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? **
After replacing the cap on the pressure-valve tube, removing the vertical tube from my mouth caused the water to return to the bottle. I expected the water to remain in the vertical tube since the pressure in the bottle was being kept constant, and no change in the pressure-indicating tube.
I lowered the pressure on the open end of the tube, that's why the water moved upward. It also gained potential energy. Placing the cap on the pressure-valve tube kept the pressure inside the bottle at atmospheric pressure. The pressure on the water column is the sum of atmospheric pressure and the product of water density, acceleration due to gravity and the height of the water column. Opening the upper end of the water column to atmospheric pressure was resulted in a higher pressure at the top than at the bottom of the water column, since I was no longer maintaining the lower pressure with my mouth. The water moved toward the lower pressure. I did not see any change in the pressure-indicating tube.
** What happens when you remove the pressure-release cap? **
Removing the cap from the pressure-valve tube did not result in any change to the system. I didn't expect any more air to escape from the bottle when I remove the cap from the pressure-valve tube since the vertical tube is already open to atmospheric pressure.
** What happened when you blew a little air into the bottle? **
Blowing air into the vertical tube caused the air column at the end of the pressure-indicating tube to shorten in length.
When I removed the vertical tube from my mouth, there was no change in the pressure-indicating tube, and there was water drawn into the vertical tube. I had expected the change in the pressure-indicating tube to reverse itself.
Blowing air into the vertical tube increased the volume of air, and therefore the pressure in the bottle. The pressure-indicating tube showed this in the shortening of the air-column. Releasing the pressure on the end of the vertical tube caused the water at the bottom of the tube to move toward the lower (atmospheric) pressure. The volume of air increased as the water moved up the tube, so the pressure-indicating tube showed no change.
** 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: **
1030
.105 m
1%
I found the change in pressure by multiplying 103000 Pa by .01
I used the relationship p-p_0 = `rho * g * `dh with 1030 Pa as the change in pressure and solved for `dh to calculate the change in height. Air temperature and pressure are inversely proportional if volume is constant, so an increase of 1% in one quantity corresponds to a 1% decrease in the other.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3
1030
.105
I multiplied the 300 K temperature by .01 to calculate 1% of the value. The change in pressure is inversely proportional to the change in temperature, so the 1% calculation above is repeated. The change in height also corresponds to the previously calculated values, as just above 10 cm.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
3 K
.3 K
After reading the next section, I re-calculated and found that I had dropped a zero in my calculator. My estimates are now further off than my original 3K /cm calculation. I've reviewed the information, and haven't found the error in reasoning that I must have made.
A 1% change in pressure is associated with a 10 cm change in column height and a 3 K change in temperature. So 1 cm corresponds to a .1% change, or .3 K, and 1 mm to .03 K.
** water column position (cm) vs. thermometer temperature (Celsius) **
0, 21.1
0, 21.1
0, 21.0
0, 21.5
+5, 21.3
+5, 21.1
+5, 21.2
+5, 21.1
+5, 21.1
+5, 21.0
+3, 21.0
+2, 21.0
+1, 20.9
+5, 20.8
0, 20.8
-1, 20.7
-1, 20.7
-1.5, 20.7
-3, 20.6
-2,20.5
-3, 20.5
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature initially increased. I think it was due to how close I was in my attempt to get accurate measurements. Without outside influence, the thermometer and bottle both reflect minor changes. The bottle is more responsive than the thermometer. I think the maximum deviation in temperature is about .4 `degrees C, based on the values toward the end of the measurement-when I had backed away from the system.
** Water column heights after pouring warm water over the bottle: **
Initial temperature 21.8
The height of the column rose immediately to 22cm, then decreased at a decreasing rate. The final height after 10 minutes was 2.5 cm above the column height before the system was warmed.
** Response of the system to indirect thermal energy from your hands: **
The column height increased by about 3.7 cm, continuing to rise for a few seconds after I moved my hands away. Once maximum height was reached, it immediately began to decrease.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
0, 21.8
0, 21.6
0, 21.7
0, 21.6
0, 21.6
0, 21.5
0, 21.5
0, 21.5
0, 21.5
-.2, 21.5
0, 21.5
This time, the tubing is in contact with the countertop. I think that is keeping the temperature more constant.
There is little gas volume in the tubing and water doesn't expand or contract much with these temperature differences, so modest temperature changes in the tubing are pretty much irrelevant.
The pressure and volume changes in the gas in the bottle are responsible for almost all the effect.
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
This time, placing my hands near the bottle caused the height of the water column to increase above the end of the tube by about 30 cm (based on the difference in the column heights before warming and after cooling). I believe the water was moved about 95 cm. (65 cm to end of tube and 30 cm before and after difference).
** 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: **
30
3
-9
-18
.003m * .01m = 30 cm^3
Assuming 1 liter (1000 cm^3) of air, 30 cm^3 is 3%
Assuming constant pressure, this corresponds to a 3% reduction in temperature 300 K (.03) = 9K
Doubling the temperature doubles the amount that corresponds to 3%
** Why weren't we concerned with changes in gas volume with the vertical tube? **
I thought a while about this one, and I'm really not sure. The difference between raising the water in a vertical tube and moving it along a horizontal tube have to do with the conservation of energy. If the water remains in the tube above the initial height, there is energy due to position stored in the water. I can't quite see the connection between that energy and the thermal energy in the bottle.
The PE change of the water in the tube is at the expense of thermal energy in the gas occupying the bottle. However the PE change of that thin column of water is small compared to the thermal energy changes in the gas.
** 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: **
5886
+17
+.07
I used the relationship p = p_0 + `rho * g * h to calculate the pressure change required for the change in height. Considering constant volume, the ratio of pressure to temperature is constant. I used that property to calculate the temperature required to increase the air pressure from 103000 to 108886. I considered pressure constant for the last calculation, but I'm not certain I should have. With the constant ratio of volume to temperature, changing volume from 3000 cm^3 to 3000.7 cm^3 beginning at 300 K requires a temperature increase of .07 K.
6 cm is .06 m and is associated with pressure different around 600 Pa, not 6000 Pa.
There is a considerable difference in the magnitudes of the quantities involved. Changing pressure by a factor of 10^3 requires a temperature change of a factor of 10^-2
** 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: **
** Optional additional comments and/or questions: **
6 hours
** **
I am having a very hard time synthesizing the smaller ideas into a real understanding of this subject. I don't know if it's one small connection I'm missing or simply that I need to spend more time on it (not that I have any left).
You're doing pretty well, though order-of-magnitude errors sometimes get in the way.
The key ideas:
When heated the gas in the bottle can expand and/or change pressure.
If the tubing is vertical then any expansion results in an increase in pressure, so the vertical tube minimizes expansion and maximizes pressure change.
If the tubing is horizontal then the pressure remains constant and the gas is free to expand, which maximizes the expansion.
The system is very sensitive to changes in temperature, especially if the tube is at or near horizontal.
The energy required to increase the PE of the water in the tube is small compared to the thermal energy changes in the gas.