Abstract

Objective

The objective of this physics science fair project is to determine how different colors absorb and re-emit radiant (light) energy, and to calculate the rate of energy flow.

Introduction

From where does the energy around you come? Most of the processes that are critical for our day-to-day lives are driven by energy provided by the Sun. Energy from the Sun warms the planet and keeps the global temperature within a range that allows life to flourish. The energy stored in the food you eat can be traced back to the Sun, through the conversion of solar energy to chemical energy in the process of photosynthesis. You might think that the gasoline that makes your car run or the oil that heats your house are sources of energy that are not derived from the Sun, but in fact, the energy in gasoline and oil is really a form of stored solar energy, since the chemicals in gasoline and oil were created by plants, thousands of years ago, using the Sun as a source of energy.

There are other important sources of energy in our world, including radioactive decay, which heats the interior of Earth, and gravitational energy, which powers the tides. But in terms of the energy needed for life, it is the Sun that "makes the world go round"!

The focus of this science fair project is how the color of an object affects its absorption of radiant energy. The absorption of light involves an interaction between photons, which are packets of light energy, and electrons, which are the negatively charged particles that whirl around atomic nuclei. When a photon is absorbed by an electron, the electron jumps to a higher energy level. For example, a photon of green light carries energy in the range of 2.2 electron volts (eV). See Table 1, below. An electron volt is a unit of energy. When an electron absorbs the energy of a 2.2-eV green light photon, the energy of the electron increases by 2.2 eV. Energy is not lost or gained. It is transmitted from the photon, which disappears, to the electron, which now has increased its energy by 2.2 eV. When sunlight hits a surface, such as your skin, the energy of many trillions of photons is changed from radiant energy to a form of electronic energy, in the sense of energy stored in electrons. What happens to the energy next? The electrons tend to return to the energy level they were at before they absorbed the photons, called the ground state. In the process, they re-emit the energy as infrared photons. Infrared photons are invisible to our eyes, but can be detected as heat, and measured with an infrared thermometer. Table 1 shows the wavelengths and energies of photons in the visible part of the electromagnetic spectrum.

Table 1. Approximate wavelengths and energies of visible light photons. Visible light is part of the electromagnetic spectrum.

The visible part of the electromagnetic spectrum can be separated into different colors: red, orange, green, blue, and violet. Objects that do not absorb any of these colors look white, and those that absorb all of these colors appear black. Colored objects absorb some colors and not others. Their color is determined by the color of light they do not absorb, since that is the light that hits your eye. In this science fair project, you will determine how color affects the absorption of radiant energy. You will use an infrared thermometer to measure the emission of energy as infrared photons.

Imagine sunlight hitting a checkerboard made up of white and black squares. The squares absorb different amounts of the solar energy. Solar energy that reaches Earth is composed of visible light (about 50 percent), infrared radiation (about 47 percent), and ultraviolet radiation (about 3 percent). The black squares will start to warm up; then after a while, they will reach a new stable temperature that is higher than the nearby white squares. This is because the black region is absorbing more energy in the form of photons than the white region is absorbing. The black squares absorb photons in the visible and the infrared part of the spectrum and then re-emit the energy as infrared photons.

You might be surprised to learn that everything around you is emitting radiation. In fact, all objects at any temperature above absolute zero will radiate to some extent. The wavelength of the radiation depends on a number of factors, but the most important is the object's temperature. The hotter the object, the shorter the wavelength of the light emitted (shorter wavelength mean higher energy). This type of radiation is called blackbody radiation. For objects near room temperature, such as the checkerboard, the wavelength of the radiation is in the infrared part of the spectrum. In the case of the checkerboard, the black squares are both absorbing and emitting radiation at a higher rate than the surrounding white squares. Very hot objects, such as the Sun, emit radiation in the visible part of the spectrum.

Based on the temperature increase caused by the light absorption, you will calculate the rate at which energy is being absorbed and re-emitted using the Stefan-Boltzmann equation. When you use the infrared thermometer to measure the temperature of the colored paper, you are measuring the heat transfer by the emission of electromagnetic waves that carry energy away from the hot object. The radiation is in the infrared region of the electromagnetic spectrum. The relationship governing radiation from hot objects is called the Stefan-Boltzmann law.

Equation 1:

The Stefan-Boltzmann equation relates the power, P, produced by a heated object to its temperature, T. Power is energy per unit time. In the example of the black region under sunlight, the power produced by the black region equals the amount of radiant energy emitted (absorbed) every second. The unit of energy is the joule (J). Note that the power varies with the fourth power of the temperature. If you double the temperature, the power is increased by 2×2×2×2 = 16 fold.

Consider one of the black squares on the checkerboard. It will be hotter than its light-colored surroundings. You can use the Stefan-Boltzmann equation to calculate the rate at which it is radiating energy. If a hot object (temperature = T) is radiating energy to its cooler surroundings at temperature T_s, the net radiation loss rate takes the form shown in Equation 2:

Equation 2:

P = σ A(T4 - Ts4)

P is power (energy per unit time), in watts (W). A is the area, in square meters (m2). T is the temperature of the hot object in kelvin. Add 273 to the temperature in °C to get kelvin. Ts is the temperature of the surroundings, in kelvin. σ Stefan-Boltzmann constant (sigma) = 5.67 × 10-8 W/m2K4.

Let's look at an example. Say the black square on the checkerboard has an area of 0.01 m² (about 4 inches by 4 inches). And let's say that it heats up to 35°C (308 K) when placed in the sunlight. Finally, assume the surrounding temperature is 20°C (293 K). Plugging these values into Equation 2, you get:

Equation 3:

P	= 5.67 × 10-8 W/m2 K4 × 0.01 m2 × (3084 - 2934) = 0.92 W

(See the Hyperphysics website in the Bibliography for an online Stefan-Boltzmann equation calculator).

The black square is absorbing and re-emitting 0.92 J per second (0.92 W). This is roughly 2 percent of the output of a 60-W lightbulb.

Once you know the energy that is being emitted per unit time, you can calculate the approximate number of photons that are emitted. For the example above, the power output is 0.92 W, which is the same as 0.92 J/sec. To convert joules to electron volts, use the equation below:

Equation 4:

1 electron volt =

1.6 × 10-19 J

For the purpose of getting a rough approximation of the photon count, assume that the average infrared photon energy is 0.000124 eV. Now you need to convert 0.92 J into electron volts, and then divide by 0.000124 eV to get the number of photons that are emitted per second:

Equation 5:

(0.92 J) × (1 eV/1.6 X 10-19 J) × (1 photon/0.000124 eV) = 4.6 × 1022 photons/sec

J is joules 1 eV = 1.6 × 10-19 J Energy of each infrared photon that is emitted = 0.000124 eV

Equation 5 is set up to convert the energy in joules (remember that 1 J/sec = 1 W) to the number of photons emitted per second. Note that the units are arranged so that electron volts and joules cancel, leaving photons as the remaining unit.

This is a very rough approximation, but gives you an insight into the large number of photons involved.

In the experimental procedure, you will compare the number of infrared photons that are emitted by different colors. Remember that once the temperatures are stable (which is true after a minute or so in the sunlight), the rate at which energy is absorbed equals the rate at which it is emitted. It's amazing how much you can discover from just the temperature difference of objects sitting in the sunlight!

Terms, Concepts, and Questions to Start Background Research

• Energy
• Radiant energy
• Photon
• Light energy
• Electron
• Energy level
• Electron volt (eV)
• Electronic energy
• Ground state
• Infrared photons
• Infrared thermometer
• Wavelength
• Electromagnetic spectrum
• Infrared proton
• Absolute zero
• Blackbody radiation
• Stefan-Boltzmann equation
• Stefan-Boltzmann law
• Watt (W)
• Power
• Joule (J)
• Vibrational ground state

Questions

• Based on your research, how is the wavelength of a photon related to its energy?
• What happens to a photon of visible light when it is absorbed by an electron?
• Scientists sometimes speak of energy being "degraded" when it is converted into infrared energy. Why do they use this term?
• Based on your research, what kinds of energy levels are most associated with the emission of infrared photons by a molecule?
• What is the approximate power output of the Sun, based on the Stefan-Boltzmann equation? Look up the temperature and area online.
• What fraction of the Sun's output hits Earth? Hint: The area of a sphere is A = 4pr² , where r = 93,000,000 miles for the orbit of Earth.

Bibliography

• Nave, R. (n.d.). Stefan-Boltzmann Law. Retrieved January 20, 2009, from http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
• Science Daily. (2009.) Radiant Energy. Retrieved January 12, 2009, from http://www.sciencedaily.com/articles/r/radiant_energy.htm
• Clark, Jim. (2007). UV-visible Absorption Spectra. Retrieved January 27, 2009, from http://www.chemguide.co.uk/analysis/uvvisible/theory.html#top

Materials and Equipment

• Scissors
• Colored construction paper
• Infrared thermometer; available online from stores such as www.amazon.com
• Lightly colored surface
• Optional: If you don't have a lightly colored surface, use Styrofoam^TM plates (5)
• Lab notebook
• Graph paper

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Experimental Procedure

1. Cut out a 4-inch construction paper square of each of the following colors: white, yellow, blue, red, and black.
2. Place the squares in a location where they are in the sunlight, not touching each other.
   1. Place the squares on a lightly colored surface, such as white or tan carpet.
   2. Don't put the squares on a hot surface. If you can't find a lightly colored surface, use the Styrofoam plates to isolate the squares from the surface.
   3. There should not be any shadows over the paper.
   4. The squares should be protected from any breezes.
3. Make a note of the time and date in your lab notebook.
4. Wait for several minutes so that the temperatures of the squares become stable.
5. Take the temperature of each square with the infrared thermometer, three times over a time period of about 1 minute. Record the data in a data table in your lab notebook.
6. Average the results for each colored square.
7. Now that you have the data, calculate the energy flow and the energy carried by the visible and infrared photons.
8. Answer the following questions:
   1. What is the power output from the squares? Calculate the power that each square is producing, using Equation 2 from the Introduction.
      • Power equals energy (joules or electron volts) per unit time (seconds).
      • Use the temperature of the white paper as the "surrounding" temperature (_s) in Equation 2.
      • Use the temperature of the colored or black square as "T" in Equation 2.
   2. How many photons are being emitted by the heated squares? Calculate the number of infrared photons that are being emitted by the squares, assuming each photon has an energy of 0.000124 eV.
   3. How does the power output and the number of photons that are emitted depend on the color?
9. Graph the temperature of each square, with color on the x-axis.
10. Graph the power output of each square. Since power depends on the fourth power of the temperature, a small difference in temperature can cause a big difference in power output.
11. Graph the number of photons emitted per second (use an energy of 0.000124 eV for each infrared photon).
12. Graph the number of photons emitted relative to the black square. That is, graph the black square as 100 percent, and the other squares relative to this standard.

Variations

• Try other colors. Avoid shiny surfaces, such as aluminum foil, since the infrared thermometer may not get a good reading from the shiny surface.
• Try heating water colored with different dyes, using solar energy. Calculate the power output based on the temperature and surface area.
• Estimate the number of photons absorbed per second for each color. The situation is more complex when looking at absorption than it is for emission, since there are different wavelengths involved, from violet to infrared. You may need to make some simplifying assumptions (for example, assume that all of the visible photons have energy of 2.2 eV).
• Based on your results, estimate the relative energy captured by a forest and an equivalent area of desert or of asphalt. Make a figure based on your estimates.

• For more science project ideas in this area of science, see Physics Project Ideas.

Credits

David B. Whyte, PhD, Science Buddies

• Styrofoam^TM is a registered trademark of The Dow Chemical Company.

Last edit date: 2009-04-06 09:35:00