bottle thermometer

#$&*

PHY 232

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

1 hour

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

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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 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 below what happens and what you expected to happen.

Also indicate why you think this happens.

****

The water will lower slightly in the tube but stabilize.

This is because there is a slight variation in the pressure

outside of the water column and in the tube.

#$&*

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 below.

****

A small amount of air escaped from the bottle

#$&*

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?

****

There were more air molecules which means there was more

pressure in the bottle. The molecules of air and water are

pushing outward because the volume of the bottle is not

large enough.

#$&*

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 air 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:

****

1000

It would change by 1 kPa which is 1000 N/m^2

p_1 + 'rho g h_1 = p_2 + 'rho g h_2

'rho g (h_1 - h_2) = p_2 - p_1

'rho g (-'dh) = 'dp

'dh = (-'dp) / ('rho g)

'dp / 'dh = 'rho g = const

'dp and 'dh are directly proportional

So the height would change by 1%

PV = nRT

P / T = nR/V = const

P and T are directly proportional so the temperature would

have to change by 1%

#$&*

@& Absolute temperature change would be 1%, because when volume is constant the absolute pressure and absolute temperature are directly proportional.

The change in height is related to the change in pressure, but unlike temperature and pressure, which can be expressed in absolute terms, the height has an arbitrary reference point and cannot be expressed absolutely.

p_1 + rho g h_1 = p_2 + rho g h_2 does not imply any ratio between the pressures or the heights.

A 1% change in pressure corresponds to 1 kPa, as you say. This pressure difference implies a specific difference in the height of the water column.*@

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:

****

0.33%

3K / 3 = 1K, thus one third of the change. So if the

pressure change of 1% results in a 1% change in Temperature

which is also 3K, then the pressure would have to change by

a third of that to get 1K, which is a 1/3% change = 0.33...%

The same goes for the height since it is directly

proportional to pressure and thus temperature change

#$&*

@& Good, but your argument does not work for the heights, per my previous note.*@

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:

****

10% or 30K

1% or 3K

1 cm change is 10% of the 10cm, thus it would mean a 10%

change in temperature, which is 30K

1 mm change is 1% of the 10cm, thus it would mean a 1%

change in temperature, which is 3K

#$&*

@& See my previous notes. You can't apply the ratios to the heights. However rho g `dh is equal to the change in pressure, which can be then be used to find the change in the temperature.*@

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 below.

****

21, 0

21.5, 0

21.5, 0.2

21, 0.2

21, 0

21.5, 0.2

22, 0.2

21.5, 0

21.5, 0.2

21.5, 0

21, 0

#$&*

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):

****

The temperature and the height of the water column did not

fluctuate much. At most 1 degree and about 2 mm at most. But

the measurements could have a small margin of error to them.

#$&*

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.

****

18, 13, 10.5, 7.7, 6, 5, 4.3, 3.6, 3.4, 3.0, 2.8, 2.6, 2.3,

2.0, 1.8, 1.5, 1.4, 1.2, 1.0, 0.9, 0.7, 0.6, 0.5, 0.4, 0.3,

0.2, 0.1, 0.0

#$&*

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:

****

it rose about 10 cm

#$&*

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 out 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:

****

22, 0

22, 0.1

22, 0

22, 0.1

22.5, 0.1

#$&*

Repeat the experiment with your warm hands near the bottle.

Report below what you observe:

****

2.8

2.2

1.8

1.5

1.3

1.1

0.8

0.6

0.5

0.4

0.3

0.2

0.1

#$&*

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

below. Starting in the sixth line, explain how you reasoned

out these results:

****

20%

0.71cm^3

100%

Unsure

Since the volume increased so much, about 20%, the pressure

must have changed similarly (20%)

'pi(0.15cm)^2 * 10cm = V_2 = 0.71cm^3

V_1 = 0, therefore 'dV = 0.71cm^3

The original distance was 10cm and it moved another 10cm

which is 100% incease

PV = nRT

V/T = nR/P = const

Directly proportional... this isnt making much sense that

the temperature increased 100%

#$&*

@& You did well to note that this doesn't make sense. My previous notes should help you correct this.*@

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?

****

We dont have to worry about that because we can consider

other variables using bernoullis theorem.

#$&*

@& The point is that the volume of the tube is a very small percent of the volume of the air in the system, so the volume change corresponding to a small rise in the vertical water column is insignificant.*@

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 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. 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. 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.

****

????I started to get confused right about here, I think I

screwed up in my algebra way back near the beginning and my

premise is wrong. What equations should i be using?

#$&*

@& You need to rework the questions about the height of the water column, per my previous notes. That should help you get this part straight, but if not we'll deal with it after you resubmit.*@

*#&!*#&!

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: **

@& Good, but your argument does not work for the heights, per my previous note.*@

@& See my previous notes. You can't apply the ratios to the heights. However rho g `dh is equal to the change in pressure, which can be then be used to find the change in the temperature.*@

@& You did well to note that this doesn't make sense. My previous notes should help you correct this.*@

@& The point is that the volume of the tube is a very small percent of the volume of the air in the system, so the volume change corresponding to a small rise in the vertical water column is insignificant.*@

@& You need to rework the questions about the height of the water column, per my previous notes. That should help you get this part straight, but if not we'll deal with it after you resubmit.*@

Your optional message or comment: