of 300 K, temperature change required to increase 3 L volume by .7 cm^3: **
2%
306K
70K
If the pressure increases then the altitute increase proportionally but when the tube is slanted the force required to move the water is less because less gravity is resisting. P1/P2= T1/T2 to find T2 is 306K. V1/V2 = T1/T2 to when T1 is 300K to find the required change in temperature to be +70K to increase the volume of the gas by .7 cm^3.
A temperature change of 70K at near-constant volume would increase the pressure in the system by over 20,000 Pa. That would support a water column 2 meters high. If kept at constant pressure a 70 K rise would increase the volume of the system by over 20%; the system contains about 1.5 liters of water, or 1500 cm^3, so that expansion would amount to around 300 cm^3, not .7 cm^3.
** The effect of a 1 degree temperature increase on the water column in a vertical tube, in a horizontal tube, and the slope required to halve the preceding result: **
.01L
.01L* cos(45)
cos(67.5)
I'm not sure how to convert from liters to cm but the volume changes in L are as shown where if the temp. increases by 1% the volume does as well.
It isn't, but it should be general knowledge that a liter is 1000 cm^3, and also that a m^3 os 1000 liters. In any case it is expected that you know the basic units of measure and how they are related. You can always refer to the inside front cover of the text (which you should simply learn), but you should know these units well enough that you never have to do so. You can't think about these quantities if you don't know the units.
** Optional additional comments and/or questions: **
5
** **
You are making errors related to the units and to Bernoulli's equation. You pretty much understand the gas laws but the units are getting in your way.
You have a good concept of the effect of slope in the tube.
Did you cap the tube as instructed? You might have to trim the top so it's 'square' to the tube, rather than angled. If the tube is leaking around the bottlecap you can always pull a little more of it through, in either direction, to tighten it.
See my notes.
&#Let me know if you have questions. &#