Phy 232
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 expect that the water would fall back down because when I inhale with the tube, I am lessening the pressure inside the bottle, which as soon as I release the vertical tube, the pressure inside the bottle will try to make itself even with the atm pressure. This is what happened. For the pressure indicating tube, I'd expect that there would be less pressure and less air pushing on the stopper. After inhaling air from the vertical tube, I'd think the pressure indicating stopper would be even sealed tighter because there's less pressure in the bottle than the air outside the bottle.
** What happens when you remove the pressure-release cap? **
I think that when I remove the cap from the pressure valve tube, air will rush in from the outside because there's more pressure on the outside of the bottle than the inside. The water will fall back into the bottle because the bottle will then have the same pressure as the outside, and there would be no pressure keeping the water up unless I continue to inhale in the vertical tube.
** What happened when you blew a little air into the bottle? **
When I blew air into the vertical tube, the water went down and then when I removed the tube from my mouth, the water came back up with some force. This is because I increased the pressure in the bottle and when I removed my mouth the pressure in the tube was trying to get out by way of the vertical tube. The question asks what the air column in the pressure indicating tube did, I dont really know what they are asking, I of course cant see air, so they must be asking how the water is moving but the only water in the tubes is the water in the vertical tube. And I have answered that question above in the first couple sentences.
** 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
The pressure would change 1000N/m^2 because that's 1% of 100000N/m^2. The height would change 10cm because 1000/100=10cm. Air temp would change 1% because temp has direct relationship with pressure and the pressure changed 1%.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
.27
1000
10
We made the asumption that temp is at 27degrees celcius, and 1% of that is .27 degrees celcius. Pressure has a direct relation relationship with temp so if we change temp 1%, we change pressure 1% so thats 1000N/m^2. The 10cm is the same as the question above.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
1/10
1/100
If 1 degree celcius change changes height by 10cm. The 1cm changes height 1/10 of a celcius degree. The second number is just the conversion from cm to mm. 10mm = 1cm.
** water column position (cm) vs. thermometer temperature (Celsius) **
25.9, +.7
26.3, +1.3
26.1, +.8
25.9, +.5
25.7, 0
25.2, -.3
24.9, -.7
24.8, -.8
24.7, -1.0
24.5, -1.3
24.6, -1.0
24.8, -.5
25.0, -.2
25.4, .3
25.8, .8
25.8, 1.0
25.9, 1.2
26.3, 1.6
26.5, 2.0
26.3, 1.6
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperatures first go up, then down, then up again. The heights support this statement. I'd say for every 1cm of celcius change, the height changes 2cm. Above in a question they state that the change in height should be a little over 3cm for every degree change. So my data probably isnt completely accurate but it is very difficult reading a thermometer and tape measure to the nearest tenth of a degree or cm. I have the jist of the experiment though.
** Water column heights after pouring warm water over the bottle: **
+5.7
+4.0
+3.1
+2.3
+1.8
+1.5
+1.4
+1.1
+.9
+.5
+.4
+.5
+.4
+.2
+.4
** Response of the system to indirect thermal energy from your hands: **
My hands warmed the bottle slightly. The height increased about .5cm. It'd difficult to tell if this is because of my hands or fluctuations in room temperature. The human body is warmer than room temperature so I'm sure the bottle must have been heated slightly.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
.25.5, .5
25.2, 0
25.7, 1.0
25.8, 1.3
26.0, 1.9
26.2, 2.4
26.1, 2.0
25.9, 1.4
25.4, .7
.25.3, .3
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
25.3, 1.2
25.5, 1.9
25.7, 2.5
26.1 , 3.7
26.3, 4.2
26.2, 4.1
26.2, 4.2
26.0, 4.0
25.9, 3.9
26.1, 4.0
This is a little hard because you cant move your hands from the bottle even when you take measurements. That's why I dont think my measurements were as accurate as my previous measurements.
** 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: **
3846
.25
12.5
3.3
200%
First #: Temp rose 1 degree, 26/27=100000/P2, P2=103846, dP=3846Pa. Second #:, Temp must have risen 3.3degrees celcius. V2=2L*29.3/26=2.25L. dV=.25L. Third #: .25/2 * 100% = 12.5%. Fourth #: 3.3 degrees. Fifth #: If temp doubles, volume doubles.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
I believe the volume wouldn't change much so there wouldn't be a significant difference in estimates of temperature change.
** 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: **
5880
1.5
6
First number: dP=rhogy=1*980cm/s/s*6cm=5880Pa. Second number: T2=26*105880/100000=27.5, 27.5-26=1.5 degrees celcius. Third number: T2=26*3.7/3=32, dT=32-26=6 degrees celcius.
Right relationships but you have errors in some details.
Your units in the first relationship don't work out to Pa. You need to use absolute temperatures in the third.
** 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: **
3.9
4.8cm.
First number: P2=100,000*27/26=103846Pa. 3846=rhogy=1*980*y. y=3.9cm. Second number= In the above experiment, the tube was at 45 degrees and when the water moved 10cm, it was a change in height of 6cm. From ratios, we see when the water moves 10cm, the change in height is 8cm. 10/8=y/3.9, y=4.88cm. I dont know about the third question, there is no greater slope than 90 degrees, and thats what we used in the first question. When the tube is in a horizontal position, it is easier for water travel across horizontal then it is for water to travel a vertical tube because gravity's full force is bearing down on the water as it goes straight up.
** Optional additional comments and/or questions: **
3 hours
** **
You understand this well but there are some errors in details. No need to resubmit but check the link:
&#Please see the following link for more extensive commentary on this lab. You should read over all the commentary and not anything relevant. Give special attention to any comments relevant to notes inserted into your posted work. If significant errors have occurred in your work, then subsequent results might be affected by those errors, and if so they should be corrected.
Expanded Commentary
Please respond by submitting a copy of this document by inserting revisions and/or self-critiques and/or questions as appropriate. Mark you insertions with #### and use the Submit Work Form. If a title has been suggested for the revision, use that title; otherwise use an appropriate title that will allow you to easily locate the posted response at your Access Page. &#
Phy 232
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 expect that the water would fall back down because when I inhale with the tube, I am lessening the pressure inside the bottle, which as soon as I release the vertical tube, the pressure inside the bottle will try to make itself even with the atm pressure. This is what happened. For the pressure indicating tube, I'd expect that there would be less pressure and less air pushing on the stopper. After inhaling air from the vertical tube, I'd think the pressure indicating stopper would be even sealed tighter because there's less pressure in the bottle than the air outside the bottle.
** What happens when you remove the pressure-release cap? **
I think that when I remove the cap from the pressure valve tube, air will rush in from the outside because there's more pressure on the outside of the bottle than the inside. The water will fall back into the bottle because the bottle will then have the same pressure as the outside, and there would be no pressure keeping the water up unless I continue to inhale in the vertical tube.
** What happened when you blew a little air into the bottle? **
When I blew air into the vertical tube, the water went down and then when I removed the tube from my mouth, the water came back up with some force. This is because I increased the pressure in the bottle and when I removed my mouth the pressure in the tube was trying to get out by way of the vertical tube. The question asks what the air column in the pressure indicating tube did, I dont really know what they are asking, I of course cant see air, so they must be asking how the water is moving but the only water in the tubes is the water in the vertical tube. And I have answered that question above in the first couple sentences.
** 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
The pressure would change 1000N/m^2 because that's 1% of 100000N/m^2. The height would change 10cm because 1000/100=10cm. Air temp would change 1% because temp has direct relationship with pressure and the pressure changed 1%.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
.27
1000
10
We made the asumption that temp is at 27degrees celcius, and 1% of that is .27 degrees celcius. Pressure has a direct relation relationship with temp so if we change temp 1%, we change pressure 1% so thats 1000N/m^2. The 10cm is the same as the question above.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
1/10
1/100
If 1 degree celcius change changes height by 10cm. The 1cm changes height 1/10 of a celcius degree. The second number is just the conversion from cm to mm. 10mm = 1cm.
** water column position (cm) vs. thermometer temperature (Celsius) **
25.9, +.7
26.3, +1.3
26.1, +.8
25.9, +.5
25.7, 0
25.2, -.3
24.9, -.7
24.8, -.8
24.7, -1.0
24.5, -1.3
24.6, -1.0
24.8, -.5
25.0, -.2
25.4, .3
25.8, .8
25.8, 1.0
25.9, 1.2
26.3, 1.6
26.5, 2.0
26.3, 1.6
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperatures first go up, then down, then up again. The heights support this statement. I'd say for every 1cm of celcius change, the height changes 2cm. Above in a question they state that the change in height should be a little over 3cm for every degree change. So my data probably isnt completely accurate but it is very difficult reading a thermometer and tape measure to the nearest tenth of a degree or cm. I have the jist of the experiment though.
** Water column heights after pouring warm water over the bottle: **
+5.7
+4.0
+3.1
+2.3
+1.8
+1.5
+1.4
+1.1
+.9
+.5
+.4
+.5
+.4
+.2
+.4
** Response of the system to indirect thermal energy from your hands: **
My hands warmed the bottle slightly. The height increased about .5cm. It'd difficult to tell if this is because of my hands or fluctuations in room temperature. The human body is warmer than room temperature so I'm sure the bottle must have been heated slightly.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
.25.5, .5
25.2, 0
25.7, 1.0
25.8, 1.3
26.0, 1.9
26.2, 2.4
26.1, 2.0
25.9, 1.4
25.4, .7
.25.3, .3
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
25.3, 1.2
25.5, 1.9
25.7, 2.5
26.1 , 3.7
26.3, 4.2
26.2, 4.1
26.2, 4.2
26.0, 4.0
25.9, 3.9
26.1, 4.0
This is a little hard because you cant move your hands from the bottle even when you take measurements. That's why I dont think my measurements were as accurate as my previous measurements.
** 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: **
3846
.25
12.5
3.3
200%
First #: Temp rose 1 degree, 26/27=100000/P2, P2=103846, dP=3846Pa. Second #:, Temp must have risen 3.3degrees celcius. V2=2L*29.3/26=2.25L. dV=.25L. Third #: .25/2 * 100% = 12.5%. Fourth #: 3.3 degrees. Fifth #: If temp doubles, volume doubles.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
I believe the volume wouldn't change much so there wouldn't be a significant difference in estimates of temperature change.
** 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: **
5880
1.5
6
First number: dP=rhogy=1*980cm/s/s*6cm=5880Pa. Second number: T2=26*105880/100000=27.5, 27.5-26=1.5 degrees celcius. Third number: T2=26*3.7/3=32, dT=32-26=6 degrees celcius.
Right relationships but you have errors in some details.
Your units in the first relationship don't work out to Pa. You need to use absolute temperatures in the third.
** 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: **
3.9
4.8cm.
First number: P2=100,000*27/26=103846Pa. 3846=rhogy=1*980*y. y=3.9cm. Second number= In the above experiment, the tube was at 45 degrees and when the water moved 10cm, it was a change in height of 6cm. From ratios, we see when the water moves 10cm, the change in height is 8cm. 10/8=y/3.9, y=4.88cm. I dont know about the third question, there is no greater slope than 90 degrees, and thats what we used in the first question. When the tube is in a horizontal position, it is easier for water travel across horizontal then it is for water to travel a vertical tube because gravity's full force is bearing down on the water as it goes straight up.
** Optional additional comments and/or questions: **
3 hours
** **
You understand this well but there are some errors in details. No need to resubmit but check the link:
&#Please see the following link for more extensive commentary on this lab. You should read over all the commentary and not anything relevant. Give special attention to any comments relevant to notes inserted into your posted work. If significant errors have occurred in your work, then subsequent results might be affected by those errors, and if so they should be corrected.
Expanded Commentary
Please respond by submitting a copy of this document by inserting revisions and/or self-critiques and/or questions as appropriate. Mark you insertions with #### and use the Submit Work Form. If a title has been suggested for the revision, use that title; otherwise use an appropriate title that will allow you to easily locate the posted response at your Access Page. &#