Activity: Global energy transfer


Explore the nature of EM radiation from a black body.
Show that Power/Area is proportional to T4.
State and apply Stefan - Boltzmann and Wein's laws.
Describe the interaction between EM radiation and different types of matter.
Define surface heat capacity, emissivity and albedo.
Understand the greenhouse effect.
Build an iterative simulation of the Earth - Sun system with Excel.

Energy from the sun

Energy from the fusion of hydrogen in the core of the Sun heats the gas on the outside causing it to emit light.

  • Why doesn't the core emit visible light?

  • Why doesn't fusion take place in the outer layers?

Black body radiation

When a body is heated it gives out energy in the form of EM radiation. The amount of energy emitted from a body per second depends on the temperature let's investigate the relationship.

You are going to use a tungsten light bulb as the black body. The resistivity of tungsten is proportional to it's temperature in Kelvin so can be used to measure the temperature of the filament. The following table will help you to convert the resistance to temperature but first you need a reference temp so use a multimeter to measure the resistance at room temperature (this is R300)

During the experiment you will measure the voltage and current so will be able to calculate the resistance at different temps then use the table to find the temperature.

The radiant power will be measured using a thermopile. This produces a voltage that is proportional to the radiation it absorbs. You should make sure this is only exposed to the light source for short periods of time or it will get hot. A silver shield is provided for this purpose.

  • Starting at the lowest voltage setting (2V) switch on the light and measure the voltage and current.
  • Remove the shield and measure the pd from the thermopile.
  • Enter the data into a spreadsheet and repeat for all settings on the power supply (2 V - 14 V).
  • In you spreadsheet calculate the Temperature for each setting.
  • Plot a graph of Thermopile PD (proportional to power per unit area) vs Temp.

You will notice this isn't linear, it is of the form

P over A equals k T to the power of n

To find the value of n you can plot a log - log graph

log left parenthesis P over A right parenthesis space equals space log left parenthesis k T to the power of n right parenthesis
log left parenthesis P over A right parenthesis space equals space n log left parenthesis T right parenthesis plus log left parenthesis k right parenthesis

You should find that

P = AσT4

This is the Stefan - Boltzmann law

The constant σ = 5.67 x 10-8 Wm-2K-4

Emissivity (e)

Not all bodies are perfect black bodies.

Sorry but I couldn't resist it

The emissivity is the ratio of radiation emitted to that radiated by a black body so a body with emissivity 0.5 will emit half the radiation of a black body.

  • Which surface has the highest emissivity, polished metal or brick?

P = eAσT4

  • Try measuring the radiant heat from the different coloured sides of the "Leslie's cube".

The black body spectrum

A spectrum shows the intensity (P/A) at each wavelength. Use the simulation below to investigate the change in the spectrum at different temperatures.

  • What does the area under the graph represent?
  • Set the temperature to the temperature of the sun and note the position of the peak.
  • What happens to the peak as the temperature is reduced?

Wien's displacement law

This gives the peak wavelength

lambda subscript p e a k end subscript equals fraction numerator 0.00289 over denominator T end fraction

You can see how the peak changes by tracking it in this simulation.

  • Given that the peak of the Sun's spectrum is at 500nm, use Wein's law to show that the temperature of the Sun is 5800K.

  • Given that the temperature of the Sun is 5800 K use Stefan - Boltzmanns law to show that the power per unit area from the sun is 6.42 x 107 Wm-2 .
  • Given that the radius of the Sun is 6.9 x 108 m show that the total power emitted is 3.9 x 1026 W

Inverse square law

The radiation from the sun spreads out in an ever increasing sphere.

The intensity, I at a distance r is power per unit area so if the total power emitted = P

I equals fraction numerator P over denominator 4 pi r squared end fraction

  • Give that the Earth is 1.5 x 1011 m from the Sun, show that the power per unit area at the Earth is 1400 Wm-2

EM radiation and the atmosphere

You have seen how light is absorbed when an atomic electron is excited from one energy level to another. This only happens for specific wavelengths so gases are quite transparent to visible light. The atmosphere consists of a mixture of gases most made of molecules rather than atoms. Carbon monoxide, nitrogen, oxygen, carbon dioxide, water, nitrogen dioxide and ozone are all molecular gases found in air. The way EM radiation interacts with these molecules can be investigated in the following simulation:

Starting with Carbon monoxide try shining different wavelengths of EM radiation at the molecules. Make a suitable table for your results. Note that a short time after absorbing the radiation the molecule returns to its non excited state, when this happens radiation is re emitted but in a random direction.

The following graphs show the % absorption of radiation by 3 different gases. Use the results from the simulation to explain them.

Radiation absorbed by the Earth

The electrons in solids exist in bands rather than levels, this means that solids can absorb many more wavelengths than gases. There are 3 possibilities

  1. Absorbed resulting in an increase in temp.
  2. Reflected
  3. pass through

No light passes through the earth so that leaves 2.

Absorbed radiation

The Earth absorbs energy only on the side facing the sun so the total power absorbed = 1400 x πr2

  • Why is the area used πr2 and not 1/2 x 4πr2 which would be the area of half of the Earth?

  • Show that the average power per unit area over the whole Earth is 350 Wm-2.

When energy is absorbed by a body the increase in temperature depends upon the specific heat capacity of the material and its mass. When heat is absorbed by the earth the temperature rise depends on the Surface heat capacity and the area.

capital delta T space equals space fraction numerator Q over denominator C subscript s A end fraction

The average surface heat capacity is 4 x 108 Jm-2K-1

If the intensity of radiation is 350 Wm-2 calculate the rise in temperature in 12 hours (assuming no heat lost).

Reflected radiation

The fraction of radiation reflected (or scattered) depends on the albedo of the surface (this is related to the colour)

A l b e d o space equals space fraction numerator T o t a l space s c a t t e r e d space p o w e r over denominator T o t a l space i n c i d e n t space p o w e r end fraction

  • Which surface has a higher albedo snow or asphalt?
  • The average albedo of the Earth is 0.3, show that the power reflected per m2 is 105 Wm-2.

Energy balance

The Earth absorbs energy from the Sun and which increases its temperature, this causes the earth to radiate EM radiation itself until the energy absorbed = energy radiated. At this point the equilibrium is reached and the temperature will stop rising.

  • If the albedo is 0,3 and the average intensity is 350 Wm-2 show that at equilibrium 245 Wm-2 must be radiated.

Simple model (no atmosphere)

  • Use the Stefan - Boltzmann law to show that the if the temperature of the earth needed to radiate 245 Wm-2 is 256 K.

  • Referring to the black body simulation show that the peak in the spectrum for a 256 K body is in the IR region.

Not so simple (with atmosphere)

  • You have seen that the radiation emitted by the earth is mainly in the IR region, will this be absorbed by the atmosphere?

The effect of the absorption by the atmosphere is to reduce the emissivity of the earth (not so much power is radiated out)

  • Show that with an emissivity of 0.6 the temperature of the Earth would be 291 K

The Greenhouse effect

This is the name given to the effect of the atmosphere absorbing radiation and re-emiting it back to Earth.

Fill in the gaps using some or all of the following words to describe the greenhouse effect. IR  atmosphere  rise  fall  visible   reflected  all   absorbed 

EM radiation from the sun is mainly in the   region. This passes through the  without being absorbed. The radiation lands on the Earth where some is  . The rest is  causing the temperature of the earth to  . The Earth emits radiation in the  region of the spectrum which is absorbed by the atmosphere and re- radiated in    directions, some of the radiation goes back to the Earth causing the temperature to  until the amount of energy radiated through the atmosphere equals the energy absorbed by the Earth.      

Hidden explanation (optional)


Total Score:

The following simulation shows how the radiation interacts with the atmosphere (not easy to see if you are colour blind)

The Greenhouse Effect


  • how the visible light photons pass through the atmosphere
  • some visible light is reflected
  • the radiation emitted from the Earth is IR

See the effect of

  • adding snow to the ground
  • clouds
  • changing the amount of greenhouse gases

Here is an energy flow diagram made in GeoGebra

  • Set the values as in the previous example to see if the numbers match.
  • Observe the effect of changing the variables.

Now you can build your own model in Excel follow the instructions on page 359.

EM radiation and matter multiple choice quiz

Global warming multiple choice quiz

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