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? **
The water went back down, but I though that it would continue to go up, this happens because I was able to maintain a constant pressure, where the water could not go up.
** What happens when you remove the pressure-release cap? **
Air escapted from the system because by removing the cap the pressure is decreased therefore causing the air to decrease as well.
** What happened when you blew a little air into the bottle? **
air increased in vertical tube, presure indicating tube moved up and then moved back to original position.
the air column moved back to its original position
in the verical tube air increased and produced bubbles
all of this happened because of changes in pressure, I anticipated the vertical not the air column.
** 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: **
1.000*10^5N/m^2, .1cm, 3%
100kpa= 1.000*10^5 N/m^2, then using 10cm, 1% of 10cm = .1cm, and the air temperature would be 300 * .01%= 3% change
A 1% change in pressure would be 1% of 10^5 N/m^2, or about 10^3 N / m^2.
What altitude difference `dy would make rho g y equal to 10^3 N/m^2?
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3deg, 3kpa, 3cm
300*.01= 3 deg
300K/ 100kpa= 3kpa
A 1% temperature change would be 3 deg Kelvin.
A 1% temperature change would be associated with a 1% pressure change, which would be 1% of 10^5 N/m^2, or 10^3 N/m^2.
This would be associated with the altitude difference from the preceding question.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
30K, 3K
300k corresponds to 10cm, 300/10= 30cm corresponds to 1 temp
10mm corresponds to 1mm, therefore 30/10= 3 degress corresponds to 1mm
** water column position (cm) vs. thermometer temperature (Celsius) **
23C, 30cm
25C, 19cm
27C, 20cm
22C, 21cm
21C, 23cm
20C, 23cm
23C, 25cm
27C, 25cm
25C, 27cm
28C, 28cm
29C, 28cm
30C, 29cm
30C, 27cm
32C, 28cm
36C, 29cm
38C, 30cm
39C, 33cm
40C, 34cm
42C, 39cm
43C, 40cm
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
temperature went up and down but water continously raised at a slow, but constant rate.
23.43
15.6deg max
** Water column heights after pouring warm water over the bottle: **
** Response of the system to indirect thermal energy from your hands: **
yes, my hands warmed the air in the bottle by 10degrees, this is because the heat equaled a loss therefore, the air increased.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
17c, 30cm
18c, 31cm
16c,28cm
20C, 32cm
24c, 36cm
25c, 35cm
23c, 33cm
24c, 32cm
27c, 36cm
28c, 38cm
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
it only increased a slight bit, not as much as before.
** 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: **
3.4N/m^2
22cm^3
3.2%
80K
102K
i used the appropiate formulas needed and used the numerical values based on the data that observed and found.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
We don't worry about the volume changed because we are concentrated more on whats going on with the pressure and how the system acts corresponding to the different conditions that the system is placed under.
if there was a significant volume change it would impact our results for that system; however the tube is thin and it takes very little volume change to displace enough water to fill the tube.
** 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: **
5.6atm
308K
307.47K
I used the formulas, p1/T1=P2/T2 and V1/T1=V2/T2 to find the answers needed and for atm I used 860kg/m^3 * 9.8m/s^2*.006m = 5.6atm
1 atm is about 10^5 Pa = 10^5 N / m^2. Your calculation 860kg/m^3 * 9.8m/s^2*.006m does not end up with units of atmospheres, but rather N / m^2.
Note that the density of water is 1000 kg / m^3; 860 kg/m^3 is closer to the density of alcohol.
** 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: **
3005.24cm^3
702cm^3
I used the formula V1/T1=V2/T2 in order to find the correct numbers needed according to what the question asked for
you don't say what you used for these quantities. T1 and T2 should have been absolute temperature expressed in Kelvin degrees. A 1 degree change is less than 1% of a 300 degree Kelvin temperature (in fact it's only about .3% of that temperature).
** Optional additional comments and/or questions: **
2 hours 30 minutes
** **
I struggled with the calculations.
See my notes and see if you can insert revisions and/or additional questions into a copy of this document, and submit. Mark your insertions with &&&&.