SciPub

Here’s a nice article from arxiv examining some of the reasons students find numerical problems easier than symbolic problems.

A handful of students from an intro calculus-based physics class at University of Illinois solved problems while thinking aloud. Students were first presented with a problem with only symbolic quantities and were asked to think aloud as they attempted to solve the problem and identify the correct symbolic solution amongst five choices. If students were unable to solve the correct symbolic problem they were presented with an identical problem in which all known quantities as well as answer choices were replaced by numerical values. Students then thought aloud as they solved the numerical version of the problem.

Symbolic Difficulties
- Students confuse two different quantities of the same type (ex. velocity for car 1 and velocity for car 2 or final velocity and velocity at some other moment). Often times students denote multiple different velocities simply by “v” and encounter difficulties in symbolic problems where both velocities appear in the same equation.
- Students confuse variables and specific unknown quantities (ex. final velocity vs. the general velocity variable). If a problem states that some object has a velocity “v” students are likely to consider this “v” to be equivalent to the “v” appearing in any general equation such as d = v*t + 1/2*a*t^2.
- Students fail to differentiate between known and unknown quantities (both represented by letters in symbolic problems). Students who don’t identify known and unknown quantities before beginning their algebra will often arrive at an equation which gives one unknown quantity in terms of another unknown quantity.

Suggestions
- Encourage students to use subscripts which writing equations. Subscripts can help differentiate different quantities of the same type (velocity of car 1 vs. velocity of car 2). Subscripts can also cue students to consider the meanings of the variables they are writing (students are forced to think about what velocity “v” stands for when considering whether or not to attach a subscript to “v”).
- Talk to students about identifying known and unknown quantities at the start of the problem. This is an unnecessary step for numerical problems so intro students are not accustomed to doing so.

Frederick Reif’s Milikan Lecture is another article on the core reading list.

Understanding and teaching important scientific thought processes, Am. J. Phys. 63, 17 (1995).

Reif proposes that the abilities to interpret scientific concepts and describe and organize scientific knowledge are all prerequisites to effective problem solving.

The ability to interpret scientific concepts and principles
Reif’s concept of interpreting a scientific concept in a variety of different scenarios is similar to McDermott’s idea of an operational definition. Students often misinterpret a concept (such as misidentifying direction of acceleration for a pendulum at different points in its trajectory) because they “retrieve remembered or plausible knowledge fragments which are often incorrect and which are rarely checked against a definition of the concept.” Accurate interpretation requires a procedure for interpreting a concept’s definition. In the case of the definition of acceleration as “the rate of change of velocity” this procedure involves explicit instructions for of how to determine each part of a definition (ie. how to define delta_v and delta_t in terms of velocity at two neighboring moments in time). These procedures should be taught explicitly rather than hoping that the students will infer them with practice. The following an interpretation procedure can be very tedious. Efficient interpretation requires an accumulation of special cases so that a given scenario can be matched to a special case and hence processed immediately without going through the step by step procedural interpretation.

The ability to describe knowledge effectively
Diagrams, words and mathematica symbols are all different types of descriptions. Reif defines a procedure for describing system with a “system diagram”. First, identify the motion (velocity and acceleration) then identify all interacting objects. Look at all long range forces then look at all contact forces. Lastly, break forces into vector components. Only at this point is the scenario fully described and student ready to apply a principle such as Newton’s 2nd law. Identifying interacting objects before discussing specific forces helps avoid the introduction of non-existent forces like “centrifugal force” that are not associated with a specific pair of objects.

The ability to organize knowledge effectively
Reif talks about organizing physics arguments in hierrarchical fashion rather than in a linear fashion. This is again similar to the overarching principles that modeling instruction is built around. This organization factors into Reif’s description of problem solving in terms of iteratively breaking the problem into subproblems that can be solved to build a solution to the full problem. In this case each level of subproblem can be thought of as corresponding to a different level of the knowledge hierarchy. This kind of hierarchy of organization and problem solving procedure could also be useful on a smaller scale. In Hands-on-Science we could think about problems involving the small particle model as involving a hierarchy starting with macroscopic characteristics that can be measured, moving to microscopic characteristics that can be inferred and finally to a conceptual model of what causes pressure at the microscopic level. At the moment we approach these problems in a very linear manner but thinking in terms of a hierarchy might help students recognize the importance and organization of each step in the solution.

Problem solving
The first step in problem solving is to analyze the problem. Problem analysis involves identifying known and desired quantities and likely drawing a diagram. An important part of problem analysis is making the connection between real world language and physics terms. It is non-trivial for a student to determine that “the reading on the scale” corresponds to “the normal force exerted by the scale on the person”. The need to analyze the problem is often not obvious to students who want to jump to constructing a solution. Having students analyze a problem and state a solution strategy but not actual solve the problem can help convince them of the importance and usefulness of analyzing a problem before attempting a solution.

Students are also often reluctant to spot check their answers once they solve a problem. To help encourage simple checks (units, magnitude, direction etc.) instructors should distinguish between answers that are simply wrong and answers that are nonsensical, deducting more points for the latter.

Other articles to read:
Reif and Allen, Cognition for interpreting scientific concepts: A study of acceleration, Cogn. Instruct. 9, 1-44 (1992).

Eylon and Reif, Effects of knowledge organization on task performance, Cogn. Instruct. 1, 5-44 (1984).

Schoenfeld, Beyond the purely cognitive: Belief systems, social cognitions, and metacognitions as driving forces in intellectual performance, Cogn. Sci. 7, 329-363 (1983).

I had a dream this morning right before I awoke that was rather strange. In the dream Stacy and I had taken a road trip somewhere and we stopped at a swanky bar to kill time before dinner. Stacy ordered a Coke at the bar and when I pulled out my wallet to pay the bartender informed me that the cost was $15! I told the bartender that he must be mistaken. This is a plain Coke with no alcohol. He assured me that he was not mistaken and that because they were a local establishment in the middle of the city that their prices were a little bit higher. I proceeded to become very exasperated and began insisting to Stacy that we leave immediately. I did not want to support an establishment that charges $15 for a Coke. Stacy said that she liked the atmosphere of the bar and didn’t want to leave because she didn’t know where else to go. I continued to argue with Stacy until Zach jumped on the bed waking me up.

When I awoke I looked at my alarm clock to find that I had overslept 45 minutes! I distinctly remember setting my alarm last night so I must have turned it off and fallen back to sleep this morning. I can’t help but wonder if this dream was my subconscious telling me “Get out of here! It’s time to go!” I’m not usually one to try to interpret my dreams but I vaguely remember feeling like I was in a dream when it was happening and thinking, “This is a strange dream. Why am I making a Coke cost $15?”

Of course I did watch Inception the other day so that could be influencing my post-dream recollection/interpretation.

Finally getting back to the PER core reading list. Last week I read a series of two papers by Patricia Heller et al. on group work in intro physics classes. These papers give some good practical information for how to form effective groups as well as a nice example of the kinds of research questions one can ask about group work.

Teaching problem solving through cooperative grouping parts 1 and 2.
Am. J. Phys. 60, 627 (1992) and Am. J. Phys. 60, 637 (1992).

Forming Effective Groups
400+ students were observed and their work evaluated in various group settings over the course of 3 years. Initially groups solved textbook problems and groups engaged in usual the “equation hunting” and reported finding group work and problem solving strategies tedious and unnecessary. Context rich problems involving extraneous information, less explicit questions and implicit assumptions and estimations were developed to encourage more expert-like problem solving. Context rich problems improved student response to group work. Groups of 3 were found to be optimal. With 4 students someone is usually left out and with 2 there are not enough different ideas/no way to resolve disagreements. Heterogeneous groups (different skill levels in each group) outperformed homogeneous groups, especially for lower performing students. Same sex groups or groups of 2 females and 1 male had better group dynamics and better communication than groups of 1 female and 2 males.

Once groups were formed and problems assigned group dynamics was further improved by introducing group roles of Manager, Skeptic and Checker/Recorder. Students were provided with example phrases for each role. Students also discussed what about the group worked well and what did not at the end of each class. One group problem was added to each test along with 2-4 individual problems (group problem 1 day, individual problems the next day). This helped encourage all students to be involved in group work in the weekly recitation sessions. Students maintained the same groups in both recitation and lab, however groups were reassigned after each test.

Related reading: Cooperation and Competition: Theory and Research, Circles of Learning: Cooperation in the Classroom.

Evaluating Group Work
Students were given an explicit problem solving strategy to use on context-rich problems. The strategy involved a series of steps translating the problem between different representations similar to modeling instruction. Problem solving ability was evaluated based on six criteria: evidence of conceptual understanding in initial sketches, usefulness/completeness of problem description, equations match description, reasonable plan for solution determined before algebra was undertaken, math progresses logically from general to specific, appropriate math (solution and assumptions). This rubric was used to score both individual and group solutions. (A recent paper argues that explicit problem solving strategies do not automatically increase students’ conceptual engagement.)

Individual vs. Group Problem Solving
“Equivalent” problems were given on group and individual sections of tests so that group vs. individual problem solving could be examined. Over several tests researchers found that scores on the group problems were statistically higher than scores on the individual problems, even for the best students. Thus, group work does not simply represent the work of the best student in the group. There is a nice discussion of possible sources of error and how these challenges were addressed in the experimental design.

To compare individual vs. group problem solving in more detail researchers compared scores on each of the 6 scoring criteria for individual and group problems. They found that group scores for each criteria are almost ways higher than individual “best in group” scores and that the group scores were most improved for qualitative/conceptual aspects of problem solving.

How would you estimate the likelihood of a satellite killing someone when it falls back to earth?  Treat this as a Fermi problem and try to come up with an equation to estimate this probability.  What more basic probabilities or ratios would be important to know?  What simplifying assumptions could you make?

Once you come up with your own equation compare yours with the FAA’s equation.

This is a great real life example of a Fermi problem to share with your students.

In the last post I estimated that the temperature of the filament in an incandescent light bulb was about 4500 Kelvin. This was a bit of an over estimate with the true value being closer to 3300 K. That’s still more than 5000 degrees Fahrenheit! That’s still more than half as hot as the surface of the sun! How can the filament possibly get that hot?

The more massive something is the more energy it takes to increase its temperature (a small pot of water reaches boiling faster than a larger pot of water). If something is very tiny, like our filament, you can increase its temperature drastically with only a small amount of energy. According to Wikipedia the current through the filament stabilizes in about 0.1 seconds. Let’s see if this agrees with our temperature of 3300 K.

As the filament heats its resistance increases from about 17 Ohms to about 248 Ohms. Taking the average value of 133 Ohms and plugging this in power = V^2/R I get an average power, during the heating, of 108 Watts. The energy the filament receives during first 0.1 seconds is thus approximately 108 Watts*0.1 seconds = 10.8 Joules. I can use the heat capacity of tungsten to figure out how much 10.8 Joules of energy would heat the filament. But first I need to figure out the mass of the filament.

Wikipedia tells me that the filament of a 60 Watt bulb is approximately 580 mm long (when uncoiled) with a diameter of approximately 0.046 mm. Using volume = Pi*R^2*L I get a volume of 0.0009639 cubic centimeters. Solid tungsten has a density of 19.25 grams/cm^3 so our filament should have a mass of approximately 0.01856 grams. Let me convert this to moles since that’s how heat capacities are often stated. The molar mass of tungsten is 183.85 g/mol so 0.01856 grams corresponds to 0.0001009 moles of tungsten.

Normally I just look up the heat capacity of a substance on Wikipedia but here the temperature of our filament is changing significantly and heat capacity depends on temperature. Luckily I found an article from NIST that gives the heat capacity of tungsten at several different temperatures. The heat capacity varies between 30 J/mol*K and 40 J/mol*K for most of our temperature range. I’ll use 34 J/mol*K for our estimate. This means that it requires approximately 34 Joules of energy to increase the temperature of 1 mole of tungsten by 1 degree Kelvin.

Putting this all together I can find the expected temperature change

ΔT = Energy/(heat capacity*moles of tungsten).

Plugging in 10.8 J, 34 J/mol*K and 0.0001009 mol I get ΔT=3453 K. Room temperature is about 300 K so our expected temperature increase is about 3000 K. Not bad. I ignored the fact that some of the 10.8 J will be dissipated to the surroundings (ie. will go into heating up the glass bulb) so it isn’t surprising that I over estimated the temperature change.

I’m still loving our recent revelation about light bulbs, resistance and brightness in series vs. parallel circuits. So much so that I decided to play around with this idea a little more…

Everything you plug into your outlets at home runs on 120 Volts (assuming you live in the U.S.). This suggests that all of the outlets in your house a connected in parallel. In this case, based on our previous logic, a 100 Watt light bulb should have a smaller resistance than a 60 Watt light bulb. Let’s check.

If I plug both light bulbs in side by side the 100 Watt bulb is indeed brighter.

I can use my handy Kill-A-Watt meter to measure the (rms) voltage and current of each bulb.

In order to measure resistance I need to break the bulbs open. Then I can use a multimeter to measure the resistance of the filament.

As expected for a parallel circuit the the 100 Watt bulb does indeed have the lower resistance.

Temperature dependent resistance
I can use the voltage and current values above to calculate the power output of each bulb. For a resistive AC circuit Power = V_rmsI_rms. This gives 61.2 Watts and 101.7 Watts for the 60 Watt and 100 Watt bulbs, respectively. The power can also be calculated using Power = I_rms²R. However, this equation gives much smaller values for power. The reason is that the resistance of each bulb increases as the filament gets hot.

I measured the resistance of each bulb when the filament was at room temperature. When the light bulb is turned on the filament becomes VERY hot VERY quickly thus increasing the resistance considerably. According to Wikipedia the resistance of a conductor has the following temperature dependence

R(T) = R(T₀)(1 + αΔT).

Let R(T₀) be the resistance at room temperature and R(T) the resistance when the filament is hot. α is a measure of how much the resistance changes as the material is heated. According to Wikipedia α_tungsten=0.0045 K^-1.

At room temperature T₀ = 300 Kelvin. I can use Planck’s Law to estimate the temperature of the filament when the light bulb is turned on. Planck’s Law describes how the spectrum of light emitted by a black body depends on the temperature of the blackbody. (image from Wikipedia)

The 5000 K and 4000 K lines peak near the visible spectrum so I’ll use 4500 Kelvin as the temperature of the filament when the light bulb is turned on.

Putting all this together I get the following resistances for each light bulb when the light is turned on

R_60W = 17.1(1+0.0045*4200)=340 Ohms
R_100W=10.1(1+0.0045*4200)=200 Ohms

The resistances increase by a factor of 20 when the filament heats up!

With these resistances I get power values of 85 Watts and 141 Watts for 60 Watt and 100 Watt light bulbs, respectively. Not perfect but pretty good.

The Science Channel has a new series looking at several classic science fiction authors and how their writings incorporated cutting edge science of their day as well as anticipated very accurately where that science would lead. I caught the first episode which looked at Mary Shelly’s Frankenstein and I thought it was pretty good. The episode has several short segments looking at current research which are weaved together with the story rather well. The episode also has some history talking about the science of the day and the process of writing Frankenstein. I would check the show out if you have a chance (Wednesdays at 9pm Central on The Science Channel).

Here’s a Wired article with a few more details and a short clip.

Last week we were talking about resistors in class. We drew an analogy between resistance in electrical circuits and friction in macroscopic motion. We led students towards the idea that resistance causes electrical energy to be transformed into thermal energy and that if an object, like a light bulb filament, gets hot enough it will start to glow.

One of the other instructors sent an email to all the instructors saying that he was concerned that we were misleading students. He argued that the power dissipated by an element in an electric circuit is given by P = ΔV²/R so a light bulb with less resistance will radiate energy more quickly and thus be brighter.

I mulled this over a bit and decided that he was correct. However, I argued that we were not misleading students because within a given circuit the object with the most resistance will be the object that dissipates the most power. Given to different circuits with two different light bulbs the lower resistance bulb will dissipate more power than the higher resistance bulb but within each circuit the light bulb will dissipate more power than the (very low resistance) wires.

Mark chewed on my comment for a bit and replied that agrees that it is important to distinguish between comparing two different circuits and comparing two different elements within the same circuit but that he thought I was only considering series circuits. For two light bulbs in series both bulbs receive the same current and we can use the power equation P = I²R to find that the bulb with the higher resistance dissipates more power and is thus brighter. For two bulbs in parallel both bulbs experience the same voltage drop so we should use the power equation P = ΔV²/R and find that the bulb with the higher resistance dissipates less power and is now brighter.

What an awesomely surprising result! Given two light bulbs with different resistances which bulb is brighter depends on whether you connect the bulbs in series or parallel. I was sure Mark was correct but I still had to check this for myself. I didn’t have access to different light bulbs when received Mark’s email but I did have access to the Phet Simulations. Here’s a screen capture of a Phet Simulation I ran with different light bulbs in series and parallel. The numbers next to each light bulb indicate the resistance.

If Mark and I weren’t teaching about light bulbs we would have never had this conversation and never stumbled onto this wonderful realization about brightness in series vs. parallel circuits.

The New York Times has a short article today responding to a reader’s question about what it would be like if you died in the vacuum of space. Two lines in the article caught my attention.

First, the article discusses how your lungs would explode if you tried to hold your breath (due to the air expanding under reduced pressure). The article then goes on to state that,

“the skin is sufficiently strong to prevent the body from exploding.”

This caught my attention because I remember The Simpsons Treehouse of Horror X where Homer and Bart eject from a space shuttle causing their heads to swell up and eventually explode. It seems this scenario is a bit of an exaggeration.

Here’s a short clip.

The second line that caught my attention comes from a description of a training session at NASA’s Johnson Space Center where an astronaut’s space suit developed a leak during a vacuum conditions. Repressurization occurred within 15 seconds and the astronaut was fine however,

“his last conscious memory was of the water on his tongue beginning to boil.”

That line sends shivers down my spine. Water on his tongue beginning to boil… wow.

You might wonder whether the boiling water burned his tongue. Even though it began to boil the water was still at room temperature. There was no energy input into the water to raise its temperature. So why did it boil? Because there was no atmospheric pressure in the vacuum. In order for water to be a liquid the water molecules have to be close enough together to form loose bonds between the molecules. Atmospheric pressure helps push the molecules together so that they can bond and form a liquid. If you take away the atmospheric pressure there is much less holding the water molecules together and it takes much less energy (lower temperature) for the molecules to break free and form a gas. You can see this if you follow the line separating liquid and gas in a Pressure vs. Temperature phase diagram.