Ask A Scientist Chemistry Archive

name         Brian
status       educator
location     CT

Question -   In helping my son with a science experiment dealing with
the difference of heat flow through air and a vacuum. I expected to see
better heat flow to a digital thermometer suspended inside a 5 gallon
water bottle with air inside than when it is evacuated with a vacuum
pump. I do see a ten degree temperature drop when first evacuated, but
the system returns to about room temperature after 25 minuets. then a
hair dryer is used to apply  heat to the out side of the bottle. I get a
10 degree temp rise either way over a ten minute heating period. I had
assumed convection currents would help transfer more of the heat.

Can anyone help justify these results? If there is no air does all the
energy go to radiation and if there is air some of the energy is
transferred by radiation and some by convection? conserving total energy?
-----------------
----If there is no air does all the energy go to radiation and if there is 
air some of the energy is
transferred by radiation and some by convection? ----
Well, yes.

And the geometry of the situation in the center of the bottle
is such that you have hot temperatures on one side and cool on the other,
and the center location tends to an average of the two,
regardless of the method of heat flow.

Think of it this way.
When you evacuated, sure, you thermally-insulated the heating path,
but you also thermally-insulated the cooling path.
Some fine scientific thermometers are made this way...
They don't want the air pressure to affect the thermometer reading.
(just kidding, actually)

To do a test of thermal conductivity,
either the heat-source path or the heat-sink path must have a fixed 
thermal resistance.

Another thought-  convection carries heat upwards in narrow plumes,
rather than straight across from bottle side-wall to thermometer.
So either the heat will not get to your thermometer well,
or it will get there because convection cycles are carrying it throughout 
the bottle.
Meanwhile it will be loosing heat out the sides of the bottle.
Either way, the temperature rise won't be quite as large as you expect.

You could directly compare a vacuum-or-air heating path, against an 
always-air cooling path,
by gluing the thermometer to the side of the bottle.
(Gluing inside or outside the bottle would be roughly the same.  Outside 
is easier.)
Then the cooling path (air convection, and some radiation) will be fixed, 
will not change when you evacuate the bottle.
Then, warming the far wall of the bottle,
heat will be transferred across the bottle volume to the thermometer:
    by radiation if evacuated,
    by radiation plus convection if not.
Clearly to have both methods working will transfer more calories of heat 
for a fixed temperature drop.
For this experiment,  you should read a higher temperature when the bottle 
is filled.

A fixed-resistance cooling path can be made with a length of thick copper 
wire from thermometer to outside the bottle.
It would take some trial-and-error or calculations to get the right thickness
that would have conductivity similar in magnitude to your air or 
vacuum/radiation.

By using resistor heating, you could eliminate the outside-to-inside 
heating path,
and then be measuring only the conductivity of the cooling path from 
inside to outside.
Resistors purely create known wattages of heat in one small spot, which 
then must escape to the surrounding environment.
Run two thin wires in to a resistor next to the thermometer tip,
having first surrounded the thermometer-tip with several wraps of aluminum 
foil for heat-spreading,
and later making a few more around resistor and thermometer together.
Power the resistor with enough voltage to make a 10degC temperature rise 
in air.
When evacuated the  I would expect the temperature rise to be greater, 
perhaps 30-60 C.
Maybe more.

Jim Swenson
====================================================
This is actually a complex and rich phenomenon. Thank you for sharing this 
experience with us.Now let me take a stab at explaining what happened (I 
am just arguing from first principles here and we really need to do the 
math to be sure).

First let us make sure that we are clear on the difference between heat 
and temperature. Heat is energy, while temperature is average velocity of 
particles. Normally as the energy of a substance increases, so does its 
temperature. But the two can also be independent variables. For example, a 
candle may have a very high temperature, but it does not provide a lot of 
energy. By comparison, a glass of warm water may be at a lower temperature 
than a candle, but it will melt an ice cube faster than a candle - 
indicating that it has a lot more heat than a candle.

Secondly, a thermometer (digital or spirit) measures the average velocity 
of a substance by coming to thermal equilibrium with the environment. This 
means that the expansion of the spirit (in spirit thermometers) or the 
thermocouple (in digital thermometers) will continue for so long as energy 
can be transferred to the thermometer - and energy will continue to be 
transferred until the substance being measured and the thermometer have 
the same average particle velocity.

Thirdly, I am going to assume that the vacuum you developed in the jug is 
not a very good vacuum. The plastic water bottle did not collapse. So what 
we are really comparing here is the difference in heat transfer between 
ambient pressure air versus low pressure air.

Finally, a word on heat capacity. Heat capacity is a sample property, not 
a material property. Thus, a gallon of water takes longer to get to a 
particular temperature then a cup of the same water. Same with air.

So here is my guess:
1) If you altered the experiment such that instead of heating with a heat 
gun, you immersed the jug in equal volumes of warm water, the thermometer 
will have practically the same *final* reading as the system thermally 
equilibrates - there will be some difference due to there being less air 
in one case, but that will be negligible.

2) As far as the rate of temperature rise - this is really a kinetics and 
not thermodynamics issue and there are competing rate factors. While the 
final temperatures will be the same, the amount of heat transferred will 
be a function of the heat capacity of the air in the jug. While the case 
of ambient pressure has more air to allow better heat transfer, it also 
takes a longer time for the air in the jug to raise its own temperature 
(there is a lot more air, it has a higher heat capacity). While the case 
of low pressure air does not transfer heat very well, it can get to a 
higher temperature faster because it has a lower heat capacity - therefore 
it will have a higher temperature differential between it and the thermometer.

Without doing the math, I cannot be sure how the two functions offset - 
but you get the idea.

Greg (Roberto Gregorius)
=====================================================
Here is a "back-of-the-envelope" calculation. The thermal conductivity
of air at room temperature is: 2.56 (watts/meter x K). Assume 1 meter
(that is a large bottle, but the distance cancels out in the comparison of
conduction and radiation). So for a temperature difference of 10 C = 10 K
this is:
25.6 (watts/m). This is conduction by diffusion, not convection. Energy
transfer by radiation obeys the Stefan-Boltzman Law: E = Ks * T^4 where the
Stefan-Boltzman constant
  Ks = 5.7 x 10^-8 watts / meter^2 * K^4. Now radiation, unlike conduction,
works both ways so the net power delivered to the thermometer is the
difference of T^4, or for a 10 C = 10 K temperature difference is E =
5.7x10^-8* ( (290^4) - (300^4) = 5.7 * (71 - 81) = 57 watts / meter^2. The
fourth power of the temperature cancels out the 10^-8. This is of the same
order of magnitude as the power due to conduction, so that is an "warning
bell" that things may not be so simple. Admittedly there are some BIG
approximations here: 1. The area in the two equations enter in a different
way: in conduction it refers to the separation of the jar wall and the
thermometer, but in the Stefan-Boltzman law it refers to the areas of the
thermometer and the jar, respectively (which are different due to the
geometric factors). 2. Since you cannot keep the temperature of the jar at a
uniform temperature both conduction and radiation calculations are only very
approximate. 3. As a result of these temperature differences in the wall or
the jar there will be some convection in addition to conduction. However,
this convection is really complicated because hot air may rise at the
surface of the jar, cool, and fall due to its increasing density, but this
path may never get near the thermometer!! So the convection increases the
temperature of the temperature probe indirectly by warming air some distance
away from the probe followed by either a new cycle of convection or by
conduction -- in either case a real mess to model.
     The common perception is that radiative heating is negligible compared
to conduction and/or convection; however, that is not so because to power
enters as T^4 which gets very large for temperatures at and above ambient.
To convince yourself of this use the palm of your hands as thermometers. Put
one hand above the burner on an electric stove. Put the other the same
distance away, but beside the burner. Turn the burner on (low will do, for
safety purposes). The hand above the burner is receiving power both by
convection, conduction, and radiation. The hand at the side is only
receiving power by radiation since it is at right angles to the burner
surface. Both hands will respond to the heat fairly equally -- at least
equally enough to be convincing that radiation is important.

Vince Calder
=====================================================