MASSACHUSETTS INSTITUTE OF TECHNOLOGY …web.mit.edu/8.02t/www/mitxmaterials/ProblemSolving/solving11.pdfblackbody is in thermal equilibrium with its surroundings the ... averaged

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.02

Problem Solving 10: The Greenhouse Effect Section ______Table and Group _____________________________

Names ____________________________________

____________________________________ ____________________________________ ____________________________________

Hand in one copy per group at the end of the Friday Problem Solving Session.

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Introduction During this problem solving session we will introduce two simple models of the earth-atmosphere system, The Bare Rock Model and the Single Layer Model, to demonstrate how the greenhouse effect works. These simple models will help us to develop a basic understanding of the effect of solar radiation and atmospheric gasses on the temperature of Earth. First, we will develop the Bare Rock Model to calculate the temperature of the surface of the Earth with an atmosphere that is completely transparent to all incoming and outgoing radiation. In this model the predicted temperature will be lower then the observed average surface temperature of Earth. Then we will add to the Bare Rock Model a single layer of atmosphere that absorbs radiation emitted by Earth and calculate the temperature of the surface of Earth. In this model our calculation will be higher then the observed average surface temperature of Earth. A more detailed of atmospheric physics is necessary in order to obtain a more accurate predicted value. To build these models, we need to introduce two new concepts: radiative equilibrium and black body radiation . Radiative equilibrium

The Earth absorbs part of the solar energy carried by the electromagnetic waves emitted by the Sun. As a result, the Earth heats up. In order to keep its temperature constant, the Earth emits electromagnetic waves carrying energy back to space in such a way that the rate of energy emitted equals the rate of energy absorbed. The continuous process of absorption and emission creates a state of thermal equilibrium called radiative equilibrium.1. 1 Radiative equilibrium refers to a thermal equilibrium in which radiation is the only mean by which the object loses energy. In general, heat can also flow by conduction and

Rate of absorption = Rate of emission

Sun

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Electromagnetic radiation: Any object with temperature above absolute zero emits electromagnetic radiation. For example, the red-hot electric burner of a stove or the coils of a toaster emit electromagnetic radiation that you can see. Even when the stove is at room temperature the electric burner radiates electromagnetic energy. The difference is that the stove is emitting electromagnetic waves at frequencies that the human eyes can’t detect, below the infrared. Blackbody radiation: The sun and Earth both emit electromagnetic radiation. In order to quantify the emitted electromagnetic radiation we introduce the concept of blackbody radiation. A blackbody refers to an opaque object that emits thermal radiation. A perfect blackbody is one that absorbs all incoming light and does not reflect any. At room temperature, such an object would appear to be perfectly black (hence the term blackbody). If the blackbody is in thermal equilibrium with its surroundings the emitted electromagnetic radiation is called blackbody radiation2 The intensity I of the electromagnetic radiation emitted by a prefect blackbody depends on the temperature and is given by the Stefan-Boltzmann law3:

I =σT 4 , where σ = 5.67 ×10−8 W ⋅m−2 ⋅K−4 is the Stefan-Boltzmann constant and T is the absolute temperature of the object. Temperature and Wavelength: Wien’s Displacement Law The temperature of a blackbody not only determines the intensity of the electromagnetic radiation but also is related to the wavelength of the radiated electromagnetic waves. The larger the temperature of the object the shorter the wavelength of the emitted radiation. A plot of the intensity versus the wavelength of the emitted waves is called a spectrum. convection, or by work done on or by the system. Thus radiative equilibrium applies when the object is not doing work and conduction and convection losses are small or absent. 2 Feymman Lectures on Physics, Volume 1, Chapter 41 3 http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan2.html - c1

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The shape of the curves depends on the temperature. Note that as the temperature goes up the maximum of the curves increases. Also, the maximum of the curve occurs at a shorter wavelength. The relationship between the wavelength for which the curve is a maximum and the temperature is a maximum is called the Wien's Displacement Law4,

λmax =

14.965

hckT

,

where h = 6.63×10−34 m2 ⋅kg ⋅s−1 is Planck’s constant, c = 3.00×108 m ⋅s−1 is the speed of light, k = 1.38×10−23 kg ⋅m2 ⋅s−2 ⋅K−1 is Boltzmann’s constant, and T is the absolute temperature. This law is often written as

λmax =

bT

,

where the constant of proportionality is called Wien’s displacement constant equal to 2.90×10−3 m ⋅K .

The intensity of the sun’s electromagnetic radiation has a maximum around wavelengths corresponding to visible light 400 nm < λ < 700 nm . The effective temperature of the Sun is 5778 K. Using Wien's law, this corresponds to a maximum wavelength of about 500 nm in the green portion of the spectrum, which is near the peak sensitivity of the human eye. On the other hand, the average temperature of the earth is about 300 K and the intensity of the earth’s electromagnetic radiation has a maximum at wavelength

4 http://hyperphysics.phy-astr.gsu.edu/hbase/wien.html - c2

500 1000 1500 2000[nm]λ

T=5500 K

T=5000 K

T=4500 K

T=4000 KT=3500 K

2

4

6

8

Powe

r den

sity

[1

013 W

/m3 ]

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corresponding to infrared radiation λ ! 10 µm . (Note: To show the Earth's spectrum, the curves in the figure below have different scales.)

The wavelength of the electromagnetic radiation is important because the gases in the atmosphere absorb electromagnetic waves of certain wavelengths. For example, the ozone absorbs the ultraviolet solar radiation of wavelength λ ! 200− 300 nm . This is fortunate because ultraviolet radiation is highly damaging to living organisms. On the other hand, water vapor and carbon dioxide absorb the infrared radiation emitted by the Earth. These gasses are the main contributors to the greenhouse effect.

8 x 107

2 x 107

4 x 107

6 x 107

0.4 0.5 0.6 0.7

The Sun6000K

Wavelength

The Earth288K

5 10 15 20

16800

λmax

( m)μ ( m)μ

λmax

Inten

sity (

W/m

2 / m

)μ

Wavelength

Solar Radiation:Higer T - shorter (visible region)

λEarth Radiation:Lower T - longer (infrared region)

λ

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Problem 1: The Solar Power at the top of Earth's Atmosphere The goal of this problem is to relate the time averaged power of the electromagnetic energy radiated by the sun, Ps , and the intensity of the electromagnetic waves at the top of Earth's atmosphere, known as the solar constant, S0 , which is the time average of the Poynting vector at the top of Earth's atmosphere.

a) Assume that the sun radiates electromagnetic waves uniformly in all directions as if it were a point source, calculate S0 the intensity of the electromagnetic waves at the top of Earth's atmosphere in terms of Ps and R0 the mean radius of Earth's orbit around the Sun.

Answer:

b) The averaged measured value is S0 = 1.36×103 W ⋅m−2 [1]. The mean radius of

the Earth's orbit around the sun is R0 = 1.50×108 km . The averaged power emitted by the sun is:

1. Ps = 3.8×1020 W

2. Ps = 3.8×1026 W

3. Ps = 3.8×1017 W

4. Ps = 3.8×1036 W Answer:

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Problem 2: Averaged Incident Power at the top of Earth's Atmosphere

a) Average incident solar power at the top of Earth's atmosphere: In the previous problem we considered the sun to be a source of electromagnetic waves radiating uniformly in all directions. Because the radius of Earth, RE = 6.37 ×103 km , is

much smaller than the mean radius of Earth's orbit around the sun, R0 = 1.50×108 km , the electromagnetic waves emitted by the sun can be assumed to be planar by the time they reach Earth.

Calculate the time averaged power of solar radiation at the top of Earth's atmosphere, Ptop

. Express your answer in terms of S0 and RE . (You can neglect the depth of the

atmosphere of 30 km with respect to the radius of Earth, RE = 6.37 ×103 km ). Answer:

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b) Incoming average solar power absorbed by Earth:

When the electromagnetic radiation from the sun reaches the top of Earth's atmosphere part of it is reflected back into space and the other part is absorbed by Earth. The fraction of energy reflected by a particular surface that arise from a variety of effects is called the albedo and it is denoted by the letter α . The yearly average albedo of Earth is within a few percent α ! 0.30 ([1]). Calculate, Pin , the time averaged rate of incoming solar radiation absorbed by Earth in terms of α , S0 , and RE . Answer:

c) Incoming intensity of the solar radiation absorbed by Earth. From our experience, we know that the daytime side of Earth is warmer than the nighttime side, and it is even warmer at the equator. In our simple model, we will assume Earth is a uniform spherical blackbody that perfectly absorbs incoming solar radiation and radiates it back into space at a temperature TE . Under this assumption, calculate Iin the intensity of the incoming solar radiation absorbed by Earth. Express your answer in terms of α and S0 . Answer: [1] The albedo varies with the amount of coverage of clouds, ice, and snow. On average, Earth's albedo is 0.30: for clear sky, α ≈ 0.15 , when fully covered by clouds or snow α ≈ 0.60 . It also depends on the particles suspended in the atmosphere. For example, the albedo of Venus is α ≈ 0.70 due to the thick layer of sulfuric acid in its atmosphere - that is the reason for why it appears so bright in the sky.

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Problem 3: The Bare Rock Model Earth absorbs the incoming solar radiation and it warms up. In order to keep the same temperature TE at all times, Earth must be in radiative equilibrium with its surroundings. Therefore, The solar radiation energy absorbed by Earth must equal the radiation emitted by Earth. We will first make the assumption that the radiation emitted by Earth does not interact with the gasses of the atmosphere; Earth's radiation escapes out to space as if there were no atmosphere. In addition, we assume that Earth radiates as a perfect spherical blackbody. This model is called the Bare Rock Model.

As a result, the intensity of the energy radiated by Earth can be calculated using the Stefan-Boltzmann law:

Iout =σTE4 ,

where σ = 5.67 ×10−8 W ⋅m−2 ⋅K−4 is the Stefan-Boltzmann constant and TE is the surface temperature of Earth.

a) Using the expression for the intensity of the solar radiation absorbed by Earth obtained in Problem 2,

Iin =

(1−α )S0

4,

calculate the surface temperature of Earth, TE , in terms of S0 , α , and σ .

Answer:

Earth

Incoming Solar Radiation Outgoing Earth radiation

Iin

= (1 )4

S0

Iout

= TE4α

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b) Using the numerical values of the solar constant S0 = 1.36 kW ⋅m−2 , Earth's

albedo α ! 0.30 , and the Stefan-Boltzmann constant σ = 5.67 ×10−8 W ⋅m−2 ⋅K−4 , calculate the surface temperature of Earth.

Answer: Predicted Earth Temperature: If you did your calculation correctly, your value for TE is too low compared to the actual

observed averaged surface temperature of Earth, 288 K, which is equal to 15 !C . Our model was very simple and did not consider the fact that the atmosphere does absorb the infrared radiation of Earth. In the next problem we will add a single layer of atmosphere to the Bare Rock model and evaluate its effect on the predicted temperature. Bare Rock Model Assumptions: Below we summarize the assumptions made in the Bare Rock Model

1. The atmosphere transmits all the solar radiation (it is transparent in wavelengths characteristic of the sun's radiation).

2. The Earth behaves as a perfect spherical blackbody at constant temperature TE . It absorbs all the incoming solar radiation and it emits infrared radiation following the Stefan-Boltzmann law: Iout =σTE

4 .

3. The atmosphere transmits all of Earth's radiation. (It is transparent to Earth's infrared radiation, which is longer in wavelength than the wavelength of visible light.)

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Problem 4: One Layer Model The Bare Rock Model predicts too low a temperature because it does not consider the absorption of radiation from the atmosphere. The greenhouse effect is the result of the absorption of the infrared radiation emitted by the surface of Earth by the gases in the atmosphere (water vapor, carbon dioxide, and methane). We can make a simple model of the impact that this absorption has on the temperature of the surface of Earth by modeling the atmosphere as a single layer with the following assumptions: 1. The atmosphere is transparent to the incoming solar energy. 2. The surface of Earth radiates as a perfect spherical blackbody at a temperature TE . 3. The single layer atmosphere absorbs all the infrared radiation emitted by Earth. 4. The atmosphere then radiates as a perfect blackbody at TA , with an outward radiation away from the surface of Earth and an inward radiation towards the surface of Earth. 5. Radiative equilibrium is the condition that the incoming solar radiation is equal to the outward radiation from the atmosphere. 6. Radiative equilibrium on the surface of Earth is the condition that the incoming solar radiation and the downward atmospheric radiation is equal to the radiation from the surface. The different types of radiation involved in the Single Layer Model are indicated in the energy diagram below. Note that the single layer of atmosphere emits radiation downward from its bottom surface and radiation upward back to space from its top surface.

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a) Consider the single layer of atmosphere to be at a constant temperature TA . Apply the radiative equilibrium condition to the outer layer of atmosphere and obtain an expression for TA

4 . Express your answer in terms of α , S0 , and σ .

Answer: b) Consider the surface of Earth to be at a constant temperature TE . Apply the

radiative equilibrium condition at the surface of Earth to obtain an expression for TE4 .

Express your answer in terms of α , S0 , TE , and TA as needed. Answer:

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c) Define TBRM to be the Earth's temperature predicted by the Bare Rock Model you

calculated in the previous problem, TBRM =

(1−α )S0

4σ⎛⎝⎜

⎞⎠⎟

1/4

. What is the ratio TE / TBRM ?

1. TE / TBRM = 22. TE / TBRM = 1/ 2

3. TE / TBRM = 21/4

4. TE / TBRM = (1/ 2)1/4

Answer: d) Calculate the temperature of the surface of Earth predicted by the Single Layer model. Enter your answer in K. Answer: Conclusions: The temperature that you calculated is too high. This is due to overly simplifying assumptions that we have made about the atmosphere. The next step is to consider a more dynamic model of the atmosphere that includes the study of radaitive-convective equilibrium, different rates of absorption of radiation by the different gasses, and a host of other effects that requires a more detailed understanding of atmospheric physics.