Atmospheric Pressure Differences and Water Phase Changes

(aka - Can Crushing!)

Purposes:

1) To illustrate the presence of significant ambient atmospheric pressure and the consequences when pressure differences develop.

2) To show how water vapor can change phase to liquid extremely rapidly when there is significant supersaturation.

3) To illustrate the significance of the latent heat that is released as water condenses (in the supplementary discussion of dew formation on a cold can of pop).

Equipment:

Hot plate (and probably an extension cord to plug it in)

Empty, clean soda cans (three or more is recommended to allow for repeat demonstrations)

Glass water dish

Fabric safety glove

Procedure

1) Fill dish mostly full with COLD water - this is where you will actually crush the can, so it's helpful to have enough depth to increase safety and the chance for success.

2) Heat up the hot plate using a very high setting. (This make take 10 minutes or more depending on the hot plate. Test the one you plan to use prior to your section, if possible.)

3) Place a SMALL amount of water in the can. You want enough to cover the bottom of the can, but not so much that boiling will take more than a few minutes. Cover the bottom of the can with up to 1 cm of water. Add enough so the water doesn't boil off immediately, but not so much that it will take too long to boil or create a scalding hazard when inverted.
4) Place the can on the hot plate and allow the water to boil.

5) Using the glove as protection, QUICKLY grab the can off the hot plate and invert it in the cold water dish. If the opening to the cans's opening is competely submerged, it should dramatically implode.

(A movie

of the "can crushing")

** You may wish to heat an empty can and invert it in the dish to demonstrate the importance of the boiling/condensing water to the experiment's success. However, the coating / paint on empty soda tends to burn quickly, emitting a foul odor that is probably not good for you. If you can heat and invert the dry can quickly, you may be able to avoid the problem. Otherwise, avoid the dry can "control" run!

Explanation

-Why did the can implode?

The pressure on the inside of the can was less than the pressure outside of the can after the dramatic temperature reduction. The dramatic pressure reduction in the can results from slower molecular motions associated with the dramatic cooling in addition to the phase change described below. The phase change from gas to liquid implies a release of latent heat (the latent heat required to boil the water is supplied by the hot plate). When pressure differences exist across a surface, a force is being applied from the zone of high pressure toward the low pressure zone.

-How did we "create" the low pressure inside the can?

Boiling changed the liquid water at the bottom of the can into water vapor. The water vapor filled much of the can's interior, replacing the well-mixed distribution of air which existed previously. Placing the can upside-down and partially submerged in the cold water does two things:
1) it isolates the interior of the can from the outside air, prohibiting any exchange as conditions evolve, and
2) it rapidly cools the interior of the can, causing much of the water vapor to condense on the inner walls of the can. As the water vapor returns to the liquid phase, the saturation vapor pressure (and therefore the total pressure) drops dramatically. This pressure change can be thought of as a dramatic shift toward liquid water in the closed system of the can's interior, leaving far fewer water molecules in the gas phase to exert pressure on the inside walls of the can.

Phase Change Questions - A Supplementary Activity

1) Two 12-oz cans of Coke are placed outside in summer, one in Denver and one in Baltimore. The outside temperature is 90 F. The cans both come out of the refrigerator at 45 F. (Neither can is opened, and nothing is drunk from either can.) Which can is likely to collect the most dew?

The one in Baltimore will have more dew, based on climatology.

2) What is the maximum value that the dew point can be in Denver if no water is condensing on the outside of the can?

It must be less than 45 F.

3) Guess the relative humidity when T_d = 45, T= 90.

(21%)

[Do not describe the math used to get this; I approximated it as the ratio of actual and saturation mixing ratios from a skew-T as 6.5/31]

3) Suppose a total of 1/10 ounce of water condenses on the outside of the 12-oz Coke can in Baltimore after 5 minutes and there is nothing condensed on the can in Denver. In which city is the Coke warmer?

Baltimore, because of latent heat released from the condensation onto the Baltimore Coke can.

[Note that 1/10 ounce of water = 2.8 gm = 2.8 cm^3 of water. Surface area of a 12-oz can is roughly 286 cm^2, so this amounts to a 0.1 mm film of water uniformly covering the top and sides of the can.]

4) Guess how much warmer the latent heat released by the 1/10 ounce of condensed water makes the 12-oz Coke in Baltimore?

(about 9 F)

[Again no math for the student. I compute 1/10 oz = 2.8 gm x 600 cal/gm =1860 cal 1860 cal/(340 gm = 12 oz) yields 4.9 cal/gm which will produce a 5 C or 9 F temperature rise. For the guessing part, you could ask for a show of hands once a few numbers are volunteered; i.e., after 4 suggestions, have them all pick the one they think is closest by a show of hands.]

5) In the above we assume all the latent heat released goes into heating the Coke, how good is this assumption?

It's a relatively good one; heat conductivity is fast through an aluminum can and there is little mass in the aluminum. Some latent heat lost to outside air. Net effect of cold can is to cool the air, but the latent heating must reduce the cooling somewhat in the moist case. It would be interesting to try this as an actual experiment with a can in humid weather.

6) At what temperature would the Baltimore Coke stop warming up due to latent heating?

When its temperature rose above the dew-point. The dew point in Baltimore might be 65 F; it could be 75 on a really muggy day. T_d = 65, T=90 corresponds to RH of 48%; T_d = 75, T=90 corresponds to RH of 61%.

Webpage composed by Justin Wettstein, Joe Casola, and Professor Dale Durran.
Last Updated March 23, 2005.