Phy202
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? **
What happened is exactly what I thought was going to happen. The level of the water dropped once I removed my mouth and put the pressure-valve back on the tube but then stopped demonstrating that air was quickly removed and lowered the water level but then was stationary once the airway was blocked by the pressure-valve.
** What happens when you remove the pressure-release cap? **
Yes, air was removed from the tube and the water dropped down back into the 2-liter like I thought would happen once I removed the pressure valve. The air escaped the system because there was no pressure-valve holding it in place.
** What happened when you blew a little air into the bottle? **
The air column moved back to its original position when I removed the tube from my mouth. Water rushed out of the vertical tube after blowing air into it because of more pressure added inside of the 2-liter is excess air and needs to escape through the vertical tube.I anticipated the movement of the air column but not the water coming out of the vertical tube. Thats cool.
** Your estimate of the pressure difference due to a 1% change in pressure, the corresponding change in water column height, and the required change in air temperature: **
1000 N/m^2
10cm
303K
1% of 100kPa equals 1 kPa which is 1000 pascals or 1000 (n/m^2), the water column would change in a heigh of 10cm based on the above experiment and the formula of (P1/T1) = (P2/T2) , would then be 100kPa/300K = 101kPa / T2, then would be T2 = 300K (101kPa/100kPa) = 303K
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3K
1kPa
3cm
These numbers above were found by using the similar equation used in the previous problems of .01*300 for the first question = 3K, then by solving 100kPa (303K/300k) = 101kPa which would be a change of 1kPa. The height column would change about 3 cm based on the kelvin scale of around 3 cm change per degree change.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
.33K
.033K
If they are proportional to each other (roughly 3:1 ratio) in temp and height then .33K temp degree change would equal 1 cm in change, and .033 temp degree change would be 1mm in change.
** water column position (cm) vs. thermometer temperature (Celsius) **
26.1,0
26.0,-.5
26.1,0
26.0,-.5
26.2,.5
26.1,0
26.2,.5
26.0,-.5
26.0,-.5
26.1,0
26.1,0
26.1,0
26.2,.5
26.1,0
26.1,0
26.0,-.5
26.0,-.5
26.0,-.5
26.0,-.5
26.0,-.5
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The trend seemed to stay relatively stationary with a couple fluctuations up and below the clear tape on the water column (only changing at the extremes to .5cm from the tape line). The maximum deivation would have to be 1cm/20 intervals = 0.05
** Water column heights after pouring warm water over the bottle: **
the rise imediatley went up to around 10 cm above the tape once the warm water was poured then slowly went down back to the tape after a period of about 4 min and then remained there until it was back to room temp.
** Response of the system to indirect thermal energy from your hands: **
Yes, but only by maybe .3cm or less...not enough to really tell. It did move a little though.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
26.1,0
27.1,3
27.7,4.4
27.5,4.1
27.5,4.1
27.1,3
26.6,1.6
26.5,1.2
26.4,1
26.3,.8
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
It moved up more above the tape than the vertical tube did. Around 3 cm but still not very much.
** 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: **
50 N/m^2
70mL
7%
6C
15C
The 50 N/m^2 was found by multiplying 1000(3/60) = 50, then 70 mL was found by the formula for volume of (3.14)(1.5^2)(10) = 70mL, then 7% was found dividing that 70/1000 of the total to get .07*100 = 7%, then the 6 C was found by 300K(70mL/1000mL) = 21C, 300K-273K = 27C - 21C = 6 C, that 300 substituted with 600 would give you 15C.
You didn't use units in your volume calculation. 10 cm of tube corresponds to much less than 70 mL; to get 70 mL into a 10 length of tube it would have to have diameter greater than that of your thumb.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The change was not significant to the pressure and temp so the volumes were held at constant with each other and would not have made a difference.
** 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: **
60kPa
120K
70K
To find 60kPa take 100(6/10) = 60kPa, then to find 120K take 300K (60kPa/ 100kPa) = 180 and 300-180 = 120K, then to find 70K you take 300K (.7L/3L) = 70K
** 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: **
1cm
4cm
25 (1:4)
This is seen from the equations before where V1/T1 = V2/T2 and that the ratio of horizontal is 4:1 making a 25 degree slope
** Optional additional comments and/or questions: **
2 1/2 hrs.
** **
You appear to have an excellent understanding of the system, as well as good data.
You did have one pretty big error at one point in your analysis, which resulted from a failure to use units properly. Be sure to review my notes and let me know if you have any questions.