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.

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. 

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.

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:

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.

***********

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.

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:

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:

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:

    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:

    ***********

    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.

     

    It is common for a student to state that the energy from your hands goes directly into the air and doesn't get to the bottle.:

    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?

    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.

    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.

    ***********

    If `dy = 6 cm then

     

    T2 / T1 = V2 / V1, provided n and P are constant. So the fractional change in temperature and volume must be the same

     

    So the fractional change in temperature is .7 / 3000.

    .7 / 3000 * 300 K = .07 K.

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

    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.

    In a completely vertical orientation the volume of the gas in the bottle would change very little.  The pressure increase would be proportional to the temperature increase.  Assuming temperature about 300 K, a 1 degree increase would be about 1/300 of the absolute temperature, so the pressure increase would be about 1/300 of the absolute pressure.  The absolute pressure is about 1 atmosphere, so the increase would be about 1/300 atmosphere or about 330 Pa.  This would correspond to the additional pressure of a water column of height about 3.3 cm. 

    If the tube is perfectly horizontal, there will be no increase in pressure and the gas will expand by factor 1/300.  Assuming 3 liters of gas, this corresponds to about .01 liter or 10 cm^3.  If the tube volume is .7 cm^3 per 10 cm length of tubing, then 10 cm^3 would require about 140 cm of tubing.  The meniscus would therefore move about 140 cm.

    The very last question will be left open here.  However observe the following: