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? **
I noticed that when I replace the cap on the pressure valve, the water in the vertical tube stayed at the same height. I thought that it would run back down into the container once I stopped applying pressure with my mouth, but it did not. I ran two trials of this part of the experiment. The first time, I did not notice any noticeable difference with the air column in the pressure valve. The second time, when I started to pull water up the tube, the water in the pressure valve was pulled back into the container.
** What happens when you remove the pressure-release cap? **
I predicted that once I remove the cap for the pressure-valve tube, the water will escape from the tube. This is true. When I have the cap on the tube, the meniscus does not move in the tube. But as soon as I remove the cap, the meniscus starts to move and water will escape into the container if I am not careful.
** What happened when you blew a little air into the bottle? **
I noticed that water rose up in the vertical tube and then returned to the container. I did not notice anything with the pressure indicating tube though. The level of the air column remained pretty status quo. I did not anticipate the water to rise in the vertical tube as high as it did. I think that the water rose in the vertical tube because I changed the pressure in the container by blowing in the tube.
** 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: **
.01
The height would decrease by 1%
2%
I used bernoulli's equation and the ideal gas law to answer the questions. The height and the pressure are inversly proportional, because velocity is constant. Therefore it follows that if gas pressure changes by 1% it would cause a decrease of 1% in the height of the bottle. In the ideal gas law, if volume is constant so we have P1/T1=P2/T2. Solving for T2, I found that the temp would increase to 306K which is a 2% change in temp for a 1% change in pressure.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3 degrees
.5 % pressure change
.05 meters
If 6 degrees corresponded to a 2% change in temp above,then it follows that 3 degrees would cause a 1% change. From ideal gas law, it says that a .5% change in pressure would cause a 1 degree change in temp. Temp is inversely related to pressure which is inversly related to height, thus the pressure will change by .5 meters with a 1 degree change in temp.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
2%
3%
I tried to use my answers in the last few questions to gauge my answers to these questions...not too sure of my accuracy though.
** water column position (cm) vs. thermometer temperature (Celsius) **
27.0, neutral
27.5 neutral
27.5, neutral
28.0, neutral
28.0, neutral
27.5, neutral
27.8 neutral
27.8, neutral
28.0 neutral
28.0 neutral
28.0 neutral
28.0, neutral
28.0 neutral
27.8, neutral
28.0, neutral
27.8 neutral
28.0 neutral
28.0 neutral
I saw virtually no movement with the height of my water column either above or below my mark.
28
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
My temps seem to hover between 27.0-28.0. The maximum deviation in temp seems to one degree above 27.0 or below 28.0 degrees Celsius. The basis may be too simple, but I looked at my trend of data which I took over the 10 minute period and noted that flucuations.
** Water column heights after pouring warm water over the bottle: **
As I poured the single cup of warm water of the sides of the bottle, I noticed that the height in my water column increased approx. 2.5 cm. As soon as I finished pouring the water, the water column decreased to about 1 cm below the initial reading.
** Response of the system to indirect thermal energy from your hands: **
It does not appear that my hands warmed the air in the bottle measurably. I did not notice a change in the water column at all.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
28.0, neutral
28.0, neutral
28.1, neutral
28.0, neutral
28.1, neutral
28.0, neutral
28.0 neutral
28.0, neutral
28.5, neutral
Again, no change in the height of my horizontal-tube meniscus.
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
I did not notice a difference in the water column with my hands when I placed them near the bottle. The only time I have noticed a difference so far is when I poured the warm water over the sides of the bottle.
** 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: **
As the height increased, all other things being constant, the pressure would have decreased.
.314 cm^3
.00003%
400K
800K
By using the initial height of 30 cm, and the final height of 40(30+10 cm that the water moved along the horizontal tube), with a diameter of .30 cm, I was able to reason the volume by using V=pi(r^2)L. Then I plugged those values into the ideal gas law to find the final temp changes.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
I'm not sure why we did not consider the volume change in the air in the case with the vertical tube. I don't think it would have made a significant difference in our estimates of temp. change, otherwise we would have included it :). But I'm not exactely sure how.
** 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: **
62.5 kPA
487 K, 187.98 temp change
300.07 K
I used the Ideal Gas Law to solve answer these questions. In the first question, I assumed that h1=10, and h2=16. Then I used the formula P2=h1/h2(P1)=P2=10/16(100kPa)= 62.5 kPa. For the next question, I plugged p1/T1=P2/T2 using P1 as 100kPa, and T1 as 300K, and P2 as 162.5, to solve for T2. I used the ratio V1/T1=V2/T2 for the third problem.
It seems that it takes much less change in temp to increase the volume of the gas then it does to increase the height of the water column.
** 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 cm
.03 cm
33(??)
Not sure about the accuracy of my results in this portion of the experiment, especially the answer to the second question. The first question I obtained from your explanation half way through the experiment. For the second question I know that volume is equal to Length(Area) in this case Area is the same for both parts so I used H1/T1=H2/T2. I solved for H2 and used H1=10, T1=300K and T2=301 K. I am not sure if this was right because my answer seems to rather low. It seems that by having the tube in a truly horizontal position would result in a higher height involving less pressure required and lower volume. Not sure if my results prove this.
** Optional additional comments and/or questions: **
Your work here is OK.