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: **
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
When I blow on the short tube, water rises in the vertical column and air compresses in the barometric tube because I am adding air pressure into the system. When I stop blowing and cap the short tube while preventing air from escaping, the excess air pressure is maintained in the system and the other tubes hold their positions, as I would expect.
** What happens when you remove the pressure-release cap? **
When I uncap the short tube, air escapes from it to return to atmospheric pressure and the other two tubes return to their previous states, as I would expect.
** What happened when you blew a little air into the bottle? **
When I blow air through the vertical tube, it bubbles through the water, increases the air pressure of the system, and compresses air in the barometric tube. When I stop blowing, water rises in the vertical tube in response to the increased pressure but air is not able to escape, so the barometric tube remains unchanged. The only thing I wasn't expecting was the force with which water would be ejected from the vertical tube when I stopped blowing and removed it from my mouth. I know now what it feels like to attempt suicide with a water pistol.
I hadn't thought of the potential danger; I should probably put a warning note in the experiment. I hate to lose the element of surprise, which can be a valuable learning tool, but I really don't want any half-drownings.
** 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: **
1N/m^2
1cm
3degK
1% of 100kPa is 1kPa, if water raises 10cm for a 10kPa change, it would raise 1cm for a 1kPa change, at constant volume, pressure and temperature are inversely proportional, so temperature would change by 1% or 3degK.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3degK
1kPa
1cm
These equivalent values follow from the knowledge that at constant volume, pressure and temperature are inversely proportional
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
3degK
.3degK
1cm is the same height I already used in the calculations above. .3 is to 3 as 1cm is to 1mm.
** water column position (cm) vs. thermometer temperature (Celsius) **
20.5,24.5
20.5,24.7
20.5,24.8
20.5,24.9
20.5,25.0
20.4,25.0
20.4,25.0
20.5,25.0
20.5,25.0
20.5,25.1
20.5,25.1
20.5,25.1
20.5,25.0
20.4,25.0
20.4,25.1
20.5,25.1
20.4,25.1
20.4,25.1
20.4,25.0
20.4,25.0
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The vertical tube meniscus only differed by 1mm. After initially climbing .5deg, the thermometer only differed by .1deg during the rest of the observations.
** Water column heights after pouring warm water over the bottle: **
70
37
28
20
14
10.5
9
7.8
7.3
7.0
6.4
6.2
5.5
5.2
5.0
4.8
4.6
4.5
4.4
4.2
4.1
4.0
4.0
It took approximately another 5 minutes to return all the way to .5cm above the original mark.
** Response of the system to indirect thermal energy from your hands: **
When I held my warm hands against the bottle, the water slowly rose until stopping about 6cm above the original mark.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
24.9,100.2
25.1,100.2
25.0,100.1
25.0,100.1
24.9,100.1
25.0,100.2
25.1,100.0
25.1,100.0
24.9,100.1
25.1,100.0
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
I encountered the following problem with this part of the experiment: I was able to get the meniscus back 14in. from the end of the tube, but the vertical tube is so sensitive, that when I hold my warm hands against the bottle, within a few seconds, the fluid reaches the end of tube. To attempt to compensate for this, I tried releasing the pressure in the system and raising the meniscus to a lesser level, but it seems 35cm is approximately the limit for a stable meniscus in this particular system with a 30cm vertical component and a 90cm horizontal component. Whenever I tried to get the meniscus to stop somewhere shorter, it wouldn't stay still and kept retreating back toward the vertical part of the tube.
** 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: **
Only a very tiny amount of pressure difference caused the water to overflow in the horizontal tube versus when the tube was vertical and the heat generated by my hands only caused the water to rise a few cm. As I also explained above, I also encountered considerable difficulty with backflow in the horizontal tube, as further evidence of the horizontal tube's hypersensitivity.
.707cm^3
.07%
0.2086K
If temperature were doubled at constant pressure, volume would have doubled.
.3cm diameter=.15cm radius. Cross sectional area=`pi*.15^2=.0707cm^2. Volume=.0707*10=.707cm^3. As the bottle is approximately half full, I estimate I started with 1L of air. So .707cm^3 is 0.000707L which is .07% of the original volume. Assuming an initial temperature of 298K, .07% of 298K is 0.2086K. .07% of 600K is .42K
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The vertical tube more closely resembled a constant volume system in that the volume of air only changed slightly with temperature compared to the horizontal system in which volume changes were more dramatic.
** 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: **
.042%
.126K
.0233%
If a .07% pressure increase is required to move 10cm, then .6*.0007=.00042=.042% pressure increase would be required to raise the water 6cm. .00042*300=.126K. .7cm^3 is .0233% of 3L, so a proportional temperature change would be required for this volume change at constant pressure.
** 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: **
47.6cm
3.33cm^3
.500K
If a .126K temperature increase is required to raise the water 6cm, 1K would raise the water 47.6cm. If .7cm^3 is inside a 10cm length of tube, 3.33cm^3 is inside a 47.6cm length of tube. .5*47.6=23.8cm. This would correspond to to a temperature increase of .500K.
The importance of having the tube in a horizontal position is demonstrated by the fact that gravity interferes with the measurement whenever there is a vertical component to the section of tubing containing the meniscus and this may provide less sensitive results than may be desired. Although for my personal setup, I found that the vertical setup provided more useful measurements in that it raised the water level noticeably but did not tend to overflow like the horizontal tube when exposed to a modest temperature increase.
** Optional additional comments and/or questions: **
Excellent work, including a good discussion of the hypersensivity of the horizontal tube.