What is the estimated average annual insolation at the Earth’s surface for each location?

EOG 101 Physical Geography

Lab 3: Atmosphere and Surface Energy Balances (Credit: Based on UCSB Geography Department laboratory with major modifications by D. Fairbanks, N. Sato, and O. Bettis)

Name ANSWERS Lab Section Date

Materials and sources • a thermal radiometer (IR) • a Kestrel weather tracker • your textbook • class lecture notes

Introduction: All molecules continuously emit radiation in an amount more or less proportional to the fourth power of their temperature (the fourth-power relationship is strictly true only for perfect emitters, but most objects are close). For example, if you double the temperature of an object, the amount of radiation it emits increases by 16 times! Energy balance plays a fundamental role in controlling the surface temperature of the Earth. The primary source of emitted energy near us is the sun, which has a relatively stable output over short time spans but has shown long-term changes in temperature (it has been slowly heating over the past 4.5 billion years and will continue to do so). The solar constant describes the density of solar radiation (in watts) distributed over a fixed area (in meters squared) at the Earth’s orbit. Once this energy reaches Earth, a portion of it is reflected back to space and another portion is absorbed. Some energy is absorbed by the atmosphere, and some by the surface of the Earth. Absorbed energy causes atoms and molecules to heat, which in turn causes them to emit radiation via excited electrons.

Key Terms: Absorption Advection Attenuation Blackbody Conduction (G) Convection Emissivity Insolation Latent Heat (LE) Longwave Radiation Mean Radiant Energy Reflection Scattering Sensible Heat (H) Shortwave Radiation Solar Albedo Transmission Thermal radiation

Thermal Radiometer In this lab, we’ll use a thermal radiometer to measure the temperatures of objects. The radiometer measures the amount of radiation it’s receiving, and calculates what temperature the objects it’s pointing at would have to be to emit that amount of radiation. The radiometer is passive, and thus has an infinite range, but it also has a field of view, just like a camera. If you want to measure the temperature of, say, your hand, hold the radiometer so it’s a few centimeters away from your hand. If you try this from farther away, you’ll get the temperature of your hand, averaged with the temperatures of any objects that happen to be in the background within the radiometer’s field of view.

 

 

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In this lab, you will gain an appreciation of the various processes that control the flow of heat energy on the Earth. We call the balance between absorbed, reflected, and emitted radiant energy “net radiation.”

Law of Physics – Radiation Experiment Equipment required: • 100-watt light globe with dimmer: (varies amount of AC to bulb). • Light meter: that measures the amount of visible radiation in lux. The lux is the

international unit of measure for light intensity based on human brightness perception. For normal office work you would have 538 lux illumination (1 lux =0.00245 Watts/m2).

• IR temperature radiometer: that measures temperature in °C by measuring the amount of emitted IR thermal radiation. Note: Thermal radiation is emitted by all solid or gaseous bodies at high pressure (stars like the sun) in the universe with a temperature above 0 K.

Visible Radiation Near Infrared (IR) Far IR 0.38 – 0.780 µm 0.78 – 5.0 µm 40 – 350 µm

Human eye sees this Invisible to human eye

Invisible to human eye

We can sense this as heat. You can feel that an iron is on

even without touching it.

Emitted by very hot objects like the sun (5800 K)

Is present in the thermal radiation emitted by the sun but not the thermal radiation emitted by the Earth (300 K)

Emitted by moderately hot objects like the Earth’s surface

(300 K)

Step 1: This experiment will seek to help you understand the answers to the following questions:

• Why does the sun emit visible light? • Why doesn’t the Earth emit visible light? • What kind of radiation are the walls in this lab emitting?

Step 2: Formulate a hypothesis concerning the relationship between temperature and irradiance. Write your hypothesis in the space below. I think that as light intensity increases then heat will increase. The graph will be a straight line going up to the right. Step 3: Collect data/observations Your lab instructor will have set up the mobile lab table with a 100-watt light globe connected to a dimmer switch, which adjusts the electric current flowing through the light globe. Two measuring devices are also on the table – light meter and thermal IR sensor.

Now you will fill in the following table. Your lab instructor will help you measure the irradiance in lux using the light meter then using the IR sensor the temperature of the outer skin of the light bulb will be measured in degrees Celsius.

 

 

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Radiation Observations

Dimmer setting

Visible estimate of brightness (amount of emitted light) – very bright, moderately

bright, bright, or zero

Color of emitted light as visible to your

eye

Light meter reading – irradiance

(convert lux to Watts/m2)

IR sensor reading –

Temp in °C in front of light

bulb

0% Zero none 3 lux 0.0074 W/m2 26°C

12.5% Dim glow orange 4 lux 0.0098 W/m2 75°C

25% bright Orange-yellow 53 lux 0.1230 W/m2 99°C

50% Moderately bright Yellow-white 1168 lux 2.68W/m2 144°C

75% Very bright white 2000 lux 17 W/m2 189°C

100% Too bright to look at white 15,050 lux 37 W/m2 245°C

1 lux =0.00245 Watts/m2 Discussion of the Observations: 1. Why doesn’t the light globe emit any visible radiation when the dimmer is set to zero? There is no power going to the light bulb. Also, the light bulb is not hot enough 2. According to the observations, as the temperature of the light bulb goes up, the perceived brightness of the bulb (increases, decreases, remains constant). increases 3. According to the observations, as the temperature of the light bulb goes up, the amount of emitted thermal radiation (increases, decreases, remains constant). Increases 4. Did the color of the light bulb change as the temperature changed? Yes, it went from no color to orange to yellow to white 5. What do the observations, as presented in the graph below, tell you about the relationship between Irradiance and thermal radiation? As thermal radiation increases, irradiance (brightness) increases exponentially

 

 

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Graph of Temperature: Emitted Radiation Use the data from your observations, to graph the relationship between the temperature of the light globe as measured with the IR sensor and emitted radiation (irradiance) as measured by the light meter. Plot your graph on the figure below.

Step 4: Did you accept or reject your hypotheses My hypothesis was wrong. It is a positive, exponential relationship. Section 1: Net Radiation – Measuring our geographical environment We have learned that insolation (incoming solar radiation) is unevenly distributed by latitude and fluctuates seasonally. Most of the energy coming from the sun consists of shortwave radiation (ultraviolet light, visible light, and near-infrared wavelengths), and most of the energy

Free Multi-Width Graph Paper from http://incompetech.com/graphpaper/multiwidth/

IR Temperature of light bulb (°C)

Fl ux

/I rr

ad ia

nc e

(W at

ts /m

2 )

 

 

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leaving the Earth consists of longwave radiation (thermal infrared). Insolation arriving at the top of Earth’s atmosphere can be absorbed, reflected, or transmitted. Absorption causes the atmosphere to heat up. We have learned how different layers of the atmosphere and different molecules absorb different wavelengths of light from the sun. The part of the sun’s radiation that is not attenuated by absorption and scattering in the atmosphere is transmitted to the Earth’s surface, where it can be reflected or absorbed. The portion of the sun’s energy that is reflected by both the Earth’s surface and atmosphere is called the solar albedo. The energy that is not reflected is absorbed by the Earth’s surface, causing it to heat up. The Earth re-emits this energy at the longer wavelengths of the thermal infrared portion of the electromagnetic spectrum. The amount of energy a surface emits is dependent upon its temperature and emissivity, which is equal to the ratio of emitted energy from an object relative to a perfect emitter known as a black body. The molecules in the atmosphere (CO2, CH4, H2O vapor) that are strong absorbers of this thermal infrared energy are the greenhouse gases.

In order to determine whether a surface heats, cools or remains neutral, it is necessary to calculate the balance between energy absorbed and energy emitted. The difference between absorbed and emitted energy is called net radiation (NET R) and is calculated as:

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) where: NET R = Net radiation (Watts/m2) SW↓ = Insolation (downward short-wave radiation) LW↓ = Long-wave radiation emitted downward by the atmosphere (arriving) LW↑ = Long-wave radiation emitted by the surface (leaving) ε = Emissivity (intrinsic property of a material; ranges between 0 – 1) Don’t worry if that equation looks scary now – we’ll explore each of the components of the equation in order, and then you’ll understand how each part works. Let’s start with albedo. For this lab, you’ll need to join a group that has a thermal radiometer and go outside. Section 2: Albedo

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) 2a) Albedo describes the reflectivity of a surface. Describe a natural surface that has a high albedo (very reflective) and one that has a low albedo (not very reflective). High albedo surface: Snow, white sand, dry soil, granite rock Low albedo surface: Basalt rock, wet soil 2b) Choose two different surfaces that are very similar in all respects except for albedo (for example, a green plastic drain cover vs. a black plastic drain cover). They’ll have to be similar in all other respects: both must be in direct sunlight, and both must have the same orientation

 

 

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(are they both flat on the ground, or is one angled toward or away from the sun). They should also be of a similar material – for example, two papers of different colors, perhaps on one of the faces of the outdoor bulletin boards outside Butte Hall. Selecting grass and concrete won’t work because they’re different materials, so that would compare more than just albedo.

Some of the members of your group should describe the two surfaces in the table below while you measure their temperatures with a thermal radiometer. Hold the radiometer very close to the surfaces – a centimeter or two away – and hold down the button for a few seconds. Watch the numbers, and if they’re changing much, pick an average to record in the table. Each member of your group should follow these procedures when taking their turns at using the radiometer in the following sections of this lab.

Description (location, lighting, orientation, moisture, material)

MUST BE SIMILAR FOR BOTH SURFACES!

Name your two surfaces

(for example: a white painted line vs. a nearby

yellow-painted line)

Albedo: (write

“higher” or “lower”)

Temperature from radiometer

(°C):

Metal cars–paint

Surface 1: White car

Higher 48

Surface 2: Black Car

Lower 58

2c) Assuming that everything else in the “NET R” equation above is the same for both of your objects (and it will be if you’ve followed the instructions in 2b), explain why the object with a low albedo is hotter. Hint: incoming energy hitting an opaque object that isn’t reflected must be absorbed. Surfaces with a lower albedo are hotter because energy is absorb instead of reflected. Section 3: Insolation (downwelling short-wave radiation)

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) 3a) Recall that insolation is incoming solar radiation. Give two reasons why latitude and orientation (whether a surface is facing north or south, towards or away from the sun, etc) influences the amount of annual insolation that a surface receives. Reason 1: The higher the latitude (=further from the equator, north OR south), the less direct the sun’s rays Reason 2: In the northern hemisphere, south facing slopes (of mountains) get more direct sunlight than north-facing slopes (this is why most California ski areas are on the north side of the mountain—the snow sticks around longer. Ex: Heavenly Valley, Northstar, Boreal)

 

 

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3b) In your group, you need to choose two different surfaces that are very similar in all respects except for insolation (for example, concrete that’s been in direct sunlight for a while, and concrete that’s been in the shade for a while). They’ll have to be similar in all other respects, such as albedo, orientation, and material. Use a thermal radiometer to measure the temperatures of the surfaces you’ve chosen.

Description (location, albedo, orientation, moisture,

material) MUST BE SIMILAR FOR

BOTH SURFACES!

Name your surfaces (for example: concrete in direct sunlight vs. shaded

concrete)

Insolation: (write

“higher” or “lower”)

Temperature from radiometer

(°C):

Concrete

Surface 1: Concrete in the sun

higher 57

Surface 2: Concrete in the shade

lower 47

3c) Assuming that everything else in the “NET R” equation above is the same for both of your objects (and it will be if you’ve followed the instructions in 3b), explain why the object receiving less insolation is cooler. The light hitting the shaded area is reflected from nearby surfaces. In shade caused by a passing cloud, it could be said the insolation is more diffuse. 3d) How much insolation do objects receive at night? None. (Moonlight is reflected sunlight, and not direct sunlight) 3e) The wavelength (λ) of peak emission, in micrometers (µm) for any object will be:

λmax = 2900 / T where T is the object’s temperature, in Kelvin (look in your book to learn how to convert temperatures to Kelvin). This is called Wien’s law. The sun’s surface is approximately 5800 Kelvin. Calculate its wavelength of peak emission, in micrometers (µm):

λmax = 2900 / 5800 = ____0.5 µm ______ 3f) What part of the electromagnetic spectrum contains this wavelength (3e)? (Look in your textbook for help). Green/Yellow Section 4: Downwelling long-wave radiation

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑)

 

 

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4a) All molecules in Earth’s atmosphere (oxygen, nitrogen, water vapor, liquid water in clouds) emit long-wave radiation in all directions, at all times – day and night. The portion of this radiation directed downward towards Earth (arriving) heats the planet’s surface. Use the equation above to explain why the Earth’s surface would be colder at night without an atmosphere: which term would change, and why? Well, let me help you… First, is there any “insolation” during the nighttime? I hope you say “no.” Insolation = incoming solar radiation. Thus, the insolation portion is 0.

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) Where does downwelling long-wave radiation come from? From the atmosphere? If there is no atmosphere, is there any? I hope you say “no.” Thus, the downwelling long-wave radiation is 0 if there is no atmosphere.

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) Now, you have this equation remaining:

NET R = − ε∗(LW↑) Interpret this with your lab instructor. At night, there is no incoming shortwave (SW) radiation, and with no atmosphere there would be nothing to “hold on to” heat generated from incoming longwave (LW) radiation. Thus, the only part of the equation that remains is Longwave (LW) radiation (heat) leaving the earth, so it would be a very cold planet. 4b) The following table illustrates measurements made under different sky conditions by pointing the radiometer skyward. In the question 4a above, we learned that the NET R is a negative value if there is no atmosphere. Well, if there is an atmosphere, this negative value is reduced by a positive value—energy from the sky [ε∗(LW↓)].

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑)

Temperature (°C) from radiometer: Low-altitude cloud, daytime 9

Low-altitude cloud, nighttime 2

High-altitude cloud, daytime -12

High-altitude cloud, nighttime -22

Clear blue sky, daytime -31

Clear sky, nighttime -42

Re-read the introduction to this lab, look at the table, and explain why cloudy nights are usually warmer than clear nights. Hint: the warmer the temperature of a given body, the more radiation the given body emits. This is true for the sky condition.

 

 

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Clouds at night keep longwave (LW) radiation from leaving the planet. The clouds act like a blanket What nights have generally colder sky temperatures? Note that clouds consist of small droplets of liquid water, and water has a very high emissivity. Cloudless nights are generally colder—without clouds, there is less to prevent heat from leaving the planet. Section 5: Upwelling long-wave radiation

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑) 5a) All molecules on Earth’s surface emit long-wave radiation upwards (upwelling), at all times. We’ll learn about this, and the other two mechanisms by which the surface loses heat energy, by comparing two nearby locations. First, use the radiometer to measure the temperature of a shaded portion of the lawn just south of Butte Hall. Then, use a Kestrel weather tracker to measure the air temperature five centimeters above the surface of that portion of the lawn.

Following this, you’ll walk north to the dorms and make measurements from somewhere in the parking lot near the dorms. Use the radiometer to measure the surface temperature of a patch of shaded asphalt somewhere in the parking lot. Use the weather tracker to measure the air temperature 5 centimeters above the parking lot’s surface. Record these in the table below.

Around the dormitory area use the radiometer to measure the surface temperature of some dry, shaded dirt. Use the weather tracker to measure the air temperature 5 centimeters above the surface of that shaded portion of dry dirt.

LOCATION Temperature from radiometer Air temperature 5 cm above the

surface

Shaded Lawn south of Butte Hall 25 degrees C 32.1 degrees C

Shaded Asphalt surface near dorms 39 degrees C 35 degrees C

Shaded Dry soil within the dormitory area 38 degrees C 34 degrees C

5b) Use Wien’s law from question 3e to calculate the wavelength of peak emission for the warmest surface we’ve measured in this lab, in micrometers (µm). Remember that your temperature value needs to be in unit kelvin. In order for you to convert your temperature value in degrees Celsius (°C) to kelvin (K), you need to add 273.15 to your °C value.

39 + 273.15 = 312.15 Kelvin 2900/312.15 = 9.29 micrometers

5c) What part of the electromagnetic spectrum contains the wavelength of maximum emission for the surface you calculated in question 5b? (Look in your textbook). Why don’t any of the surfaces we’ve measured in this lab emit visible light? longwave infrared radiation

 

 

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They are not hot enough. The reason we can see surfaces (besides the sun) is because they reflect visible light. Section 6: Latent heat So far we have discussed only one way in which energy is transferred from one object to another: through radiation. There are several other ways energy can be transferred. One way is by direct molecular transfer of kinetic energy, in which energetic (hot) molecules transfer energy to less energetic (cooler) molecules by molecular collisions (thus speeding up the cool molecules while slowing down the hot ones). This energy transfer mechanism, called conduction, occurs when a thermal gradient exists in an object and involves the transfer of energy from the hot end to the cold end. Convective energy transfer mechanisms involve the movement of molecules along a thermal gradient. There are two convective energy transfer mechanisms, sensible heat, in which hot molecules move upwards or downwards towards a cooler surface and latent heat, in which energy is transferred by a phase change (gas/liquid/solid). Energy is required to transform water from one phase to another. Heat energy is drawn from the environment surrounding a quantity of liquid water and converted into latent heat as that liquid water evaporates to become gaseous water vapor. This is called latent heat of evaporation.

This latent heat is held in the gaseous water vapor and is not released until the water vapor condenses back into liquid water. The released energy is called the latent heat of condensation, which warms the environment near the place where the condensation occurred.

We are all quite familiar with latent heat exchange. For example, when we sweat, we are cooled as liquid water on our skin evaporates into the atmosphere and removes energy from our bodies. The movement of water vapor and its load of latent heat is one method of transferring energy.

Convective and conductive energy transfer must balance with net radiation. In other words, as a surface receives energy from net radiation, this energy is transferred through the system by sensible heat transfer (H), conduction (G), or latent heat transfer (LE). This can be summarized as:

NET R = H + G + LE 6a) Use the preceding explanation of latent heat to explain why, in Section 5 of this lab, the radiometer measured a lower temperature for the lawn than the dry dirt. Hint: we water the lawn regularly, and the grass blades contain plenty of liquid water. Think this way… If there is no water to evaporate, energy is not allocated for latent heat of evaporation. If so, the equation is re-written as:

NET R = H + G + LE = = H + G A given amount of total energy available (NET R) is allocated for H and G only. On the other hand, if there is water on the surface (e.g., wet surface), then NET R is allocated for all three (H, G, and LE). Since H (sensible transfer) is responsible for the temperature change, under which condition do you have a higher H value?

Conditions where there is less water to “hold” heat

 

 

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Section 7: Putting it all together 7a) Use the concepts of latent heat, albedo, and our measurements from Section 5 in this lab to explain in the summer, why are vegetated landscapes usually cooler than the nearby developed (urban) areas with paved streets and parking lots? The moisture in the landscape “holds” the heat (Latent heat) and makes it more resistant to temperature change. 7b) The Earth, as a whole (including the atmosphere), radiates away as much energy as it receives from the sun. This is a state of equilibrium, in which NET R = 0. If the Earth suddenly received more energy than it emitted, the planet would heat up. This would cause the Earth’s surfaces and atmosphere to emit more radiation (because the amount emitted increases with increasing temperature), and the Earth would move back to equilibrium.

Conversely, if the Earth suddenly received less energy than it emitted, the planet would cool down. This would cause the Earth’s surfaces and atmosphere to emit less radiation (because the amount emitted decreases with decreasing temperature), and the Earth would move back to equilibrium. One important determinant of the planet’s albedo is the size of its ice caps. Use section 2 of this lab to explain why the planet would heat up, and reach equilibrium at a higher average temperature if the ice caps melted.

If the ice caps melted, there earth would be less reflective overall (less snowy ice

caps = less high-albedo surfaces). So more incoming solar radiation would be absorbed by the planet rather than reflected.

7c) Re-read the introduction to this lab. Do any of the land surfaces at your latitude emit radiation for 24 hours per day? Yes! This can be demonstrated using a thermal radiometer. 7d) Use your answers to questions 3d and 7c to explain why “NET R” is usually negative at night. At night, there is no incoming shortwave radiation coming in (no sunlight). With this no longer part of the equation, there is more radiation leaving the earth and so the “NET R” is no longer zero. 7e) One of the main concerns associated with a rise in atmospheric CO2 levels is that CO2 is good at absorbing long-wave radiation in the atmosphere. With more CO2, the atmosphere would absorb more of the energy that Earth’s surface is emitting, and consequently heat up. As a result, the atmosphere would then emit more energy, some of which would warm the Earth’s surface. (We measured the energy we’re receiving from the atmosphere in section 4).

Re-read section 1 of this lab, and use the Rnet equation to explain why an increase in CO2 levels could warm the Earth’s surface, if everything else in the equation remains constant. Which part of the equation would increase (circle it), and why?

NET R = (1 − albedo) × (SW↓) + ε∗(LW↓) − ε∗(LW↑)

 

 

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Incoming longwave (LW) radiation would increase, because with more CO2 there is more matter in the atmosphere that can emit thermal radiation.

Now finally collect some radiation data for your cities that you used in lab 1.

Your Locations Latitude Longitude Degrees Minutes Seconds Degrees Minutes Seconds

1

2

3

Use the following figure to determine the monthly insolation values (W/m2) for each location. Place the table below and then graph the data (each location gets a different color).

 

 

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Location Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Average

 

 

 

 

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Note that you will have three separate curved lines plotted for your three cities, these lines will show that insolation is highest in June-July for all three cities, and that the city with the lowest latitude will have the highest insolation.

8a)Explain the causes of any similarities or differences in the insolation patterns for your three cities insolation values that you have, and make sure to include the values in your answer. What is the estimated average annual insolation at the Earth’s surface for each location?

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