A neat stop-motion movie made by manipulating individual atoms.
This is a great spark-the-imagination video because you can use it to talk about the physics of atoms and molecules, and what is temperature — they had to cool the atoms down to 4 Kelvin to keep them from moving too much.
Microsoft Excel, like most graphical calculators and spreadsheet programs, has the built in ability to do linear regression of measured data using certain types of functions — lines, polynomials, logarithms, and exponents for example. However, you can get it to do any type of function — sinusoidal, natural log, whatever — if you work through the spreadsheet and can use the iterative Solver tool.
This more general approach is quite useful in teaching pre-Calculus, because the primary purpose of all the functions they have to learn is to create mathematical models (functions) based on data that can be used for predictions.
The Data
I started this year’s pre-calculus class by having them collect some data. In a simplification of the snow-melt experiment I did with the middle school last year, I had them put a beaker of water (about 300ml) on a hot plate and measure the temperature every minute as warmed up.
To make the experiment a little more interesting, I had each student in each group of four take just three consecutive measurements and try to find the equation of the straight line that best fit their data, and could be used to try to predict the other measurements of their peers in their group.
It did not quite work out as I’d hoped. Since you only need two points to find the equation of a straight line, having three points produced a little confusion. I’d hoped to produce that confusion, but hadn’t realized that I’d need to do a review of how to find the equation of a straight line. A large fraction of the class was a little bit rusty after hot months of summer.
So, we pooled all the data and reviewed how to find the equation of a straight line.
Table 1: The Data
Time (minutes)
Measured Temperature (°C)
0
22
1
26
2
31
3
36
4
40
5
44
6
48
7
53
8
58
9
61
10
65
11
68
12
71
Finding the Equation for a Straight Line using Two Points
The general equation for a straight lines is:
(1)
and we need to determine the coefficients m and b. m is the slope, which can be calculated from two points using the equation:
(2)
using the points at t=6 and t=11 — the points (x1, y1) = (6,48) and (x2, y2) = (11,68) respectively — for example, gives a slope of:
so our general equation becomes:
to find b we substitute either one of the points into the equation for x and y. If we use the first point, x = 6, and y = 48, we get:
and the equation of our line becomes:
(3)
Now, since we’re actually looking at a relationship between temperature and time, with temperature on the y-axis and time on the x-axis, we could relabel the terms in the equation with T = temperature and t = time to have:
(4)
While this equation is more satisfying to me, because I think it better describes the relationship we have, the more vocal students preferred the equation in terms of x and y (Eqn 3). These are the terms they are more familiar with in the context of a math class, and I recall seeing some evidence that students seem to learn better with the more abstract representations sometimes (though I can’t quite remember the source; I should have blogged about it).
Plotting the Data and the Modeled Straight Line
The straight line equation we came up with (Eqn. 4) is our model of the data. It’s not quite perfect. All the data do not lie on the line, although, if we did everything right, only the points (6, 48) and (11, 68) are guaranteed to be on the line.
I showed the class how to plot the scatter graph using MS Excel, and how to draw the line to show the modeled data. The measured data are represented as points since the measurements were made at discrete points in time. The modeled equation, however, is a continuous function, hence the straight line. The Excel sheet below (Resource 1) illustrates:
The Excel spreadsheet (Resource 1) was set up so that when I entered the slope (m) and intercept (b) values, the graph would quickly update. So I went through the class. Everyone called out their slope and intercept values, I plugged them in, and they could all see how the modeled line changed slightly based on the points used to calculate it. So I put the question to them, “How can we figure out which model equation is the best?”
That’s how I was able to introduce the topic of error. What if we compared the temperature predicted by the model for each data point, to the actual value. The smaller the difference in modeled versus measured temperatures, the better the fit of the model. Indeed, if we sum all the differences, or better yet take the average of the differences, we could get a single number, we’ll call the average error (ε), that could be used to compare the different models. I used this opportunity to introduce sigma notation, which the pre-calculus students had not seen much of before.
As a first pass (which, as we’ll see below, has a major problem), the error (ε) for each point (i) is:
The average error is the sum of all the errors divided by the number of points (n) (we have 12 points so n=12 in this example):
(5)
Now this works, but there is one problem. I was quite pleased and a little bit surprise that one of my students recognized what it was without any coaxing and also suggested a solution: by simply taking the difference to calculate the error, a point that is offset above the modeled line can be canceled out by a point offset by the same amount below the line. So what we really need is to use the absolute value of the error.
(6)
This works, and is what we went with, but I did also point that what’s usually done is to use the square of the error instead of the absolute value. Squaring makes any number positive, so it accomplishes the same goal as the absolute value, and is the approach we’ll use when I go into linear regression later on.
Setting up the Excel spreadsheet to calculate the average error is fairly straightforward as shown in Resource 2:
So once again, we went through the class and everyone called out their slope and intercept values and cheered when I plugged the numbers in and they saw if they had the lowest value.
It is important to remember, though, that the competition gives a somewhat random result: students’ average error is a function of the points they happened to pick, not how well they did the math (assuming everyone did the math correctly).
Though it might not sound much more interesting than watching paint dry, the relationships between phase changes, heat, and temperature are nicely illustrated by melting a beaker of snow on a hot plate.
This week’s snowfall created an opportunity I was eager to take. We have access to an ice machine, but closely packed snow works much better for this experiment, I think; the small snowflakes have larger surface-area to volume ratio, so they melt much more evenly, and demonstrate the latent heat of melting much more effectively.
Instructions
My instructions to the students are simple: collect some snow, and observe how it melts on the hot plate.
I also ask them to determine the mass and density of the snow before (and after) the melting, so I could show that throughout the phase changes and transformations the mass does not change (at least not a lot) and so they can practice calculating density1,2.
Procedure
I broke up my middle school students into groups of 2 or 3 and had them come up with a procedure and list of materials before they started. As usual I had to restrain a few of the over-eager ones who wanted to just rush out and collect the snow.
I guided their decision-making a little, so they would use glass beakers for the collection and melting. Because I wasn’t sure what the density of the packed snow would be, I suggested the larger, 600 ml beakers, which turned out to work very well. They ended up with somewhere between 350 and 400 grams of snow, giving densities around 0.65 g/ml.
When they put the beakers on the hot-plate, I specifically asked the students to observe and record, every minute or so, the changes in:
temperature,
volume
appearance
I had them continue to record until the water was boiling. This produced the question, “How do we know when it’s boiling?” My answer was that they’d know when they saw the temperature stop changing.
They also needed to stir the water well, especially when the ice was melting, so they could get a “good”, uniform temperature reading.
Results
We ended up with some very beautiful graphs.
Temperature Change
The temperature graph clearly shows three distinct segments:
In the first few minutes (about 8 min), the temperature remains relatively constant, near the freezing/melting point of water: 0 ºC.
Then the temperature starts to rise, at an constant rate, for about 20 minutes.
Finally, when the water reaches close to 100 ºC, its boiling point, the temperature stops changing.
Volume Change
The graph of volume versus time is a little rougher because the gradations on the 600 ml beaker were about 25 ml apart. However, it shows quite clearly that the volume of the container decreases for the first 10 minutes or so as the ice melts, then remains constant for the rest of the time.
Analysis
To highlight the significant changes I made copies of the temperature and volume graphs on transparencies so they could be overlain, and shown on the overhead projector.
Melting Ice: Latent Heat of Melting/Fusion
The fact that the temperature only starts to rise when the volume stops changing is no coincidence. The density of the snow is only about 65% of the density of water (0.65 g/ml versus 1 g/ml), so as the snow melts into water (a phase change) the volume in the beaker reduces.
When the snow is melted the volume stops changing and then the temperature starts to rise.
The temperature does not rise until the snow has melted because during the melting the heat from the hot plate is being used to melt the snow. The transformation from solid ice to liquid water is called a phase change, and this particular phase change requires heat. The heat required to melt one gram of ice is called the latent heat of melting, which is about 80 calories (334 J/g) for water.
Conversely, the heat that needs to be taken away to freeze one gram of water into ice (called the latent heat of fusion) is also 80 calories.
So if we had 400 grams of snow then, to melt all the ice, it would take:
400 g × 80 cal/g = 32,000 calories
Since the graph shows that it takes approximately 10 minutes (600 seconds) to melt all the snow the we can calculate that the rate at which heat was added to the beaker is:
32,000 cal ÷ 10 min = 3,200 cal/min
Constantly Rising Temperature
The second section of the temperature graph, when the temperature rises at an almost constant rate, occurs after all the now has melted and the beaker is now full of water. I asked my students to use their observations from the experiment to annotate the graphs. I also asked a few of my students who have worked on the equation of a line in algebra to draw their best-fit straight lines and then determine the equation.
All the equations were different because each small group started with different masses of snow, we used two different hot plates, and even students who used the same data would, naturally, draw slightly different best-fit lines. However, for an example, the equation determined from the data shown in the figure above is:
y = 4.375 x – 35
Since our graph is of Temperature (T) versus time (t) we should really write the equation as:
T = 4.375 t – 35
It is important to realize that the slope of the line (4.375) is the change in temperature with time, so it has units of ºC/min:
slope = 4.375 ºC/min
which means that the temperature of the water rises by 4.375 ºC every minute.
NOTE: It would be very nice to be able to have all the students compare all their data. Because of the different initial masses of water we’d only be able to compare the slopes of the lines (4.375 ºC/min in this case, but another student in the same group came up with 5 ºC/min).
Furthermore, we would also have to normalize with respect to the mass of the ice by dividing the slope by mass, which, for the case where the slope was 4.375 ºC/min and the mass was 400 g, would give:
4.375 ºC/min ÷ 400 g = 0.011 ºC/min/g
Specific Heat Capacity of Water
A better alternative for comparison would be to figure out how much heat it takes to raise the temperature of one gram of water by one degree Celsius. This can be done because we earlier calculated how much heat is being added to the beaker when we were looking at the melting of the ice.
In this case, using the heating rate of 3,200 cal/min, a mass of 400 g, and a rising temperature rate (slope from the curve) of 4.375 ºC/min we can:
The amount of heat it takes to raise the temperature of one gram of a substance by one degree Celsius is called its specific heat capacity. We calculated a specific heat capacity of water here of 1.8 cal/ºC/g. The actual specific heat capacity of water is 1 cal/ºC/g, so our measurements are a wee bit off, but at least in the same ballpark (order of magnitude). Using the students actual mass measurements (instead of using the approximate 400g) might help.
Evaporating Water
Finally, in the last segment of the graph, the temperature levels off again at about 100 ºC when the water starts to boil. Just like the first part where the ice was melting into water, here the water is boiling off to create water vapor, which is also a phase change and also requires energy.
The energy required to boil one gram of water is 540 calories, which is called the latent heat of vaporization. The water will probably remain at 100 ºC until all the water boils off and then it will begin to rise again.
Conclusion
This project worked out very well, and there was so much to tie into it, including: physics, algebra, and graphing.
Place a little hot water (400 ml at 94-100°C) into a plastic, gallon-sized, milk jug. Give it a moment to warm the air in the jug, then put the cap on and seal tightly (hopefully airtightly).
As the air in the jug cools the gas inside with shrink, reducing its pressure, and causing the atmospheric pressure to push in the sides of the jug.
Admittedly, this experiment is a little more dramatic if you use a metal tin, but it works well enough with the milk jug to surprise and impress.
In an interesting application of thermodynamics, BAE Systems has developed panels that can be placed on a tank to mask what it looks like to infra-red goggles, or help it fade into the background.
The panels measure the temperature around them and then warm up or cool so they’re the same temperature and therefore emitting the same wavelength of infrared light. So someone looking at the tank with infra-red goggles would have a harder time distinguishing the tank from the background.
The panels are thermoelectric, which means they use electricity to raise or lower their temperatures, probably using a Peltier device.
Peltier devices, also known as thermoelectric (TE) modules, are small solid-state devices that function as heat pumps. A “typical” unit is a few millimeters thick by a few millimeters to a few centimeters square. It is a sandwich formed by two ceramic plates with an array of small Bismuth Telluride cubes (“couples”) in between. When a DC current is applied heat is moved from one side of the device to the other – where it must be removed with a heatsink. The “cold” side is commonly used to cool an electronic device such as a microprocessor or a photodetector. If the current is reversed the device makes an excellent heater.
Talk about evoking conflicting emotions. The image is astoundingly beautiful – I particularly like the rich, intense colors – but the subject, global warming, always leaves me with sense of apprehension since it seems so unlikely that enough will be done to ameliorate it.
The source of the image, Global Warming Art has a number of excellent images, diagrams and figures. The National Oceanic and Atmospheric Administration also has lots of beautiful, weather-related diagrams. I particularly like the seasonal temperature change animation I made from their data.
Energy cannot be either created or destroyed, just changed from one form to another. That is one of the fundamental insights into the way the universe works. In physics it’s referred to as the Law of Conservation of Energy, and is the basic starting point for solving a lot of physical problems. One great example is calculating the average temperature of the Earth, based on the balance between the amount of energy it receives from the Sun, versus the amount of energy it radiates into space.
The Temperature of Radiation
Anything with a temperature that’s not at absolute zero is giving off energy. You right now are radiating heat. Since temperature is a way of measuring the amount of energy in an object (it’s part of its internal energy), when you give off heat energy it lowers your body temperature. The equation that links the amount of radiation to the temperature is called the Stefan-Boltzman Law:
where:
ER = energy radiated (W/m-2)
T = temperature (in Kelvin)
s = constant (5.67 x 10-8 W m-2 K-4)
Now if we know the surface area of the Earth (and assume the entire area is radiating energy), we can calculate how much energy is given off if we know the average global temperature (the radius of the Earth = 6371 km ). But the temperature is what we’re trying to find, so instead we’re going to have to figure out the amount of energy the Earth radiates. And for this, fortunately, we have the conservation of energy law.
Energy Balance for the Earth
Simply put, the amount of energy the Earth radiates has to be equal to the amount of energy gets from the Sun. If the Earth got more energy than it radiated the temperature would go up, if it got less the temperature would go down. Seen from space, the average temperature of the Earth from year to year stays about the same; global warming is actually a different issue.
So the energy radiated (ER) must be equal to the energy absorbed (EA) by the Earth.
Now we just have to figure out the amount of solar energy that’s absorbed.
Incoming Solar Radiation
The Sun delivers 1367 Watts of energy for every square meter it hits directly on the Earth (1367 W/m-2). Not all of it is absorbed though, but since the energy in solar radiation can’t just disappear, we can account for it simply:
Some if the light energy just bounces off back into space. On average, the Earth reflects about 30% of the light. The term for the fraction reflected is albedo.
What’s not reflected is absorbed.
So now, if we know how many square meters of sunlight hit the Earth, we can calculate the total energy absorbed by the Earth.
With this information, some algebra, a little geometry (area of a circle and surface area of a sphere) and the ability to convert units (km to m and celcius to kelvin), a student in high-school physics should be able to calculate the Earth’s average temperature. Students who grow up in non-metric societies might want to convert their final answer into Fahrenheit so they and their peers can get a better feel for the numbers.
What they should find is that their result is much lower than that actual average surface temperature of the globe of 15 deg. Celcius. That’s because of how the atmosphere traps heat near the surface because of the greenhouse effect. However, if you look at the average global temperature at the top of the atmosphere, it should be very close to your result.
They also should be able to point out a lot of the flaws in the model above, but these all (hopefully) come from the assumptions we make to simplify the problem to make it tractable. Simplifications are what scientists do. This energy balance model is very basic, but it’s the place to start. In fact, these basic principles are at the core of energy balance models of the Earth’s climate system (Budyko, 1969 is an early example). The evolution of today’s more complex models come from the systematic refinement of each of our simplifications.
Advanced Work
If students do all the algebra for this project first, and then plug in the numbers they should end up with an equation relating temperature to a number of things. This is essentially a model of the temperature of the Earth and what scientists would do with a model like this is change the parameters a bit to see what would happen in different scenarios.
Feedback
Global climate change might result in less snow in the polar latitudes, which would decrease the albedo of the earth by a few percent. How would that change the average global temperature?
Alternatively, there could be more snow due to increased evaporation from the oceans, which would mean an increase in albedo …
This would be a good chance to talk about systems and feedback since these two scenarios would result in different types of feedback, one positive and one negative (I’m not saying which is which).
Technology / Programming
Setting up an Excel spreadsheet with all the numbers in it would give practice with Excel, make it easier for the student to see the result of small changes, and even to graph changes. They could try varying albedo or the solar constant by 1% through 5% to see if changes are linear or not (though they should be able to tell this from the equation).
A small program could be written to simulate time. This is a steady-state model, but you could assume a certain percent change per year and see how that unfolds. This would probably be easier as an Excel spreadsheet, but the programming would be useful practice.
Of course this could also be the jumping off point for a lot of research into climate change, but that would be a much bigger project.
References
Yochanan Kushnir has a page/lecture that treats this type of zero-dimesional, energy balance model in a little more detail.