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: **
Part 2 (Part 1 was submitted yesterday)
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
** What happens when you remove the pressure-release cap? **
** What happened when you blew a little air into the bottle? **
** 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: **
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
** water column position (cm) vs. thermometer temperature (Celsius) **
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
** Water column heights after pouring warm water over the bottle: **
** Response of the system to indirect thermal energy from your hands: **
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
** 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
0.7065 cm^3
0.024 m^3
0.008478 K
0.016956 K
1) verry little
2) pi r^2 h so 3.14 * 0.15mm^2 * 10 = 0.7065
3) PV=RT 100Kpa, R=8.34 T=300K V=RT/P so V= 8.34*300/ 100 * 10^3 = 0.025m^3 is initial volume. V2 = 0.7065 * (10^-2)^3 so v2= 7.065 * 10^-7 so v1-v2= 0.0249 m^3
4) pv=rt v1/t1 = v2/t2 (radial gas eqn) p1v=rt1 p2v2=rt2, t2=t1v2/v1 so 300 * 7.065 * 10^-7 / 0.025 = 0.008478
5) 600 * 7.065 * 10^-7 / 0.025 = 0.16956 K
** Why weren't we concerned with changes in gas volume with the vertical tube? **
Because the volume rising has no significant role in the temperature. So even if volume rises there is no significant changes in temp.
** 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: **
588pa
1176kPa
7*10^-5 degrees c
1) pressure = rho g h = 1000 * 9.8 * 6 * 10^-2 = 588 pa
2) 588 pa t=300 pv=rt, p1/p2 = t1/t2, p2 = 588 *600 / 300 = 1176 kpa
3) p1= 588, p2 = 1176, V1 = 3 v2=0.07 * (10^-2)^3 so t1v2/v = 7*10^-5
** 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: **
0.1cm
Decrease
1
1)We did that earlier, right?
2)The pressure will decrease if fluid is moving in horizontal direction. In verticle pressure is more
3) slope is one
If the tube is horizontal the pressure decreases considerably. Any deviation from that changes this.
** Optional additional comments and/or questions: **
Very hard. Almost impossible. This took a long time. I have to power through these experiments and get to exam 2 very soon. This is not my best work but it was very time consuming. Is it true there are no experiments in section 4 (part leading up to exam 4?)
Good work, but be sure to see my notes on your previous submission.
You are through most of the experiments, though there are still a few with Part 3. There are no experiments in Part 4, which most students find to be not too difficult.