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? **
As I pulled water up the tube, the length of the air column increased. this is what I expected to happen and made sense because I caused a vacuum inside the tube, which then caused the length to increase.
** What happens when you remove the pressure-release cap? **
I expect that air would move into the system becuase the vacuum I caused by raising water up the vertical tube. When I removed the pressure valve, this is what happened.
** What happened when you blew a little air into the bottle? **
When air was blown into the tube the length of the air column in the pressure indicating tube shrank slightly.
The length of the air column in the pressure indicating tube change when I blew air into the system because I increased the pressure of the system. If we increase the pressure and keep everything else constant, the volume must shrink.
Water moved up the verticle column when I removed my mouth from the tube.
I anticipated all of these things to happen. It all follows the ideal gas law. If we increase the air pressure in the system, the volumn of the pressure indicating tube changes. At the same time, there air was greater than atmospheric in the system and just atmospheric outside, so when I removed my mouth from the tube the pressure in the system pushed water up the tube.
** 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: **
If the gas pressure in the bottle changed by 1%, it would change by 1000 Pa.
The height of the water above the bottle would change by about 10 cm.
The temperature would only have to change by 1 % or by 3 Kelvin.
1% of 1 atm is about 1000 Pa. I took this value and used P=rho*g*h to find the change in the height of the column. I realized that pressure and temperature are directly preportional to eachother, so a change in one would equal the same percent change 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 Kelvin
About 1000 Pa.
About 10 cm.
Since the temperature was 300 K, I calculated 1 % of that. I then realized that if everything else is constant, T and P are directly preportional to eachother. So a 1% change in temp. will result in a 1% change in pressure. I then used P=rho*g*h to find the change in height.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
about 0.33 Kelvin.
About 0.033 Kelvin.
I used the ideal gas law and just realized that if there was percent change in the temperature, it would relate to the same percent change in pressure. I then took this pressure change and plugged it into P=rho*g*h.
** water column position (cm) vs. thermometer temperature (Celsius) **
26.1, 11.52
26.1, 11.52
26.15, 11.55
26.15, 11.55
26.15, 11.55
26.15, 11.55
26.1, 11.52
26.1, 11.52
26.1, 11.52
26.15, 11.55
26.15, 11.55
26.1, 11.52
26.1, 11.52
26.1, 11.52
26.1, 11.52
26.15, 11.55
26.15, 11.55
26.15, 11.55
26.15, 11.55
26.15, 11.55
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature fluctuation showed that as the temperature increased the column of water rose, and as the temperature decreased the column of water fell. The temperature fluctuation over the 10-minute period showed only a fluctuation of 0.05 C and 0.03 cm.
The maximum deviation in temperature over for the alcohol thermometer was 0.05 C.
The maximum deviation in temperature for the bottle thermometer was 0.01 C.
My estimate for the bottle termometer is based on the fact that for a 1 C temperature change, the water column changes about 3 cm. Since my column only changed by 0.03 cm, the temperature change that it estimated was 0.01 C.
** Water column heights after pouring warm water over the bottle: **
26.15, 11.55
11.67
11.67
11.67
11.67
11.66
11.66
11.65
11.65
11.64
11.62
11.62
11.62
11.62
11.62
11.61
11.61
11.6
11.6
11.6
11.6
11.59
11.59
11.59
11.59
11.59
11.58
11.58
11.58
11.58
11.58
11.57
11.57
11.57
11.57
11.57
11.57
11.57
11.57
11.57
11.57
** Response of the system to indirect thermal energy from your hands: **
I didn't read that the water column rose at all. I don't think it would make a difference putting my hands around the container to try and warm it without touching because the heat from my hand, which isn't very much to begin with, is going straight to the air and not sending very much to the water.
Your hands do exchange energy with the air by convection, and this energy will be unlikely to make it to the bottle.
However your hands also radiate thermal energy directly through air, with very little loss. Just as you feel the energy from the Sun, the bottle will 'feel'energy from your hands.
Of course the Sun is at a much higher temperature than your hands.
In addition, especially in the warm months, your hands might not be at a temperature much different than the room you are in. Since the room is also radiating thermal energy to the bottle, your hands might not make much of a difference.
However if your hands are at, say, 30 degrees Celsius when the room and the contents of the bottle are at 20 Celsius, it is not uncommon to be able to detect the effect of the energy radiated from your hands.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
26.1, 3.5
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
I see no change in the height. Like before, I don't think that the hands will produce enough heat.
** 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: **
As the water moves 10 cm, the pressure changes by 1000 Pa
The pressure would be raised by 1000 Pa
1% change
A 1% temperature change would be needed
0.5% change
All of these answers were reasoned out using the ideal gas law. The first and second question were calculated previously. If the pressure changed by 1000 Pa and the water was raised by 10 cm, then the volume of air was calculated to change by 1% because it is directly proportional to temperature that must change by 1% too. If temperature was double then you would only have to change the volume by 1/2% instead because temperature and volume are directly proportional so, if temperature is increased by double then in order to get the same pressure in the tub you would only want to change the volume by 1/2.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
It was harder to push the air up the vertical tube because gravity was acting against it; thereofore, it wasn't affected as much by the pressure and temperature changes.
** 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: **
1000 Pa
303 K
510 K
We calculated above that if the water moved along the tube 10 cm, that there would be a 1000 Pa pressure difference. It would be a 1% Pressure change as calculated above; therefore 1% of 300 is 3K. By using the ideal gas law, PV = nRT you can find the temperature change. I said that the P/T = P/T and solved the ratio for the new temperature.
** 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: **
The postion would change 3.7 cm
The postion would change 14.3 cm.
I would think at a slope of about 45 degrees.
I used the fact that temperature change is directly preportional to volume or pressure change. Keeping that it mind, I did all my calculations the same way as was done earlier in the lab. If the tube is truely horizontal, it does not take nearly as much pressure change to move the water.
** Optional additional comments and/or questions: **
3.5 hours
** **
Good data, and good answers on many questions. However some responses need some revision.
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