Blackbody

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Contents    

Atif M. Khokhar Assistant Professor Department of Mechanical Engineering HITEC University, Taxila Cantt, Pakistan

  

Importance of Black Body History Black-Body (Definition) Black-Body Radation Laws 1- The Planck Law 2- The Wien Displacement Law 3- The Stefan-Boltzmann Law 4- The Rayleigh-Jeans Law Application for Black Body Conclusion Summary 2

Importance • •

•

History

The black body is importance in thermal radiation theory and practice.. The ideal black body notion is importance in studying thermal radiation and electromagnetic radiation transfer in all wavelength bands. The black body is used as a standard with which the absorption of real bodies is compared.

 The term "black body" was introduced by

G. Kirchhoff in 1860 1860.. The light emitted by a black body is called black-body radiation..

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Definition of a black body

Black Body

A black body is an ideal body which allows the whole of the incident radiation to pass into itself ( without reflecting the energy ) and absorbs within itself this whole incident radiation (without passing on the energy). This propety is valid for radiation corresponding to all wavelengths and to all angels of incidence. Therefore, the black body is an ideal absorber of incident radaition.



 Since the radiation in such an environment

has a spectrum that depends only on temperature,, the temperature of the object temperature is directly related to the wavelengths of the light that it emits. 5

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A black body at temperature T emits exactly the same wavelengths and intensities which would be present in an environment at equilibrium at temperature T, and which would be absorbed by the body.

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Wavelengths and Colors Emitted

Black Body  At room temperature, black bodies emit

infrared light, light, but as the temperature increases past a few hundred degrees Celsius,, black bodies start to emit at visible Celsius wavelengths,, from red, through orange, yellow, and white before ending up at blue, beyond which the emission includes increasing amounts of ultraviolet ultraviolet..

4000 Å

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7000 Å

Different colors of light emitted correspond to different wavelengths wavelengths..

Black Body Emission Spectrum

The Planck Function

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths.

• Blackbody radiation follows the Planck function

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Basic Laws of Radiation

Basic Laws of Radiation

1) All objects emit radiant energy.

1) All objects emit radiant energy. 2) Hotter objects emit more energy than colder objects.

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Basic Laws of Radiation

Basic Laws of Radiation

1) All objects emit radiant energy.

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object.

2) Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object raised to the fourth power.  This is the Stefan Boltzmann Law

F =  T4 F = flux of energy (W/m2) T = temperature (K)  = 5.67 x 10-8 W/m2K4 (a constant)

Basic Laws of Radiation

Basic Laws of Radiation

1) All objects emit radiant energy.

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object.

2) Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object.

3) The hotter the object, the shorter the wavelength () of emitted energy.

3) The hotter the object, the shorter the wavelength () of emitted energy. This is Wien’s Law

max  3000 m T(K)

 Stefan Boltzmann Law.

We can use these equations to calculate properties of energy radiating from the Sun and the Earth.

F =  T4 F = flux of energy (W/m2) T = temperature (K)  = 5.67 x 10-8 W/m2K4 (a constant)

 Wien’s Law

max  3000 m T(K)

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6,000 K

300 K

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Ultraviolet Catastrophe

Black--Body Radiation Laws (1 Black ( 1) 1- The Rayleigh-Jeans Law. *Based on Kinetic Theory of Gases. (A Classical Theory) *It agrees with experimental measurements for long wavelengths.. wavelengths * It predicts an energy output that diverges towards infinity as wavelengths grow smaller. smaller. • The failure has become known as the ultraviolet (short wavelength)catastrophe wavelength )catastrophe..

I ( , T ) 

2ckT 4

I ( , T ) 





2ckT 4

This formula also had a problem. The problem was the  term in the denominator. For large wavelengths it fitted the experimental data but it had major problems at shorter wavelengths.

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Comparison between Classical and Quantum viewpoint

Black--Body Radiation Laws (2 Black (2 ) 2- Planck Law -

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We have two forms. As a function of wavelength. I ( ,T ) 

2 hc 2 5

1 hc e  kT  1

And as a function of frequency I ( , T ) 

2h c2

3

1 h e kT  1

The Planck Law gives a distribution that peaks at a certain wavelength, the peak shifts to shorter wavelengths for higher temperatures, and the area under the curve grows rapidly with increasing temperature.

There is a good fit at long wavelengths, but at short wavlengths there is a major disagreement. Rayleigh-Jeans ∞, but Black-body 0. 21

Black--Body Radiation Laws (4) Black

Black--Body Radiation Laws (3 Black (3 )  max 

3- Wein

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b T

Displacement Law

- It tells us as we heat an object up, its color changes from red to orange to white hot. - You can use this to calculate the temperature of stars. The surface temperature of the Sun is 5778 K, this temperature corresponds to a peak emission = 502 nm = about 5000 Å. - b is a constant of proportionality, called Wien's displacement constant and equals 2.897 768 5( 5(51 51)) × 10–3 m K = 2.897768 5( 5(51 51)) × 106 nm K. 23

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Comparison of Rayleigh-Jeans law with Wien's law and Planck's law, for a body of 8 mK temperature.

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Application for Black Body

Application for Black Body

- The

area of Earth's disk as viewed from space is, Area = πr2. - The total energy incident on Earth is, Incident energy = (πr2)So. - The energy absorbed by the Earth/atmosphere system, as viewed from space is Absorbed energy = (πr2)So(1 - A). As we know that bodies must be in radiative equilibrium. equilibrium. The solar energy striking Earth's disk as viewed from space is rere-emitted as thermal radiation by the surface of the entire globe, globe, as described by the StefanStefanBoltzmann Law,

Emitted energy = (4 (4πr2)σT4. - Set the absorbed energy equal to the emitted energy: (πr2)So(1 - A) = (4 (4πr2)σTE4, Solving for T yields: TE = [So(1 - A)/( A)/(4 4σ)](1/4) = [1370 [1370•( •(1 1-0.3)/( )/(4 4•5.67 67xx10 10--8)](1/4) = 255 K. 25

The Sky in Different Wavelength Bands

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Which of the following types of electromagnetic radiation has the highest photon energies?

1. 2. 3. 4. 5.

Radio Waves Visible light g-rays Infrared X-rays

Radio Visual Gamma-Rays Infrared X-Rays

0% 1

0%

0%

0%

2

3

4

We can use these equations to calculate properties of energy radiating from the Sun and the Earth.

T (K) 6,000 K

5

300 K

Sun

6000

Earth

300

max (m)

region in spectrum

F (W/m2)

0% 5

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Electromagnetic Spectrum T (K)

max (m)

region in spectrum

F (W/m2)

Sun

6000

300

10

100

10

1

0.1

Low Energy

T (K)

max (m)

6000

0.5

region in spectrum

F (W/m2)

Visible (yellow?)

Earth

300

10

infrared • Blue light from the Sun is removed from the beam by Rayleigh scattering, so the Sun appears yellow when viewed from Earth’s surface even though its radiation peaks in the green

Sun

T (K)

max (m)

region in spectrum

6000

0.5

Visible (green)

Earth

6

300

10

infrared

F (W/m2)

0.01 High Energy

 (m)

Sun

x-rays

0.5 1000

Earth

visible light ultraviolet

infrared

microwaves



Stefan Boltzman Law. F =  T4 F = flux of energy (W/m2) T = temperature (K)  = 5.67 x 10-8 W/m2K4 (a constant)

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Solar Radiation and Earth’s Energy Balance

Sun

T (K)

max (m)

6000

0.5

region in spectrum

F (W/m2)

Visible 7 x 107 (green)

Earth

300

10

infrared

460

Planetary Energy Balance

Some Basic Information: Area of a circle =  r2

 We can use the concepts

learned so far to calculate the radiation balance of the Earth

Area of a sphere = 4  r2

Energy Balance:

Energy Balance:

The amount of energy delivered to the Earth is equal to the energy lost from the Earth.

Incoming energy = outgoing energy

Otherwise, the Earth’s temperature would continually rise (or fall).

Ein = Eout

Eout

Ein

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How much solar energy reaches the Earth?

(The rest of this derivation will be on these slides . in case anyone wants to look at them.)

How much solar energy reaches the Earth?

How much solar energy reaches the Earth?

As energy moves away from the sun, it is spread over a greater and greater area.

As energy moves away from the sun, it is spread over a greater and greater area.  This is the Inverse Square Law

So = L / area of sphere So = L / (4  rs-e2) = 3.9 x 1026 W

= 1370 W/m2 4 x  x (1.5 x 1011m)2

So is the solar constant for Earth

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Each planet has its own solar constant…

So = L / (4  rs-e2) = 3.9 x 1026 W

= 1370 W/m2 4 x  x (1.5 x 1011m)2

So is the solar constant for Earth It is determined by the distance between Earth (rs-e) and the Sun and the Sun’ luminosity.

How much solar energy reaches the Earth?

How much solar energy reaches the Earth?

Assuming solar radiation covers the area of a circle defined by the radius of the Earth (re)

Assuming solar radiation covers the area of a circle defined by the radius of the Earth (re) Ein = So (W/m2) x  re2 (m2)

re

Ein

Ein

re

How much energy does the Earth emit?

How much energy does the Earth emit? Eout = F x (area of the Earth) 300 K

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How much energy does the Earth emit?

How much energy does the Earth emit?

Eout = F x (area of the Earth)

Eout = F x (area of the Earth)

F =  T4 Area = 4 

F =  T4 Area = 4  re2

re2

Eout = ( T4) x (4  re2)

Sun

Earth

1000

100

10

1

0.1

0.01

Hotter objects emit more energy than colder objects

 (m)

Sun

Earth

Hotter objects emit at shorter wavelengths.

1000

100

10

 (m)

1

0.1

0.01

Hotter objects emit more energy than colder objects F =  T4

How much energy does the Earth emit?

max = 3000/T

Eout = F x (area of the Earth)

Sun

Earth

Eout 1000

100

10

 (m)

10

1

0.1

0.01

Hotter objects emit more energy than colder objects F =  T4

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How much energy does the Earth emit? How much solar energy reaches the Earth?

Eout = F x (area of the Earth) F =  T4 Area = 4  re2 Eout = ( T4) x (4  re2) Eout

Ein

How much solar energy reaches the Earth?

How much solar energy reaches the Earth?

We can assume solar radiation covers the area of a circle defined by the radius of the Earth (re).

We can assume solar radiation covers the area of a circle defined by the radius of the Earth (re). Ein = So x (area of circle)

Ein

re

Ein

re

Remember…

How much solar energy reaches the Earth? So = L / (4  rs-e2) = 3.9 x 1026 W

= 1370 W/m2 4 x  x (1.5 x 1011m)2

We can assume solar radiation covers the area of a circle defined by the radius of the Earth (re). Ein = So x (area of circle) Ein = So (W/m2) x  re2 (m2)

So is the solar constant for Earth It is determined by the distance between Earth (rs-e) and the Sun and the Sun’s luminosity. Ein

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re

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How much solar energy reaches the Earth? Ein = So  re2

How much solar energy reaches the Earth? Albedo (A) = % energy reflected away

BUT THIS IS NOT QUITE CORRECT!

Ein = So  re2 (1-A)

**Some energy is reflected away**

re

Ein

re

Ein

Energy Balance: How much solar energy reaches the Earth? Albedo (A) = % energy reflected away A= 0.3 today

Incoming energy = outgoing energy

Ein = Eout

Ein = So  re (1-A) 2

Ein = So  re2 (0.7) Eout

Ein

re

Ein

Energy Balance:

Energy Balance:

Ein = Eout

Ein = Eout

Ein = So  re2 (1-A)

Ein = So  re2 (1-A)

Eout =  T4(4  re2)

Eout

Ein

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Eout

Ein

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Energy Balance:

Energy Balance:

Ein = Eout

Ein = Eout

So  re2 (1-A) =  T4 (4  re2)

So  re2 (1-A) =  T4 (4  re2)

Eout

Eout

Ein

Ein

Energy Balance:

Radiation emitted by a human body

Ein = Eout

 Black Black--body laws can be applied to human

beings. For example, some of a person's energy is radiated away in the form of electromagnetic radiation, most of which is infrared.. infrared

So (1-A) =  T4 (4)

Eout

 Ein



The net power radiated is the difference between the power emitted and the power absorbed: Pnet = Pemit − Pabsorb. Applying the Stefan– Stefan– Boltzmann law, 76

Radiation emitted by a human body 



The total surface area of an adult is about 2 m², and the midmid- and farfar-infrared emissivity of skin and most clothing is near unity, as it is for most nonmetallic surfaces. Skin temperature is about 33°°C, but clothing reduces the surface 33 temperature to about 28 28°°C. Hence, the net radiative heat loss is about . The total energy radiated in one day is about 9 MJ (Mega joules joules), ), or 2000 kcal (food calories calories))

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Conclusion 





Summary

As the temperature increases, the peak wavelength emitted by the black body decreases. As temperature increases, the total energy emitted increases, because the total area under the curve increases. The curve gets infinitely close to the xx-axis but never touches it.

- A black body is a theoretical object that absorbs 100% of the radiation that hits it. Therefore it reflects no radiation and

appears perfectly black. - Roughly we can say that the stars radiate like blackbody

radiators. This is important because it means that we can use the theory for blackbody radiators to infer things about stars. - At a particular temperature the black body would emit the maximum amount of energy possible for that temperature. - Blackbody radiation does not depend on the type of object emitting it. Entire spectrum of blackbody radiation depends on only one parameter, the temperature, T.. 79

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