Frosty Mornings – Can’t Help Thinking About Physics
Our unseasonably warm spell is continuing. The mornings are frosty and icy as the temperatures that overnight fell below freezing rise into the mid 40s by noon. Its hard not to think about physics when I’m out walking or biking. I snapped a few interesting pictures, took some data with my iPhone and decided to come home and read about frost.
The amount of water vapor that can be held by the air depends on the temperature. At higher temperatures the air can hold lots of water. At lower temperatures it can hold less. Water vapor (gaseous water) has more energy than liquid water which has more than solid water (ice or frost). As the temperature of the air cools, the water vapor can undergo a change of state to liquid or solid.
The temperature at which this change occurs depends on how much vapor is in the air. If the air is quite full of water vapor, even a small drop in temperature can lead to the formation of dew. When the air is dryer, with less water vapor, the temperature must fall further before dew begins to form. So the term “dew point” refers to the temperature to which the air must fall before the formation of liquid water (dew) begins. In some ways, dew point can be thought of as a measure of the water vapor load of the air.
We used to measure the dew point in the 9th grade science class by cooling a beaker of water with ice until condensation started to be seen on the outside of the container. (An ordinary glass would have worked just as well but somehow it all seemed more official and scientific if we used a beaker.) One year I heard the teacher in the room next door having trouble with the lab. I stepped in to see what was the matter. Finally it dawned on us that following a restructuring of the course, the lab moved from August to January. The difference being that in August the air is very juicy and the dew points can easily be as high as 60 or 70 F. In January the air is very dry and the dew point actually falls below freezing. We were never going to get the temperature of the water in the beaker low enough to form condensation using just ordinary ice from the freezer. It was one of those head-slap moments that sometimes leads to more learning than a lab that goes according to plans. The teacher took a few moments with the class to explain what was going on and they all brainstormed ways to actually measure such a low dew point. Using dry ice was a common suggestion. The students thought that by adding frozen CO2 to water you could lower its temperature below freezing. Clearly they needed some additional time and experience with these concepts.
So what actually does happen when the dew point falls below freezing? Instead of dew forming in the grass the water vapor moves from it gaseous state straight to a solid on the blades of grass. Instead of forming dew, it forms frost in a process called sublimation.
My interest was piqued when I noticed that the temperatures were near freezing and the frost in the shade had not melted while the sunny spots were nearly clear of ice. There was a very narrow strip of frost that had newly been exposed, but not yet melted, as the shadows moved.
I collected some data related to the position of the sun and the weather conditions. The time was about 9:35 CST on December 5.
I wondered how quickly the sun would be able to provide enough energy to melt the frost. Solar energy calculations are pretty straight forward. The sun sends out energy at a fairly constant and well known rate. How much power falls on each point depends also upon the angle of the suns rays. The easiest way to find the angle is to compare the height of an object to the length of its shadow. Objects used in this ways are referred to as gnomons. This sign acted as a fine gnomon. I don’t need to know its actual height, only the relative lengths of the object and the shadow.
Not everything has to be high tech. The easiest way to get this ratio was to make the picture full screen and then use a ruler to measure the two distances. Sadly, all I could find was this old school ruler. For measuring ratios, one unit is as good as another.
Here are the measurements and angle calculation.
Its always fun to check the Navy or Nasa websites to see how close your measured valued are to the official tables.
This would indicate that my measurements were pretty far off. Disappointing. Too late to run back and set up a better gnomon, one that is more perfectly vertical and on a surface that is more perfectly horizontal. And then take a better picture. I think I will trust the Navy Table and call it 19 degrees. Interestingly, the azimuth as measured by pointing my phone at the sun and using the compass agrees with the table pretty well. Too bad that is not the measurement I need.
Here is the way to calculate how much solar energy is falling on each square meter of the grass.
1. The sun produces energy which spread out into the sphere that surrounds it. If you divide the total solar output by the number of square meters on the surface of an imaginary sphere with a radius of 1 AU (a sphere that would touch the earth) you get a value of 1360 Joules of energy passing through each meter of the sphere’s surface each second. That value would be measured at the top of the earth’s atmosphere.
2. Some of that energy is reflected or absorbed by the atmosphere. Variations in the atmosphere will affect how much passes through but the value is generally considered to be near 1000 Joules falling on each square meter each second. This would be true of a spot on the earth directly below the sun, where the sun’s rays would be striking the surface perpendicularly.
3. At all other locations that same amount of sunlight is spread out over a larger effective area making the energy per area less. The lower the sun is in the sky the more spread out the beam becomes. The energy is less by the sine of the angle with the ground, the angle of altitude (or cosine of the angle with perpendicular, called the zenith angle). For 19 degrees the energy available each second on each square meter is a measly 330 Joules. Little wonder it is so cold here this time of year.
I am going to assume that 330 Joules of energy is available each second on each square meter and that it can all melt frost. This will let me calculate the minimum amount of time needed to melt the frost and the actual time will be something more than this.
Well, how much energy is needed? It depends on how much water there is. Each gram of ice requires about 334 Joules to undergo that phase change from ice to water assuming the frost isn’t too cold. This is a pretty good assumption since the temperature was already 34 F. But how many grams of water are there?
To determine this I weighed a paper towel, took it to a newly melted spot and pressed it into the grassy surface to collect surface water. I then reweighed the towel. Pretty cool! I measured the length and width of the paper towel.
Using this data I estimated that the total mass of frost on one square meter of grass was about 33 grams. If each gram requires 334 Joules of energy to melt, this means 11,000 Joules are required to melt the frost on each square meter. If the sun can supply 330 Joules each second to this square meter then the minimum time required for the sun to melt the frost is….. 33 seconds.
No wonder the frost band outside the shadow is so small.
While reading about frost I learned a few interesting things and ran across some useful web sites.
Great explanation of the effect of angle on illumination (with diagrams).
No longer called the “solar constant” since we now know that it varies with the occurrence of sun spot in an 11 year solar cycle. How called total solar irradiance.
About 29 percent of incoming sunlight is reflected back to space by bright particles in the atmosphere or bright ground surfaces, which leaves about 71 percent to be absorbed by the atmosphere (23 percent) and the land (48 percent).
Remember, always use physics for good, not evil!