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? **
Removing the cup and drawing air and water through the tube will raise the level of the water in the tube. Replacing the cap and then removing suction on the tube will allow some water to remain in the tube. This happened as I thought. Some air in the system was removed from the tube while the same amount of air remained in the bottle, thus with more air overall. The same amount of water is present in the system. The air, having no place else to go, forces the water into the tube.
** What happens when you remove the pressure-release cap? **
Air will escape. Extra air is in the bottle and can escape easily through the uncapped tube allowing the water level in the tube to drope It does what I expected.
** What happened when you blew a little air into the bottle? **
The water level in the vertical tube raises, and the air column moves right and stays there. By adding air to the bottle, this increases the pressure of the air on the water in the pressure-indicating tube, causing it to move right. The air on that end of the tube also increases in pressure.
** 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: **
1000 N/m^2
.102 m
.01
Using 100,000 N/m^2 as a rough estimate of atmospheric pressure, and multiplying it by .01 (1%), yields 1000 N/m^2. Using `dP = `rho * g * h, Solving this for h, I get `dP/(`rho * g) = h. Using 1000 kg/m^3 for `rho, 9.8m/s^2 for g and solving for h, gives .102 m as the increase in h. Using PV = nRT, solving for nR = (PV)/T. Setting PV/T before equal to PV/T after will let us solve for T after to see how much the temperature has changed, which turns out to be 3 K.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3 K
100333 N/m^2
.034 m
1% temperature change: 300K * .01 = 3K
For pressure: 300K/100000N/m^2 = 301K/P2; P2 = 301K/(300K/100000N/m^2)
For hegiht: 100333 N/m^2 - 100000 N/m^2 = 1000kg/m^3 * 9.8m/s^2 * h; h = 0.034m
You would need 303 K to get a 1% increase in temperature, and this would support a column about 0.1 m high.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
0.294 K
0.0294 K
For each, P = 100000 N/m^2 = 1000kg/m^3 * 9.8m/s^2 * h; using 0.01m and 0.001m respectively and using the result in the relation T/P to find the corresponding temperature for the pressure found at that height. The differences in the temperatures are what are reported on the first two lines.
** water column position (cm) vs. thermometer temperature (Celsius) **
10.2,23
11,23
10,23
10,23
9.9,23.1
9.8,23.1
9.8,23.1
9.8,23.1
9.6,23.1
9.5,23.1
9.4,23.1
9.2,23.1
9,23.1
8.8,23.1
8.6,23.1
8.5,23.1
8.2,23.1
8.2,23.1
8.1,23.1
8,23.1
7.8,23.2
7.5,23.2
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature of the alcohol thermometer increased, while that of the bottle thermometer appeared to decrease. The standard deviation of the alcohol thermometer was 0.0526 and that of the bottle thermometer 0.921.
** Water column heights after pouring warm water over the bottle: **
30
21.5
15.9
12.4
9.5
7.6
6.3
5.3
4.6
4.1
3.8
3.6
3.4
3.3
3.25
3.2
3.2
3.2
3.2
3.2
3.2
3.3
3.3
3.3
3.3
These measurements were taken in cm from the initial water level 30cm above the water level in the bottle. Measurements are from this point and the initial water level was at 0cm.
** Response of the system to indirect thermal energy from your hands: **
Yes, by about 1 degree C.
P = 100000 N/m^2 + 1000 kg/m^3 * 9.8 m/s^2 * (0.073m - 0.035m) = 100372 N/m^2
Using T/P ratio: 295.15 K / 100000 N/m^2 = T / 100372 N/m^2, T = 296.25 K
So `dT is about 1 degree.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
6,22
6,22
6,22
6,22
6,22
6,22
6,22
6,22
6,22
6,22
These measurements are in cm from the same mark used in the last section and degrees Celsius.
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
60.3
52.1
41
34.5
27.3
21
16.2
13.5
12.7
12
10.8
10.3
9.1
8.8
8.5
8.0
7.0
5.2
3.3
2.0
1.5
1.4
1.3
1.3
1.3
1.3
1.3
1.3
1.3
** 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: **
I expect the pressure to change as much as when the tube was vertical.
.707 cm^3
.707 cm^3
7.07 * 10^-4
.21 K
.42 K
The volume is 10cm * `pi * (0.15cm)^2 = .707 cm^3
Percentage increase is .707 cm^3 / 1000 ml= 7.07 * 10^-4
Using V/T = nR/P and assuming 1000 ml of air initially in the container and an initial temperature of 300 K, T2 = 1000.707 ml /(1000 ml/300 K) = 300.21, so the temperature change is about 0.21 K. If it were 600K, it would be 600.42 or 0.42 K change.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The percentage change in volume is so small that it is really insignificant.
** 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: **
588 N/m^2
1.7 K
.07 K
`dP = 100000 N/m^3 * 9.8m/s^2 * .06m = 588 N/m^2.
T2 = 300 K / 100000 N/m^2 * 100588 N/m^2 = 301.7 K; 301.7 K - 300 K = 1.7 K
T2 = 300 K / 3000ml * 3000.7ml = 300.07 K; 300.07 K - 300 K = 0.07 K
** 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: **
0.034 m
0.099 m
30 degrees
PV/T = nR; 100000 N/m^2 * 3000 ml / 300 K = 3000ml * P / 301 K; P = 100333 N/m^2. `dP = 333 N/m^2, so h = 333 N/m^2/(1000 kg/m^3 * 9.8 m/s^2) = 0.034 m. 100333 N/m^2 * .7cm^3 = 100333 N/m^2 * 7 * 10^-7 m = 0.07J; P = F/A, so 10333 N/m^2 * (0.0015 m)^2 * `pi = 0.71 N; 0.07J / 0.71 N = 0.099 m. At a 30 degree slope 0.099 m * sin 30 degrees is about 0.049 m
** Optional additional comments and/or questions: **
Very good. See my brief notes and let me know if you have questions.