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 expected that as soon as I blew the air would push the water to rise much more easily than when I squeezed it. This did actually happen and the water skyrocketed.
** What happens when you remove the pressure-release cap? **
I expected removing the cap would cause air to escape from the system. There are two different pressures, one in the bottle and one outside the bottle. Air always tries to move from a higher pressure to a lower pressure. When I removed the cap I believe that air did escape. Now blowing air does not cause water to skyrocket.
** What happened when you blew a little air into the bottle? **
The water level will not rise as easily as it did before. I assume this is because both pressure systems have been equalized so there is no longer air trying to move from a higher pressure to a lower pressure. Air was moved because my air did apply pressure under the water causing it to rise. When I stopped blowing the air moved back to its original position. The vertical tube acted similarly, however in order to get it to move like I did the first time(when there was a much higher pressure in the bottle) I had to blow much harder. By the way I could not get it to move like it did originally. The water still moved in little packets rather than a continuous flow of water.
** 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: **
1 N/m^2
3-5cm
1%(My reasoning for temperature is that pressure is directly proportional to temperature. That being true, if pressure goes up, temperature much go up an equal amount. 1% of 300K being about an increase of 3 Kelvin.)
Good explanation of your reasoning, and the reasoning is correct.
However:
1 N/m^2 is only about .001%, not 1%, of atmospheric pressure.
An additional 3-5 cm of water column height would correspond to a pressure difference of about 300 - 500 N/m^2, which does not correspond either a 3 Kelvin temperature difference or to a 1% difference 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 Kelvin
1 kPa(We are assuming initially there is 100kPa and 300 K. So 1kpa increase is equal to an increase of 3 Kelvin)
3-5cm(this here is guesswork(
The 1 kPa is correct, and this might be what you meant in your previous response regarding pressure.
However 3-5 cm is not correct. Bernoulli's Equation, among others, can tell you what the water column height is which corresponds to a 1 kPa change in pressure.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
This is all assuming my 3-5cm raise is true.
1Kelvin
0.1Kelvin(I am using algebra assuming temperature would change 3cm for every 1 change in kPa.)
** water column position (cm) vs. thermometer temperature (Celsius) **
Celsius Relative position(cm)
24.5, 0
24.5, 0
24.6, 0.3
24.6, 0.3
24.5, 0
24.4, -0.3
24.5, 0
24.5, 0
24.5, 0
24.6, 0.3
24.7, 0.6
24.7, 0.6
24.6, 0.3
24.6, 0.3
24.5, 0
24.5, 0
24.5, 0
24.4, -0.3
24.5, 0
24.5, 0
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The air temperature in the varied by a few tenths of a degree(due to the air conditioning running). However the biggest range was about 0.3 Kelvin(so 0.9cm) There was almost a range of 0.9cm
The mean temperature is 24.5 degrees Celsius
The deviation is 0.08127 for degrees Celisus
The mean relative position from 0 in cm is .1815
The deviation is .2059.
** Water column heights after pouring warm water over the bottle: **
24.8
The temperature with the warm tap is 30 degrees Celsius
Celsius relative position (O at 24.5 degrees celsius)
30.1, 16.8
29.7, 15.6
29.3, 14.4
29.0, 13.5
28.8, 13.3
28.6, 13.1
28.4, 12.9
28.2, 12.7
28.1, 12.6
28.0, 12.5
27.9, 12.4
27.8, 12.3
27.8, 12.3
27.7, 12.2
27.7, 12.2
27.6, 12.1
27.6, 12.1
27.4, 11.9
27.4, 11.9
27.4, 11.9
** Response of the system to indirect thermal energy from your hands: **
Yes I did slightly by about 0.2cm
My hand was an outside system heating up the container. The temperature increased slightly causing the water level to rise.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
Celsius Relative position in cm (0 at 24.5)
25.0, 1.5
25.0, 1.5
25.0, 1.5
24.9, 1.2
24.8, 0.9
24.8, 0.9
24.8, 0.9
24.8, 0.9
24.8, 0.9
24.7, 0.6
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
Celsius, Relative position in cm(0 at 24.5)
24.7, 0.6
24.7, 0.6
24.7, 0.6
24.6, 0.3
24.7, 0.6
24.8, 0.9
24.8, 0.9
24.8, 0.9
24.8, 0.9
24.7, 0.6
My hands warming up the bottle caused the water to rise slightly. The temperature increased slightly but this had a very small impact on the system. More than likely, the fluctuation in room temperature had more to do with it than I did.
** 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: **
1kPa
0.03cm^3
0.0003%
0.0003%
the volume would double
Volume, like pressure, is directly proportional to Temperature. I am not sure if these percentages are correct. However, if temperature is affected, so will volume.
you don't give a basis for the .03 cm^3 result, and I don't believe this is correct. I believe the cross-sectional area of the tube is on the order of .1 cm^2, so that the volume change would correspond to more like 1 cm^3.
You also don't give a basis for your calculation of the percent change in air volume. If the volume of air in the system is in fact 1 liter, then your result does follow from your result for volume change.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
Volume change is directly proportional to temperature change. However, the change is much smaller and we would do much better to analyze changes in pressure. The answers would not be as significant if we used proportionality with volume.
The point is the volume of the vertical tube is almost completely negligible compared to the volume of the entire system.
this ** 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: **
1kPa
300 K is about a standard temperature so any increase in temperature would increase the pressure
2.1 K
You don't give a basis for your conclusions. I don't believe your conclusions are accurate.
This shows that the change in volume is not nearly as significant as change in pressure.
** 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: **
3cm^3
1.5cm^3
1/2
These answers are confusing but I am not sure how I would calculate these unless I use some trigonometrey. I know what is being asked but how am I supposed to have exact numerical answers for these last few parts?
In a vertical tube you know the pressure is increasing because the height of the water column changes. You appear to be missing the relationship between the height of the water column and the pressure. As I mentioned in previous note above, Bernoulli's Equation should allow you to determine this relationship and give an accurate numerical answer to the first question. That answer is slightly affected by the fact that the volume of the system changes slightly, and you should know how to take account of that effect, but an estimate based on constant volume would not go far wrong.
With a horizontal tube the movement of the water is almost entirely due to the expansion of the air in the system. This should be a basis for your answer to the second question.
Trigonometry is indeed required for the last question. Hint: What is it that indicates the pressure change? Think of Bernoulli's Equation.
** Optional additional comments and/or questions: **
120 minutes
** **
Many of the calculations that were not measurements were very confusing to find.
You did pretty well with most of your answers, but your answers related to pressure were generally incorrect. You need to use Bernoulli's equation.
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