Phy 202
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: **
I had a lot of trouble with the last question so please include notes.
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
When the water was pulled out of the tube, the pressure gauges water level decreased, This is because the pressure inside the bottle was decreasing as water was pulled out. When i put the cap back on, however, nothing really happened.
** What happens when you remove the pressure-release cap? **
When i removed the cap, the liquid in the tube moved quickly away from the tube, I am assuming this was to equal out the pressures. Air did not escape from the system however, because there is water covering the tube in the bottle and water blocking the pressure gauge. All tubes are blocked. I expected the water level in the pressure gauge to increase so this met my expectations.
** What happened when you blew a little air into the bottle? **
When i blew air into the system, the water level in the tube rose about a cm. This is because i am increasing the pressure in the tube by adding air. The air column did not return to original position because the air has no where to escape. The vertical tube put water into the system which bubbled up from the water. All of these things happened because i was adding pressure to the system by adding air. Since there is more air in the system that can not escape, the pressure has increased.
** 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: **
1050N/m^2
0.1m
1%
1 pa=1n/m^2. 1% of 105 is 1.05 meaning that the pressure would increase by 1.05 kpa. This would be 1050 N/m^2
Using Bernoulli's eqn, the change in height would be 0.1m
H2=(-1050N/m^2)/(1000kg/m^3)(9.8m/s^2)
Finding temperature change is easy. You could either set up the equation p1/T1=P2/T2 or use the relationship that pressure and volume are related directly so a 1% increase in P is a 1% increase in T.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3K
0.35kpa
0.04m
I got the first number 3K by actually solving the equation P1/T1=P2/T2 that I described above. My third number i found by comparing 105kpa/300K=x/301 and solved for x. I got .04m by using Bernoulli's equation to find 'dh. I used the same process as described above.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
280K
25.7K
I used Bernoulli to solve for the change in pressure and then used P2 to find T2. The pressure changes for 0.01m was to 203kpa and the pressure change for .001m was to 114.8kpa. You can then use P1 and P2 to solve for T2. I set this up using (105kpa/300K)=(203kpa/x) and (105kpa/300K)=(114kpa/x). I solved for x in each case to find the new temperature. I then subtracted to find the temperature change.
** water column position (cm) vs. thermometer temperature (Celsius) **
27.1 C, 0cm
27.1 C, 0cm
27.5 C, 0cm
28.5 C, 0cm
28.0 C, .5cm
27.3C, .5cm
27.0C, .3cm
26.9C, .1cm
26.5C, .1cm
26.5C, .1cm
26.4C, .1cm
26.2C, 0cm
26.0C, -.5cm
26.0, -.7cm
26.1C, -.5cm
26.5C, 0cm
26.5C, 0cm
26.7, 0cm
26.7, 0cm
26.7, 0cm
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
When temperature decreased substantially, the water level also began to decrease. When the water level increases, the temperature was also increasing. There was definite fluctuation. The mean temp was 26.78 so the max deviation of temperature would be 28.5. This is based off my estimates mean and the distance from this mean.
** Water column heights after pouring warm water over the bottle: **
26.0C
22
16.5
12
9
6.7
4.5
3.5
2.0
1.0
0
-1.0
-2.3
-2.0
** Response of the system to indirect thermal energy from your hands: **
There was a 4cm height increase. We can use Bernoulli's equation to solve for the temperature increase by finding the change in pressure. Using Bernoulli's equation we find that a .04m increase in height is equal to 192pa in pressure change. Converting this to kpa, we find that 105kpa/299k(room temp)=105.392kpa/T2. T2 turns out to be 300.1 meaning that I heated the gas by 1.1K.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
0cm, 27.0C
0cm, 27.0C
0cm, 27.0C
0cm, 27.0C
0cm, 27.4C
0cm, 27.4C
.5cm, 27,8C
1cm, 27.5C
1.5cm, 27.8C
1.5cm, 27.9C
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
The water rose much quicker horizontally. Just a few moments of heat and the water level went up 8.5cm.
** 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: **
66.75N.
300pa
70.65cm^3
1%
0.78K
2K
I got the force required for the first bottle experiment from my previous experiment.
I would guess that much like my previous answer, the pressure was only and change of about 300pa. It would be slightly less than the vertical tube.
I found 70.65 by finding the area of the tube and multiplying it by the length.
I found the percent volume by finding V1 using P1V1=P2V2 and comparing them
I found the T2 by using V1/T1=V2/T2
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The change of the air volume was most likely relatively constant throughout our experiment, so it wouldn't have really skewed the results. Also it was so minimal.
** 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: **
58.8kpa
168K
0.7K
I assumed that P1 was 105kpa as before. I then used Bernoulli's eqn to find the change in pressure since I knew the height change.
I then used this pressure to find the change in temperature that would be required using P1/T1=P2/T2.
To find the amount needed to raise the volume, i converted cm^3 to ml and also l to ml. I then found the temp needed to raise the volume 0.7ml using V1/T1=V2/T2.
** 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: **
.00035m
There would be no height difference so I don't know how you could use bernoullis equation.
?
V1/T1=V2/T2 so I used this to find V2 which was 3.01. I then used this volume to find the change in pressure which is 0.35kpa. I then found the change in height using bernoullis eqn which was .000035m.
Having the tube horizontal is important because it is more sensitive to temperature changes.
** Optional additional comments and/or questions: **
2 hours
** **
I would appreciate notes on the last question. I am really confused about the horizontal vs. vertical.
I've added some commentary at the end of the document linked below, related to the last question.
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