bottle thermometer

Phy 202

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?

What happened was I got the water up to the top with a given amount of sucking pressure then i took the stopper off and the water became much easier to suck.

What happens when you remove the pressure-release cap?

If I remove the cap i think it will be easier, like drinking soda through a straw. Instead when i did it it was harder to suck the water up.

What happened when you blew a little air into the bottle?

THe bubble in the pressure indicating tube moved about a centimeter. Once the pressure was taken off the vertical tube water spewed out of it. It moved because there was pressure behind it. All these things happened because we increased the normal pressure in the system and it had to equilibriate by pushing water out of the vertical tube.

In order to lower the pressure the air in the bottle had to expand, displacing water into 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:

approximately +/- 10 N/m^2

approximately 1 cm

the air temperature would have to change by 20 percent to get a resulting change.

Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change:

20-25 degrees of temperature change. We would get a resulting pressure of 15 kPa with a 1 degree change in temperature. THe water coulumn would rise half a centimeter with a 1 degree change.

The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change:

10 degree difference in temp would result in a 1 cm rise in the water tube. A 1 degree difference would give us a 1 mm change in the vertical tube.

water column position (cm) vs. thermometer temperature (Celsius)

23.1 degrees celsius, 9 cm initially, -.1, -.1, -.2, -1cm, -2cm, -2.2cm, -2.6cm, -3cm, -3.8cm, -4.1cm, -5.0cm, -5.6cm, -5.9cm, -6,8cm, - 7cm, -7.7cm, -8.3cm, -9cm.

Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer.

The temperature did not change at all during the ten minutes. There was no deviation of the temperature. The basis for my estimate is the fact that the temperature never changed.

I suspect the system wasn't well sealed (no fault of yours, unless you left the cap loose, which is unlikely). A well-sealed system would be sensitive to small temperature changes you can't detect with the thermometer.

Water column heights after pouring warm water over the bottle:

Ok i did this

Response of the system to indirect thermal energy from your hands:

My hands did not change the temperature in the air of the bottle by anything significant. My hands were not touching the bottle and air is a poor conductor of heat. If i had been touching the container then i may have seen a difference.

Unless your hands are cold, the will radiate thermal energy into the system, and the amount is typically sufficient to change the temperature in the bottle by a few tenths of a degree; a well-sealed system would be sensitive to a change of this magnitude. An alcohol thermometer would not respond quickly enough to detect the temperature change.

position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals

23.1 degrees celsius during the whole experiment the meniscus did not move at all. It was lying horizontal with no pressures acting upon it.

What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container?

Again nothing happened. I would suppose if it had gone the way it was supposed to the water would have risen in the horizontal tube maybe a centimeter or two.

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:

I think there was a small increase in the pressure inside the bottle. The volume of the air inside the system would be .1 cm cubed.

The percent change in the system would be less than 1 percent.

About a 4 degree temp change would give us this resulting pressure.

The water would rise out of the tube because of the resulting temperature change.

It was stated earlier that a 1 degree change i celsius would result in about a 3 cm change in the vertical tube. I did my estimations based on this assumption.

Why weren't we concerned with changes in gas volume with the vertical tube?

It has no bearing on the pressure. No mater the air volume in the tube the atmospheric pressure will always be the same.

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:

The pressure would have to change by less than 10 kPa.

A 2 degree increase

a 6 degree increase.

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:

It would rise approx 3 cm.

there would be 10 cm rise.

At about 45 degrees.

Optional additional comments and/or questions:

1.2 hours

Overall you understand what's going on here and what you would have expected to see. See my notes.

A lot of your results are undocumented, meaning that you didn't show how you got them, and many of them appear to be inconsistent with the given conditions.

Compare your results with those given in the following commentary, and let me know if you have questions. Note that analysis of this sort does appear frequently on Test 1.

bottle thermometer

Bottle Thermometer

You can use the bottle, stopper and tubes as a very sensitive thermometer.

This thermometer will have excellent precision, clearly registering temperature

changes on the order of .01 degree. The system will also demonstrate how a

very basic thermal engine and its thermodynamic properties.

Set up your system with a vertical tube and a pressure-indicating tube, as in

the experiment on measuring atmospheric pressure. There should be half a

liter or so of water in the bottom of the container.

Refer back to the experiment 'Measuring Atmospheric
Pressure' for a detailed description of how the pressure-indicating tube is
constructed for the 'stopper' version of the experiment.

For the bottle-cap version, the pressure-indicating tube is the
second-longest tube. The end inside the bottle should be open to the gas
inside the bottle (a few cm of tube inside the bottle is sufficient) and the
other end should be capped.

The figure below shows the basic shape of the tube; the left end extends
down into the bottle and the capped end will be somewhere off to the right.
The essential property of the tube is that when the pressure in the bottle
increases, more force is exerted on the left-hand side of the 'plug' of
liquid, which moves to the right until the compression of air in the
'plugged' end balances it. As long as the liquid 'plug' cannot 'leak'
its liquid to the left or to the right, and as long as the air column in the
plugged end is of significant length so it can be measured accurately, the
tube is set up correctly.

If you pressurize the gas inside the tube, water will rise accordingly in the

vertical tube. If the temperature changes but the system is not otherwise

tampered with, the pressure and hence the level of water in the tube will change

accordingly.

When the tube is sealed, pressure is atmospheric and the system is unable to

sustain a water column in the vertical tube. So the pressure must be

increased. Various means exist for increasing the pressure in the system.

You could squeeze the bottle and maintain enough pressure to support, for
example, a 50 cm column. However the strength of your squeeze would vary
over time and the height of the water column would end up varying in response
to many factors not directly related to small temperature changes.

You could compress the bottle using mechanical means, such as a clamp.
This could work well for a flexible bottle such as the one you are using, but
would not generalize to a typical rigid container.

You could use a source of compressed air to pressurize the bottle.
For the purposes of this experiment, a low pressure, on the order of a few
thousand Pascals (a few hudredths of an atmosphere) would suffice.

The means we will choose is the low-pressure source, which is readily

available to every living land animal. We all need to regularly, several

times a minute, increase and decrease the pressure in our lungs in order to

breathe. We're going to take advantage of this capacity and simply blow a

little air into the bottle.

Caution: The pressure you will need to exert and the amount of air you
will need to blow into the system will both be less than that required to blow
up a typical toy balloon. However, if you have a physical condition that
makes it inadvisable for you to do this, let the instructor know. There is
an alternative way to pressurize the system.

You recall that it takes a pretty good squeeze to raise air 50 cm in the

bottle. You will be surprised at how much easier it is to use your

diaphragm to accomplish the same thing. If you open the 'pressure valve',

which in this case consists of removing the terminating cap from the third tube,

you can then use the vertical tube as a 'drinking straw' to draw water up into it.

Most people can easily manage a 50 cm; however don't take this as a challenge.

This isn't a test of how far you can raise the water.

Instructions follow:

Before you put your mouth on the tube, make sure it's clean and make sure
there's nothing in the bottle you wouldn't want to drink. The bottle and
the end of the tube can be cleaned, and you can run a cleaner through the tube
(rubbing alcohol works well to sterilize the tube). If you're careful you
aren't likely to ingest anything, but of course you want the end of the tube to
be clean.

Once the system is clean, just do this. Pull water up into the tube.
While maintaining the water at a certain height, replace the cap on the
pressure-valve tube and think for a minute about what's going to happen when you
remove the tube from your mouth. Also think about what, if anything, is
going to happen to the length of the air column at the end of the
pressure-indicating tube. Then go ahead and remove the tube from your
mouth and watch what happens.

Describe in the box below what happens and what you expected to happen.

Also indicate why you think this happens.

***********

Now think about what will happen if you remove the cap from the

pressure-valve tube. Will air escape from the system? Why would you

or would you not expect it to do so?

Go ahead and remove the cap, and report your expectations and your

observations in the box below.

***********

Now replace the cap on the pressure-valve tube and, while keeping an eye on

the air column in the pressure-indicating tube, blow just a little air through

the vertical tube, making some bubbles in the water inside the tube.

Blow enough that the air column in the pressure-indicating tube moves a little,

but not more than half a

centimeter or so. Then remove the tube from your mouth, keeping an eye on

the pressure-indicating tube and also on the vertical tube.

What happens?

Why did the length of the air column in the pressure-indicating tube
change length when you blew air into the system? Did the air column move
back to its original position when you removed the tube from your mouth?
Did it move at all when you did so?

What happened in the vertical tube?

Why did all these things happen? Which would would you have
anticipated, and which would you not have anticipated?

***********

Place the thermometer that came with your kit near the bottle, with the bulb

not touching any surface so that it is sure to measure the air temperature in

the vicinity of the bottle and leave it alone until you need to read it.

Now you will blow enough air into the bottle to raise water in the vertical

tube to a position a little ways above the top of the bottle.

Use the pressure-valve tube to equalize the pressure once more with
atmospheric (i.e., take the cap off). Measure the length of the air column in the
pressure-indicating tube, and as you did before place a measuring device in the
vicinity of the meniscus in this tube.

Replace the cap on the pressure-valve tube and again blow a little bit of air
into the bottle through the vertical tube. Remove the tube from your mouth
and see how far the water column rises. Blow in a little more air and
remove the tube from your mouth. Repeat until water has reached a level
about 10 cm above the top of the bottle.

Place the bottle in a pan, a bowl or a basin to catch the water you will soon
pour over it.

Secure the vertical tube in a vertical or nearly-vertical position.

The water column is now supported by excess pressure in the bottle.

This excess pressure is between a few

hundredths and a tenth of an atmosphere.

The pressure in the bottle is probably in the range from 103 kPa to 110 kPa,

depending on your altitude above sea level and on how high you chose to make the

water column. You are going to make a few estimates, using 100 kPa as the

approximate round-number pressure in the bottle, and 300 K as the approximate round-number air

temperature. Using these ball-park figures:

If gas pressure in the bottle changed by 1%, by how many N/m^2 would it
change?

What would be the corresponding change in the height of the supported
water
column?

By what percent would air temperature have to change to result in this
change in pressure, assuming that the container volume remains constant?

Report your numbers in the first three lines below, one number to a line,

then starting in the fourth line explain how you made your estimates:

1% of 100 kPa is 1 k Pa, or 1000 Pa, meaning 1000 N / m^2.

The pressure required to support a vertical water column is rho g `dy.
In this case rho = 1000 kg / m^3 (the density of water), g = 9.8 m/s^2 (the
acceleration of gravity) and `dy is the height of the water column, so we have

rho g `dy = 1000 N / m^2, so that

`dy = ( 1000 N / m^2 ) / (rho g)

= (1000 N / m^2) / (1000 kg / m^3 * 9.8 m/s^2)

= (1000 N / m^2) / (9800 N / m^3)

= .102 m, approx., or about 10.2 cm.

We apply the gas laws to determine the temperature difference.

Remember that the gas laws apply only to Kelvin temperatures.

In this situation n and V are constant or very nearly so, so P2 / P1 =
T2 / T1, and a pressure change of 1% implies a temperature change of 1%.

Both initial and final temperatures are close to 300 Kelvin, and .01 *
300 K is about 3 K.

We conclude that this pressure difference could be achieved by a
temperature difference of about 3 degrees Kelvin (a temperature difference
of around 5 degrees Fahrenheit).

A 1 cm change in water column height implies a pressure change of

`dP = rho g `dy = 1000 kg/m^3 * 9.8 m/s^2 * .01 m = 98 Pa.

This is about (98 Pa / (100,000 Pa / atmosphere) = .001 atmosphere and would
correspond to a temperature difference of about .001 * 300 K = .3 K.

Continuing the above assumptions:

How many degrees of temperature change would correspond to a 1% change in
temperature?

How much pressure change would correspond to a 1 degree change in
temperature?

By how much would the vertical position of the water column change with a 1
degree change in temperature?

Report your three numerical estimates in the first three lines below, one

number to a line, then starting in the fourth line explain how you made your

estimates:

As seen previously a 1% change in temperature is about 3 degrees Kelvin.

So a temperature change of 1 degree Kelvin would constitute a change of
about 1/3 of 1%, a factor of about .003.

A pressure change of 1/3 of 1% would be about .003 * 100 kPa = . 3 kPa = 300
Pa.

This would change the height of the vertical water column by about 3 cm.

How much temperature change would correspond to a 1 cm difference in the

height of the column?

How much temperature change would correspond to a 1 mm difference in the

height of the column?

Report your two numerical estimates in the first two lines below, one

number to a line, then starting in the third line explain how you made your

estimates:

We have seen that a 1% temperature change, or about 3 Kelvin, would
correspond to a 10 cm change in the vertical column. So a 1 cm change
being about 1/10 of this, the temperature change would be about 1/10 of the 3
Kelvin change, i.e., a change of about .3 degrees Kelvin.

A 1 mm difference is about 1/10 of this, corresponding to a change of
about .03 degrees Kelvin.

We see that the water column height is very sensitive to temperature
changes.

A change in temperature of 1 Kelvin or Celsius degree in the gas inside the

container should correspond to a little more than a 3 cm change in the height of

the water column. A change of 1 Fahrenheit degree should correspond to a

little less than a 2 cm change in the height of the water column. Your results should be consistent with these

figures; if not, keep the correct figures in mind as you make your observations.

The temperature in your room is not likely to be completely steady. You

will first see whether this system reveals any temperature fluctuations:

Make a mark, or fasten a small piece of clear tape, at the position of the water
column.

Observe, at 30-second intervals, the temperature on your alcohol
thermometer and the height of the water column relative to the mark or tape
(above the tape is positive, below the tape is negative).

Try to estimate the temperatures on the alcohol thermometer to the
nearest .1 degree, though you won't be completely accurate at this level of
precision.

Make these observations for 10 minutes.

Report in units of Celsius vs. cm your 20 water column position vs. temperature data, in the form of a

comma-delimited table in the box below.

A fluctuation of 1 cm would indicate a temperature change of about 1/3 of a
degree; a fluctuation of 1 mm would correspond to about .03 degrees.

Due to some expansion of the bottle with temperature, the actual
fluctuations in the water column will be a bit less than this, but the
column should still be sensitive to changes on the order of .1 degree.

In the box below describe the trend of temperature fluctuations. Also

include an estimate (or if you prefer two estimates) based on both the alcohol thermometer and the 'bottle thermometer' the maximum deviation in temperature over the 10-minute period.

Explain the basis for your estimate(s):

For every cm of deviation the temperature change is about .3 degrees Kelvin.

Now you will change the temperature of the gas in the system by a few degrees

and observe the response of the vertical water column:

Read the alcohol thermometer once more and note the reading.

Pour a single cup of warm tap water over the sides of the bottle and
note the water-column altitude relative to your tape, noting altitudes at
15-second intervals.

Continue until you are reasonably sure that the temperature of the
system has returned to room temperature and any fluctuations in the column
height are again just the result of fluctuations in room temperature.
However don't take data on this part for more than 10 minutes.

***********

If your hands are cold, warm them for a minute in warm water. Then hold

the palms of your hands very close to the walls of the container, being careful

not to touch the walls. Keep your hands there for about a minute, and keep

an eye on the air column.

Did your hands warm the air in the bottle measurably? If so, by how

much? Give the basis for your answer:

Unless your hands are cold, the will radiate thermal energy into the system,
and the amount is typically sufficient to change the temperature in the bottle
by a few tenths of a degree; a well-sealed system would be sensitive to a change
of this magnitude.

An alcohol thermometer would not respond quickly enough to detect the
temperature change.

Now reorient the vertical tube so that after rising out of the bottle the

tube becomes horizontal. It's OK if some of the water in the tube leaks o ut during this

process. What you want to achieve is an open horizontal tube,, about 30 cm

above the level of water in the container, with the last few centimeters of the

liquid in the horizontal portion of the tube and at

least a foot of air between the meniscus and the end of the tube.

The system might look something like the picture below, but the tube running

across the table would be more perfectly horizontal.

Place a piece of tape at the position of the vertical-tube meniscus (actually

now the horizontal-tube meniscus). As you did earlier, observe the alcohol thermometer and

the position of the meniscus at 30-second

intervals, but this time for only 5 minutes. Report your results below in

the same table format and using the same units you used previously:

The position of the meniscus in the horizontal tube should be much more
sensitive to temperature changes than the position in the vertical tube.

Repeat the experiment with your warm hands near the bottle. Report below

what you observe:

The greater sensitivity of the horizontal configuration should result in a
significant change in the meniscus position.

When in the first bottle experiment you squeezed water into a horizontal

section of the tube, how much additional pressure was required to move water

along the horizontal section?

By how much do you think the pressure in the bottle changed as the water
moved along the horizontal tube?

If the water moved 10 cm along the horizontal tube, whose inner diameter
is about 3 millimeters, by how much would the volume of air inside the
system change?

By what percent would the volume of the air inside the container
therefore change?

Assuming constant pressure, how much change in temperature would be
required to achieve this change in volume?

If the air temperature inside the bottle was 600 K rather than about 300
K, how would your answer to the preceding question change?

Give your answers, one to a line, in the first 5 lines of the box below.

Starting in the sixth line, explain how you reasoned out these results:

If the tube was horizontal, the vertical level of the water would not
change, and no change in pressure inside the bottle would be required.

The cross-sectional area of the tube is about pi r^2 = pi (.15 cm)^2 =
.07 cm^2, approx.. A 10 cm section of the tube therefore has volume
(10 cm) * (.07 cm^2) = .7 cm^3.

A 10 cm change in meniscus position therefore corresponds to a .7 cm^3
change in the volume of the gas in the container.

Assuming container volume 1500 cm^3 (this is 3/4 the volume of a 2-liter
container, about what we would expect if the container was 1/4 full of
water), the .7 cm^3 volume of the 10 cm section of the tube is .7 cm^3
/ (1500 cm^3) = .0005 times the volume of the air, or about .05%.

A .05 % change in volume, at constant pressure, would correspond to a
.05% change in temperature. Assuming an initial temperature in the
vicinity of 300 Kelvin, this would be about .0005 * 300 Kelvin = .15 Kelvin.

If the air temperature was 600 Kelvin, then the .05% change in
temperature would correspond to .05% of 600 Kelvin, or about .3 Kelvin.

These numbers would differ proportionally for larger or smaller
containers and air volumes.

There were also changes in volume when the water was rising and falling in

the vertical tube. Why didn't we worry about the volume change of the air in

that case? Would that have made a significant difference in our estimates

of temperature change?

A 1 cm change in vertical position would be associated with an insignificant
change in volume (the change could easily be calculated; it would be 1/10 the
change calculated for a 10 cm change in horizontal position, or about .07 cm^3,
about .005% of a 1500 cm^3 volume).

As we saw previously a 1 cm change in vertical position corresponds to a
temperature change of about .3 Kelvin, while a 10 cm change in horizontal
position corresponds to a temperature change only half as great. The
horizontal configuration is therefore about 20 times as sensitive to
temperature change as the vertical.

If the tube was not completely horizontal, would that affect our estimate of

the temperature difference?

For example consider the tube in the picture below.

Suppose that in the process of moving 10 cm along the tube, the meniscus

moves 6 cm

in the vertical direction.

By how much would the pressure of the gas have to change to increase the
altitude of the water by 6 cm?

Assuming a temperature in the neighborhood of 300 K, how much
temperature change would be required, at constant volume, to achieve this
pressure increase?

The volume of the gas would change by the additional volume occupied by
the water in the tube, in this case about .7 cm^3. Assuming that there
are 3 liters of gas in the container, how much temperature change would be
necessary to increase the gas volume by .7 cm^3?

Report your three numerical answers, one to a line, in the box. Then

starting on the fourth line, explain how you obtained your results. Also

make note of the relative magnitudes of the temperature changes required to

increase the altitude of the water column, and to increase the volume of the

gas.

***********

Continue to assume a temperature near 300 K and a volume near 3 liters:

If the tube was in the completely vertical position, by how much would
the position of the meniscus change as a result of a 1 degree temperature
increase?

What would be the change if the tube at the position of the meniscus was
perfectly horizontal? You may use the fact that the inside volume of a
10 cm length tube is .7 cm^3.

A what slope do you think the change in the position of the meniscus
would be half as much as your last result?

Report your three numerical answers, one to a line, in the box. Then

starting on the fourth line, explain how you obtained your results. Also

indicate what this illustrates about the importance in the last part of the

experiment of having the tube in a truly horizontal position.

bottle thermometer

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? **

** What happens when you remove the pressure-release cap? **

** What happened when you blew a little air into the bottle? **

In order to lower the pressure the air in the bottle had to expand, displacing water into 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: **

** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **

** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **

** water column position (cm) vs. thermometer temperature (Celsius) **

** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **

I suspect the system wasn't well sealed (no fault of yours, unless you left the cap loose, which is unlikely). A well-sealed system would be sensitive to small temperature changes you can't detect with the thermometer.

** Water column heights after pouring warm water over the bottle: **

** Response of the system to indirect thermal energy from your hands: **

** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **

** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **

** 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: **

** Why weren't we concerned with changes in gas volume with the vertical tube? **

** 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: **

** 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: **

** **

Bottle Thermometer

***********

***********

***********

1% of 100 kPa is 1 k Pa, or 1000 Pa, meaning 1000 N / m^2.

The pressure required to support a vertical water column is rho g `dy. In this case rho = 1000 kg / m^3 (the density of water), g = 9.8 m/s^2 (the acceleration of gravity) and `dy is the height of the water column, so we have

rho g `dy = 1000 N / m^2, so that

`dy = ( 1000 N / m^2 ) / (rho g)

= (1000 N / m^2) / (1000 kg / m^3 * 9.8 m/s^2)

= (1000 N / m^2) / (9800 N / m^3)

= .102 m, approx., or about 10.2 cm.

We apply the gas laws to determine the temperature difference.

Remember that the gas laws apply only to Kelvin temperatures.

In this situation n and V are constant or very nearly so, so P2 / P1 = T2 / T1, and a pressure change of 1% implies a temperature change of 1%.

Both initial and final temperatures are close to 300 Kelvin, and .01 * 300 K is about 3 K.

We conclude that this pressure difference could be achieved by a temperature difference of about 3 degrees Kelvin (a temperature difference of around 5 degrees Fahrenheit).

A 1 cm change in water column height implies a pressure change of

`dP = rho g `dy = 1000 kg/m^3 * 9.8 m/s^2 * .01 m = 98 Pa.

This is about (98 Pa / (100,000 Pa / atmosphere) = .001 atmosphere and would correspond to a temperature difference of about .001 * 300 K = .3 K.

As seen previously a 1% change in temperature is about 3 degrees Kelvin.

So a temperature change of 1 degree Kelvin would constitute a change of about 1/3 of 1%, a factor of about .003.

A pressure change of 1/3 of 1% would be about .003 * 100 kPa = . 3 kPa = 300 Pa.

This would change the height of the vertical water column by about 3 cm.

We have seen that a 1% temperature change, or about 3 Kelvin, would correspond to a 10 cm change in the vertical column. So a 1 cm change being about 1/10 of this, the temperature change would be about 1/10 of the 3 Kelvin change, i.e., a change of about .3 degrees Kelvin.

A 1 mm difference is about 1/10 of this, corresponding to a change of about .03 degrees Kelvin.

We see that the water column height is very sensitive to temperature changes.

A fluctuation of 1 cm would indicate a temperature change of about 1/3 of a degree; a fluctuation of 1 mm would correspond to about .03 degrees.

Due to some expansion of the bottle with temperature, the actual fluctuations in the water column will be a bit less than this, but the column should still be sensitive to changes on the order of .1 degree.

For every cm of deviation the temperature change is about .3 degrees Kelvin.

***********

Unless your hands are cold, the will radiate thermal energy into the system, and the amount is typically sufficient to change the temperature in the bottle by a few tenths of a degree; a well-sealed system would be sensitive to a change of this magnitude.

An alcohol thermometer would not respond quickly enough to detect the temperature change.

The position of the meniscus in the horizontal tube should be much more sensitive to temperature changes than the position in the vertical tube.

The greater sensitivity of the horizontal configuration should result in a significant change in the meniscus position.

If the tube was horizontal, the vertical level of the water would not change, and no change in pressure inside the bottle would be required.

The cross-sectional area of the tube is about pi r^2 = pi (.15 cm)^2 = .07 cm^2, approx.. A 10 cm section of the tube therefore has volume (10 cm) * (.07 cm^2) = .7 cm^3.

A 10 cm change in meniscus position therefore corresponds to a .7 cm^3 change in the volume of the gas in the container.

Assuming container volume 1500 cm^3 (this is 3/4 the volume of a 2-liter container, about what we would expect if the container was 1/4 full of water), the .7 cm^3 volume of the 10 cm section of the tube is .7 cm^3 / (1500 cm^3) = .0005 times the volume of the air, or about .05%.

A .05 % change in volume, at constant pressure, would correspond to a .05% change in temperature. Assuming an initial temperature in the vicinity of 300 Kelvin, this would be about .0005 * 300 Kelvin = .15 Kelvin.

If the air temperature was 600 Kelvin, then the .05% change in temperature would correspond to .05% of 600 Kelvin, or about .3 Kelvin.

These numbers would differ proportionally for larger or smaller containers and air volumes.

A 1 cm change in vertical position would be associated with an insignificant change in volume (the change could easily be calculated; it would be 1/10 the change calculated for a 10 cm change in horizontal position, or about .07 cm^3, about .005% of a 1500 cm^3 volume).

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Your optional message or comment:

What happens when you pull water up into the vertical tube then remove the tube from your mouth?

What happens when you remove the pressure-release cap?

What happened when you blew a little air into the bottle?

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:

Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change:

The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change:

water column position (cm) vs. thermometer temperature (Celsius)

Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer.

Water column heights after pouring warm water over the bottle:

Response of the system to indirect thermal energy from your hands:

position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals

What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container?

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:

Why weren't we concerned with changes in gas volume with the vertical tube?

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:

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:

The pressure required to support a vertical water column is rho g `dy.
In this case rho = 1000 kg / m^3 (the density of water), g = 9.8 m/s^2 (the
acceleration of gravity) and `dy is the height of the water column, so we have

In this situation n and V are constant or very nearly so, so P2 / P1 =
T2 / T1, and a pressure change of 1% implies a temperature change of 1%.

Both initial and final temperatures are close to 300 Kelvin, and .01 *
300 K is about 3 K.

We conclude that this pressure difference could be achieved by a
temperature difference of about 3 degrees Kelvin (a temperature difference
of around 5 degrees Fahrenheit).

This is about (98 Pa / (100,000 Pa / atmosphere) = .001 atmosphere and would
correspond to a temperature difference of about .001 * 300 K = .3 K.

So a temperature change of 1 degree Kelvin would constitute a change of
about 1/3 of 1%, a factor of about .003.

A pressure change of 1/3 of 1% would be about .003 * 100 kPa = . 3 kPa = 300
Pa.

We have seen that a 1% temperature change, or about 3 Kelvin, would
correspond to a 10 cm change in the vertical column. So a 1 cm change
being about 1/10 of this, the temperature change would be about 1/10 of the 3
Kelvin change, i.e., a change of about .3 degrees Kelvin.

A 1 mm difference is about 1/10 of this, corresponding to a change of
about .03 degrees Kelvin.

We see that the water column height is very sensitive to temperature
changes.

A fluctuation of 1 cm would indicate a temperature change of about 1/3 of a
degree; a fluctuation of 1 mm would correspond to about .03 degrees.

Due to some expansion of the bottle with temperature, the actual
fluctuations in the water column will be a bit less than this, but the
column should still be sensitive to changes on the order of .1 degree.

Unless your hands are cold, the will radiate thermal energy into the system,
and the amount is typically sufficient to change the temperature in the bottle
by a few tenths of a degree; a well-sealed system would be sensitive to a change
of this magnitude.

An alcohol thermometer would not respond quickly enough to detect the
temperature change.

The position of the meniscus in the horizontal tube should be much more
sensitive to temperature changes than the position in the vertical tube.

The greater sensitivity of the horizontal configuration should result in a
significant change in the meniscus position.

If the tube was horizontal, the vertical level of the water would not
change, and no change in pressure inside the bottle would be required.

The cross-sectional area of the tube is about pi r^2 = pi (.15 cm)^2 =
.07 cm^2, approx.. A 10 cm section of the tube therefore has volume
(10 cm) * (.07 cm^2) = .7 cm^3.

A 10 cm change in meniscus position therefore corresponds to a .7 cm^3
change in the volume of the gas in the container.

Assuming container volume 1500 cm^3 (this is 3/4 the volume of a 2-liter
container, about what we would expect if the container was 1/4 full of
water), the .7 cm^3 volume of the 10 cm section of the tube is .7 cm^3
/ (1500 cm^3) = .0005 times the volume of the air, or about .05%.

A .05 % change in volume, at constant pressure, would correspond to a
.05% change in temperature. Assuming an initial temperature in the
vicinity of 300 Kelvin, this would be about .0005 * 300 Kelvin = .15 Kelvin.

If the air temperature was 600 Kelvin, then the .05% change in
temperature would correspond to .05% of 600 Kelvin, or about .3 Kelvin.

These numbers would differ proportionally for larger or smaller
containers and air volumes.

A 1 cm change in vertical position would be associated with an insignificant
change in volume (the change could easily be calculated; it would be 1/10 the
change calculated for a 10 cm change in horizontal position, or about .07 cm^3,
about .005% of a 1500 cm^3 volume).