Your work on bottle thermometer has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
hen I blew, bubbles started and the air column on the pressure indicating tube shortened. Once I quit, the water rushed up the vertical tube and stayed. The air in the pressure indicating tube also stayed. The things happened because blowing the air in increased the pressure in the system. The air was not able to escape through the pressure valve tube. And air could not escape through the other tubes. Thus, the pressure in the tube increased and stayed that way.
I would not think air would escape from the system because it already can out of the vertical tube. There is no water in the vertical tube so the pressure would be the same as outside the tube.
When I blew, bubbles started and the air column on the pressure indicating tube shortened. Once I quit, the water rushed up the vertical tube and stayed. The air in the pressure indicating tube also stayed. The things happened because blowing the air in increased the pressure in the system. The air was not able to escape through the pressure valve tube. And air could not escape through the other tubes. Thus, the pressure in the tube increased and stayed that way.
1* 10 ^3
.4
1
Since 100Kpa = 1 N/ M^2 one percent of 1 is .01. We raised the water to 10 cm above the bottle. From the top of the water to a point 10 cm above the tube was 40cm.
3
1
One degree would move it 4 cm.
Since 300k is what the initial temperature is one percent of that is
1.33
10.33
Since 1% on the temperature is 3 I divided 4 by 3 and got 1.33cm. I then converted cm to mm and got 10.33mm.
0, 22.8
-.2, 22.6
-.25, 22.7
-.25, 22.8
-.25, 22.8
-.25, 22.7
-.25, 22.7
-.30, 22.7
-.30, 22.7
-.35, 22.7
-.35, 22.6
-.35, 22.6
-.40, 22.6
-.40, 22.5
-.40, 22.4
-.45, 22.3
-.45, 22.1
-.50, 22.1
-.50, 22.0
-.50, 22.0
The trend appeared to be that the lower the temperature the lower the water dropped. I'm not exactly sure what you asking on the estimate question. I ran a standard deviation on both and got a standard deviation of .2726 for the alcohol thermometer and a standard deviation of .1237 on the bottle thermometer. I got a mean of 22.52 on the alcohol thermometer and a mean of .335 for the bottle deviation. So I would say the maximum deviation is 22.52 +- .2726 for the alcohol and .335+-.1237.
2.24 degrees on alcohol thermometer
3.75 cm
3.7
3.6
3.4
3.1
2.9
2.8
2.6
2.45
2.3
2.05
1.9
1.7
1.6
1.45
1.3
1.15
1.1
.9
.7
.5
.4
.35
.25
.3
.2
.15
.1
.1
0
0
.5
.25
.25
.5
.5
.4
.2
.15
Yes, it raised the water 7.1cm. This is around a 2.4 degree increase. This answer is based on the fact that 1 degree change equals a 3 cm increase.
0, 22.2
-.6, 22.1
-.6, 22.1
-.7, 22.0
-.7, 22.0
-.7, 22.0
-.6, 22.1
-.7, 22.0
-.7, 22.1
-.6, 22.1
When I warmed my hands then placed them near the bottle the results were similar to those above. It seemed to take a little longer before the meniscus started to move. After it started it moved 3.1 cm
0
.706
2.8
3.3
3.3
When we performed the previous experiment squeezing the bottle it required no additional pressure to move the fluids. We just had to keep a constant pressure. I used the formula V='pi r^2*h to get the volume in the tube. I was unsure on the third question. I used pv=nRT to find the volume before 100v=1(8.314)(300)=24.942 (.701/24.942)*100=2.83%
I was also unsure of this question. Before 3 cm change was equal to a 1 degree change. I assumed that 10/3=3.33degree change. I would assume that regardless of the temperature the change would be the same.
In this case we have gravity working against us and a change in pressure. I would not think that we would have a significant difference in our estimates of temperature change due to the change in pressure.
2
6
3.33
I really was not sure what the statement above was saying Suppose that in the process of moving 10 cm along the tube, the meniscus 6 cm in the vertical direction. Using the assumption that 1 degree equals three centimenters and the pressuer is 100kPA it would be a 2 percent increase (moving 6cm) increasing the pressure to 102kPa. So, it would be a 2 kPa change. .7cm^3 is equal to 10cm so it would be the same as the previous question 3 cm per degrees would be 10/3=3.33
3
.07
67.5
Again using the assumption that 1 degree equal 3cm change. Since 10cm equals .7cm^3 one tenth of that is .07cm^2. I assumed that between vertical and horizontal would be a 67.5 degree angle going up and out.
I had an increasingly difficult time as this lab went on. I understand a lot of the concepts but had a difficult time applying them. Any light you could shed on the problems I had would be greatly appreciated. Thanks.
You have very good data and good insight in most of your answers. Nobody this semester has yet gotten all the way through this experiment without having some difficulty.
See how much of the following makes sense of and helps you clarify the situation. You are welcome to insert answers to these questions (or if necessary more questions) and send me a copy of this note for further clarification.
A 6 cm change in the height of the water column indicates a specific change in rho g h, therefore a specific change in pressure and therefore temperature.
How much would the pressure have to change?
What would be the new and the old pressure, the ratio of pressures, the ratio of temperatures and therefore the change in temperature? This is the change in temperature necessary to raise the water 6 cm in a vertical tube.
Now if volume was to increase by .7 cm^3, without changing the pressure, how much temperature change would be required? This is the temperature required to the meniscus to move 6 cm in a horizontal tube.
Since the temperature change required to make a 6 cm change in the vertical position is to much greater than that required to make a 6 cm change in the horizontal position, we would expect that a the angle that equalizes the influence of altitude change and gas expansion would be much closer to horizontal than to vertical.