Phy 122
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 am so worn out from doing all of these experiments and assignments. I worked long and hard and just couldn't think any longer. I really tried my best, but felt lost by the time I got to the end.
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
As I pulled the water up, the air column in the other tube moved toward the bottle. When I replaced the plug, the air column maintained its position and the water
that I had pulled up descended back into the bottle. I expected the air column to move some because there was increased pressure pulling on it as I pulled water up
in the other tube. I expected the water I pulled up to return into the bottle (as long as I kept the tube vertical) once I removed my mouth because air was allowed to
return into the tube.
** What happens when you remove the pressure-release cap? **
The air column returned to its original position, so some air must have escaped. I expected the system to normalize.
** What happened when you blew a little air into the bottle? **
The air column did not seem to move much, but when I removed my mouth water flowed into the vertical tube. It was hard to tell if the air column returned to its
normal position because it moved so little. As I forced air into the bottle it created a negative vacuum and in return it forced water into the vertical tube when the
pressure from blowing was ceased.
** 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 Pa
102 m (which makes no sense!)
1%
100 kPa equal 100,000 Pa, so a 1% increase would be 100 Pa to equal a total of 101,000 Pa. To find the change in height I used the formula P2-P1 = 'rho'g (y2 -
y1). To find the air temp change I used the formula P1/T1 = P2/T2 to get 303K and then divided the difference of 3K by the initial temp and multiplied by 100%.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3K
It would change from 100,000 Pa to 100,333 Pa, which is a 0.3% increase.
33.98 m (my height calculations are apparently wrong)
300 K x 1% = 3K. I used the formula P1/T1 = P2/T2 to find the pressure change and then divided the difference by the original pressure and multiplied by 100%.
For the change in vertical position I again used the formula P2-P1 = 'rho' g (y2 - y1).
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
0.000294 K
0.0000294 K
I determined what the new pressure would be with the height change of each by using P2-P1 = 'rho' g (y2 - y1). Then I plugged that number in to P1/T1 = P2/T2 to
find the final temp and subtracted the initial temp to find the change.
** water column position (cm) vs. thermometer temperature (Celsius) **
20.0, 16
20.2, 17
20.2, 17.3
20.3, 17.5
20.3, 17.8
20.3, 18
20.3, 18
20.3, 18
20.3, 18
20.3, 18.1
20.4, 18.3
20.6, 18.3
20.8, 18.4
20.8, 18.4
20.9, 18.5
20.9, 18.8
21.0, 19
20.9, 18.9
20.8, 18.9
20.9, 19
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
As the temp increased, the water column increased. The highest temp was 21 and the lowest was 20, so the avg is 20.5 and the max deviation from that would be
21 and 20. The highest measurement was 19 and the lowest was 16, so the avg was 17.5 and the max deviation from that would be 19 and 16.
** Water column heights after pouring warm water over the bottle: **
ok
** Response of the system to indirect thermal energy from your hands: **
Yes! By about 5 1/2 cm. The increase in temp caused an increase in pressure.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
22, 0.8
22, -1.0
21.9, -1.5
21.9, -4
21.9, -6.5
21.9, -13
21.9, 14.5
21.9, -15.5
21.9, -16
21.9, -16.5
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
The water level increased along the horizontal 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: **
Very little
V = pi r^2 l = pi (0.15cm)^2 (10cm) = 0.71cm^3
0.71%
You might have had 1500 cm^3 of air in the tube, so the percent would be much less than 0.71%.
T2 = V1T1/V2
IF T / V = constant then T2 / V2 = T1 / V1 and T2 = T1 * (V2 / V1).
????
** Why weren't we concerned with changes in gas volume with the vertical tube? **
Because it was read using the pressure tube.
No
** 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: **
rho g (6cm) = 58,800
you didn't include units for any of these quantities; using units you might have gotten 58 800 kg/(m^2 s^2) * cm; however you wouldn't want m and cm to appear in the same calculation
T2 = P1T1/P2
T2 = V1T1/V2 = 3000cm^3(300K)/3000.7cm^3 = 299.93K
** 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: **
V2 = V1T2/T1 = 3000(301)/300 = 3010cm^3 = 10cm^3
????the same????
????
I just can't do anymore.
** Optional additional comments and/or questions: **
Forever. I began this on Monday and worked on it (and the other labs) all day and then off and on the rest of the week. I feel lost now and my brain is fried.
** **
You did quite well, right up to the end. See my notes. Then check out the link below for further discussion.
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