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: **
** #$&* What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
After I removed my mouth the water went down a little in the vertical tube, but stayed at around 25 cm or so. The air column in the indicating tube didnt move until I removed my mouth. Once the water went down, the air column shortened just a little.
** #$&* What happens when you remove the pressure-release cap? **
Yes air escaped and everything went back to normal. No water in the vertical tube.
** #$&* What happened when you blew a little air into the bottle? **
The air I blew into the tube ended up going in the bottle and increasing the pressure in the bottle. It wasn't able to escape back out of the tube I blew into because that tube is under water. Therefore the indicating tube's air column shortened because the pressure in the bottle increased.
When I removed my mouth the air column did lengthen and water went up the vertical tube. However, the air column did not lengthen back to its original length. It's almost like the air column and water in the vertical tube equaled out in some way. When I removed the plug from the valve everything went back to normal.
** #$&* 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
10
1
P = F/A Pascal = Newton/m^2. So 1% change in 100000 Pascals is a 1000 pascal change. So the change is 1000 Newtons/m^2
Use B equation to find change in height. Since velocity is 0, the middle term cancels: rho g h + P = 0
1000*9.8*h + 1000 = 0
h=-.102 meters or a change in 10 cm
P1/T1 = P2/T2
T2(P1/T1) = P2
T2 = P2 T1/P1
T2 = 101 (300)/100
T2 = 303 which is a 1% change
** #$&* Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
.27
.33
3.4
300-273 = 27 degrees C *.01 = .27 C
100/300 = x/301
x = 100.33 kPa
So change is .33 kPa or 330 Pa
1000*9.8h + 330 = 0
h= .0337m or 3.4 cm
The 1% change in temperature would be based on the initial temperature, which is 300 degrees. Celsius temperature is not absolute, and it is meaningful to speak of a percent change only in terms of absolute temperatures.
** #$&* The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
.294
.0294
1000*9.8*.01 + P = 0
P = 98 change
100098/x = 100000/300
x = 300.294 which is a .294 degree change
1000*9.8*.001 + P = 0
P = 9.8 change
100009.8/x = 100000/300
x = 300.0294 which is a .0294 degree change
** #$&* water column position (cm) vs. thermometer temperature (Celsius) **
19.1, 0
19.1, -.1
19.1, -.1
19.0, -.2
19.0, -.3
18.9, -.4
18.9, -.4
18.8, -.4
18.9, -.3
19.0, -.2
19.0, -.1
18.9, -.2
18.9, -.2
18.9, -.2
18.9, -.3
18.9, -.2
19.0, -.1
19.0, -.1
18.9, -.2
18.9, -.2
** #$&* Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature didn't change too much, but it did decrease a little, the greatest being .3. Surprisingly, so did the water line. It decreased by .4 cm at the most which is more than I expected to see.
I figured that every degree change should show a 3.4 cm diff in the water line. .4/3.4 = .12
So the water line showed a .12 degree change whereas I read a .3 degree change.
** #$&* Water column heights after pouring warm water over the bottle: **
18.0 C
5.1
4.7
4.6
4.2
3.8
3.3
2.9
2.5
2.0
1.6
1.2
.8
.4
0
** #$&* Response of the system to indirect thermal energy from your hands: **
Yes. The water line rose 2.0 cm
2.0/3.4 = .59 degree C increase in the temperature inside the bottle.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
17.7, -.4
17.8, -1
17.8, -1.5
17.8, -2
17.9, -2.6
17.9, -2.9
17.9, -3.4
17.8, -4
17.8, -4.3
17.9, -4.7
For some reason I do not think that the bottle holds pressure when I have the tube horizontal.
** #$&* What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
17.8, .1
17.8, .2
17.8, .1
17.9, .1
17.9, 0
17.9, 0
17.8, -.1
17.8, -.2
17.9, -.2
17.8, -.3
My hands do raise the water level noticeably.
** #$&* 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: **
It increased
.7065
70.65
.7065
It would still change by .0765 percent
10*3.14*.15^2 = .7065 cm^3
V/T = V/T
1/1 = 1.7065/T
T change = .7065 or 70.65 percent
1/600 = 1.7065/x
** #$&* Why weren't we concerned with changes in gas volume with the vertical tube? **
No T and V have a constant ratio: V/T = V1/T1
When one changes, the other does by the same percentage.
** #$&* 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
1.764
.21
1000*9.8*.06 +P = 0
P change is 588 Pascals
100000/300 = 100588/x
x = 301.764
Change is 1.764
A liter is 100 cm^3
1000/300 = 1000.7/x
x = 300.21
** #$&* The effect of a 1 degree temperature increase on the water column in a vertical tube, in a horizontal tube, and the slope required to halve the preceding result: **
3.4 cm
3.4 cm
I thought they should rise by the same amount
** #$&* Optional additional comments and/or questions: **
3 hours
** **
This was difficult
You did well. See my note(s) above.
Check out the discussion at the link below; no revision is necessary unless you have questions.
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