Lecture 5_Blackbody Radiation_Energy Balance

7/28/2019 Lecture 5_Blackbody Radiation_Energy Balance

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Lecture 5 --Blackbody Radiation/

Planetary Energy Balance

Abiol 574


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Electromagnetic Spectrum

(m)

1000 100 10 1 0.1 0.01

visible

light

0.7 to 0.4 m


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(m)

1000 100 10 1 0.1 0.01

ultraviolet

visible

light


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(m)

1000 100 10 1 0.1 0.01

ultraviolet

visible

lightinfrared


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(m)

1000 100 10 1 0.1 0.01

ultraviolet

visible

lightinfraredmicrowaves x-rays


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(m)

1000 100 10 1 0.1 0.01

ultraviolet

visible


High

Energy

Low

Energy


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Blackbody Radiation

Blackbody radiationradiation emitted by a body that

emits (or absorbs) equally well at all wavelengths


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The Planck Function

Blackbody radiation follows the Planck function


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Basic Laws of Radiation

1) All objects emit radiant energy.


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2) Hotter objects emit more energy than colder

objects.


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objects. The amount of energy radiated is

proportional to the temperature of the object.


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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 = T4F = flux of energy (W/m2)

T = temperature (K)

= 5.67 x 10-8 W/m2K4 (a constant)


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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.


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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.

This isWiens Law

max 3000 mT(K)


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Stefan Boltzmann Law.

F = T4F = flux of energy (W/m2)

T = temperature (K)

= 5.67 x 10

-8

W/m

2

K

4

(a constant)

Wiens Law

max 3000 mT(K)


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We can use these equations to calculate properties

of energy radiating from the Sun and the Earth.

6,000 K 300 K


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T(K)

max(m)

region inspectrum F

(W/m2)

Sun 6000

Earth 300


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T(K)

max(m)

region inspectrum F

(W/m2)

Sun 6000 0.5

Earth 300 10


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(m)

1000 100 10 1 0.1 0.01

ultraviolet

visible


High

Energy

Low

Energy


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T(K)

max(m)

region inspectrum F

(W/m2)

Sun 6000 0.5 Visible

(yellow?)

Earth 300 10 infrared


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Blue light from the Sun is removed from the beam

by Rayleigh scattering, so the Sun appears yellow

when viewed from Earths surface even though its

radiation peaks in the green


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T(K)

max(m)

region inspectrum F

(W/m2)


(green)

Earth 300 10 infrared


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Stefan Boltzman Law.

F = T4F = flux of energy (W/m

2

)T = temperature (K)

= 5.67 x 10-8 W/m2K4 (a constant)


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T(K)

max(m)

region inspectrum F

(W/m2)


(green)

7 x 107

Earth 300 10 infrared 460


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Solar Radiation and Earths Energy Balance


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Planetary Energy Balance

We can use the concepts learned so far

to calculate the radiation balance of the

Earth


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Some Basic Information:

Area of a circle = r2Area of a sphere = 4 r2


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Energy Balance:

The amount of energy delivered to the Earth is

equal to the energy lost from the Earth.

Otherwise, the Earths temperature would

continually rise (or fall).

E B l


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Energy Balance:

Incoming energy = outgoing energy

Ein = Eout

Ein

Eout


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(The rest of this derivation will be done on the

board. However, I will leave these slides in here

in case anyone wants to look at them.)


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How much solar energy reaches the Earth?


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As energy moves away from the sun, it is

spread over a greater and greater area.


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As energy moves away from the sun, it is

spread over a greater and greater area.

This is the Inverse Square Law


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So = L / area of sphere


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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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So = L / (4 rs-e2

) = 3.9 x 1026 W = 1370 W/m2

4 x x (1.5 x 1011m)2


It is determined by the distance between Earth (rs-e)

and the Sun and the Sun luminosity.

E h l t h it l t t


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Each planet has its own solar constant


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

Ein re


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

Ein = So (W/m2) x re2 (m2)

Ein re


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How much energy does the Earth emit?

300 K


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Eout = F x (area of the Earth)


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F = T4

Area = 4 re2


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F = T4

Area = 4 re2

Eout = ( T4) x (4 re2)


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(m)

1000 100 10 1 0.1 0.01

Earth Sun

Hotter objects emit

more energy thancolder objects


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(m)

1000 100 10 1 0.1 0.01

Earth Sun

Hotter objects emit


F = T4


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(m)

1000 100 10 1 0.1 0.01

Earth Sun

Hotter objects emit at

shorter wavelengths.

max = 3000/T

Hotter objects emit


F = T4

H h d th E th it?


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Eout

H h d th E th it?


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F = T4

Area = 4 re2

Eout = ( T4) x (4 re2)Eout


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Ein


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

Einre


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Ein = So x (area of circle)

Einre

R b


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So = L / (4 rs-e2

) = 3.9 x 1026

W = 1370 W/m2

4 x x (1.5 x 1011m)2


It is determined by the distance between Earth (rs-e)

and the Sun and the Suns luminosity.

Remember


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Ein = So x (area of circle)

Ein = So (W/m2) x re2 (m2)

Einre


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Ein = So re2

BUT THIS IS NOT QUITE CORRECT!

**Some energy is reflected away**

Einre


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Albedo (A) = % energy reflected away

Ein = So re2 (1-A)

Einre


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Albedo (A) = % energy reflected away

A= 0.3 today

Ein = So re2 (1-A)Ein = So re2 (0.7)

reEin

Energy Balance:


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gy

Incoming energy = outgoing energy

Ein = Eout

Eout

Ein

Energy Balance:


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gy

Ein = Eout

Ein = So re2 (1-A)

Eout

Ein

Energy Balance:


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gy

Ein = Eout

Ein = So re2 (1-A)Eout = T4(4 re2)

Eout

Ein

Energy Balance:


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gy

Ein = Eout

So re2 (1-A) = T4 (4 re2)

Eout

Ein

Energy Balance:


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Ein = Eout

So re2 (1-A) = T4 (4 re2)

Eout

Ein

Energy Balance:


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Ein = Eout

So (1-A) = T4 (4)

Eout

Ein

Energy Balance:


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Ein = Eout

So (1-A) = T4 (4)T4 = So(1-A)

4Eout

Ein


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T4 = So(1-A)

4If we know So and A, we can calculate the

temperature of the Earth. We call this theexpected temperature (Texp). It is the

temperature we would expect if Earth behaves

like a blackbody.

This calculation can be done for any planet,provided we know its solar constant and albedo.


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T4 = So(1-A)

4For Earth:

So = 1370 W/m2A = 0.3

= 5.67 x 10-8 W/m2K4


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T4 = So(1-A)

4For Earth:

So = 1370 W/m2A = 0.3

= 5.67 x 10-8

T4

= (1370 W/m2

)(1-0.3)4 (5.67 x 10-8 W/m2K4)


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T4 = So(1-A)

4For Earth:

So = 1370 W/m2

A = 0.3 = 5.67 x 10-8

T4 = (1370 W/m2)(1-0.3)

4 (5.67 x 10-8

W/m2

K4

)

T4 = 4.23 x 109 (K4)

T = 255 K


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Expected Temperature:

Texp = 255 K

(oC) = (K) - 273


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Expected Temperature:

Texp = 255 K

(oC) = (K) - 273

So.

Texp= (255 - 273) = -18oC

(which is about 0oF)


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Is the Earths surface really -18 oC?


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NO. The actual temperature is warmer!

The observed temperature (Tobs) is 15oC, or

about 59 oF.


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NO. The actual temperature is warmer!

The observed temperature (Tobs) is 15oC, or

about 59 oF.

The difference between observed andexpected temperatures (T):

T = Tobs - Texp

T = 15 - (-18)

T = + 33 oC


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T = + 33 oCIn other words, the Earth is 33 oC warmer thanexpected based on black body calculations

and the known input of solar energy.


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T = + 33 oCIn other words, the Earth is 33 oC warmer thanexpected based on black body calculations


This extra warmth is what we call theGREENHOUSE EFFECT.


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T = + 33 oC

In other words, the Earth is 33 oC warmer thanexpected based on black body calculations


This extra warmth is what we call theGREENHOUSE EFFECT.

It is a result of warming of the Earths surface

by the absorption of radiation by molecules in

the atmosphere.


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The greenhouse effect:

Heat is absorbed or trappedby gases in the atmosphere.

Earth naturally has a

greenhouse effect of +33o

C.


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The concern is that the amount of greenhouse warming

Lecture 5_Blackbody Radiation_Energy Balance

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