Temperature and
Radiation
3
Part I. Introduction
All objects emit electromagnetic radiation. The amount of energy that is emitted by an object
per unit surface area of the object is directly proportional to the surface temperature of the object
(T is in the numerator). This is known as the Stefan-Boltzmann Law which can be written as:
where;
(1)
E = "T4
E is the total amount of energy emitted from one square meter of the surface of the object
every second.
!
It is expressed in units of Joules of energy per square meter per second or
J
Watts 2 since a Watt is a J ,
2 or
s
m
ms
-8
2o 4
σ is the Stefan-Boltzmann constant equal to 5.67 x 10 W/(m K ), and
o
T is the surface temperature of the object in degrees Kelvin, K.
!
!
!
The wavelength at which the maximum radiation is emitted by the surface of the object is
indirectly proportional to the temperature of the surface (T is in the denominator). This is shown
by the Wien's Displacement equation, which is:
λmax =
where,
2897µmoK
....................................... (2)
T
λmax
= wavelength at which the maximum radiation is emitted;
2897µmoK = Wien's constant in µm (micrometers) degrees Kelvin. A µm (a
micrometer) is equal to one-millionth of a meter. If the wavelength of
radiation was desired in units of centimeters, the constant would be
0.2897cmoK.
T
= surface temperature of the emitting object in degrees Kelvin (oK).
Looking at equation (2), since T is in the denominator, then when T becomes a larger number
(i.e., the surface temperature of the object becomes greater), then λmax will become a smaller
number (i.e., the wavelength at which the maximum radiation is emitted by the surface of the
object becomes shorter). Thus, as the temperature increases, the wavelength at which the
maximum radiation is emitted becomes shorter.
Wavelength and frequency are related by the following equation.
49
c
f = λ ,......................................................... (3)
where,
f
=
c =
λ =
frequency of radiation; i.e. how often an event which is in periodic
motion recurs.
speed of light 3.0 X 108 m/sec
wavelength of radiation.
Thus, as the wavelength of the emitted radiation becomes smaller, the frequency of the
emitted radiation increases.
Problem 1.
The average surface temperature of Mars is approximately -81oF (-63oC or
210oK).
On the average, what is the total amount of energy that is being emitted from a
square meter of surface area of Mars every second?
Problem 2.
What is the wavelength (on the average) at which the maximum amount of
radiation is being emitted by the surface of Mars?
Problem 3.
The average surface temperature of Venus is approximately 867oF (464oC or
737oK). On the average, what is the total amount of energy that is being
emitted from a square meter of surface area of Venus every second?
Problem 4.
What is the wavelength (on the average) at which the maximum amount of
radiation is being emitted by the surface of Venus?
The average amount of radiation that the Earth receives from the Sun at the “top of the
2
atmosphere” is called the solar constant and is equal to approximately 1367 W/m of energy. In
actuality, it is not a constant because it varies slightly throughout the year. Scientists also believe
that the emission of radiation by the Sun has been gradually increasing over the last 4.6 billion
years of the Earth’s history. The change in the solar constant that reflects the change in the
amount of radiation emitted by the Sun for the last 4.6 billion years can be expressed by the
following equation:
SE =
1367
(4)
t
)]
4.6
2
where the units of SE (solar constant of Earth) are in W/m . The t in the denominator is the time
in billions of years from the assumed “start of Earth’s history.” Thus, a t = 0 means 4.6 billion
years ago and 4.6 means!the present time. By using this equation, one can estimate the value of
the solar constant at various times during Earth’s history.
[1 + 0.4(1 -
Problem 5.
Using equation (4), what was the estimated solar constant 2.3 billion years ago
(meaning t = 2.3 or 2.3 billion years from the “start of Earth’s history”?
50
Problem 6.
Using equation (4), what was the estimated solar constant 700 million years ago
(meaning 3.9 billion years from the “start of Earth’s history” or t = 3.9?
At both times for problem 5 and 6, the Earth was experiencing an “ice age.” The value of the
solar constant would only be part of the reason for the “ice age.” Other factors play a role and
perhaps a greater role than the solar constant.
This amount of radiation received at the top of the atmosphere, the Earth’s solar constant, is only
a small portion of that energy emitted by the Sun. Those planets farther from the Sun receive
less radiation per square meter than those closer. The amount decreases by the square of the
distance from the Sun. If we use the average distance from the Sun to the Earth as one
astronomical unit (AU), then we can write a simple equation that expresses the solar constant for
another planet in our solar system based on the Earth’s solar constant and the distance that planet
is from the Sun in astronomical units. The equation is below.
Planet's solar constant =
Earth's solar constant
, …………….(5)
R2
where R is the distance the planet is from the Sun in astronomical units.
Problem!7.
Mars orbits the Sun at an average distance of 1.52 AU. What would be the
current solar constant for Mars?
Problem 8.
Assume that the Sun’s radiation output will continue to increase in the future as
described equation (4). Using both equation (4) and (5) determine how many
years into the future will Mars’ solar constant equal the present Earth’s solar
2
constant of 1367 W/m ?
Part II.
Demonstration and Use of an Infrared Thermometer
The IRT is essentially a radiometer which, like the eye of a camera, intercepts
electromagnetic radiation from a remote “target” or surface. However, it is made to sense only
radiation from a remote “target” or surface. It cannot sense the temperature of the air directly as
a mercury thermometer does.
At the earth's surface, and in the atmosphere, most radiation from surfaces naturally heated
by the sun (i.e., not artificially heated by burning something) is in the infrared range (i.e., longer
wavelengths than in the visible range). Consequently, by selectively sensing radiation emitted in
the frequency band produced by objects within the general range of temperatures found on the
earth's surface and in the atmosphere, the IRT can remotely determine the temperature of these
objects. Most IRTs sense radiation over the wavelength interval of 10 - 12 µm because the
atmosphere is nearly transparent to the energy in this waveband and this is the wavelength in
which the maximum amount of radiation is being emitted by most objects on the Earth’s surface.
In addition, with a suitable lens, the instrument only intercepts radiation in a narrow cone, i.e., its
field-of-view is relatively small. Hence, the target area sampled depends on the distance
between the sensor and the target. For example, at close range, the target area may be only 1
51
cm2; but when the distance involved is, say 37,000 km (23,000 miles as with the GOES
satellite), the target may be tens of km2.
The sensing element of the IRT is called a bolometer, a sensitive and very small
resistance thermometer. The electrical resistance of such a device changes in a known way with
changes in temperature. Radiation passing through the lens of the IRT falls on a small surface in
which is imbedded the small resistance thermometer. The radiation causes the surface to become
warmer and the resistance in the thermometer changes with this increased temperature of the
surface. Through a chopping mechanism, the surface of the bolometer is alternately exposed to
radiation from the target and then to radiation emitted by an internal source which is maintained
at a known high temperature. The resulting difference between the resistance measured when
the bolometer is exposed to the target and the resistance measured when the bolometer is
exposed to the known, internal source provides a determination of the difference in temperature
between the target and the internal source. Since the internal source temperature is known and
controlled, the IRT can easily determine the target temperature.
This is the type of instrument which is mounted on satellites or on aircraft for
determining land/water surface and cloud-top temperatures. When used in this way, it is possible
to determine temperatures over a very large region in a short period of time. This is especially
critical for areas over which there are few or no other observations, e.g., the oceans. In addition,
since clouds are found at high levels in the atmosphere, hence are colder than the earth's surface,
the IRT makes it possible to delineate these “cold targets” from the warmer, surface objects. It
not only shows the extent of cloud systems, but also permits estimation of cloud heights, since
low clouds appear warmer than high clouds. Such observations are quite valuable to weather
forecasters.
The purpose of this exercise is to demonstrate the ease with which the temperature of a
surface may be determined remotely. In addition, the exercise will show how surface
temperatures vary in space in the microscale domain. Since the atmosphere receives most of its
energy secondhand from the ground surface, not directly from the sun (i.e., primarily by
conduction from the ground surface to air molecules close to the ground which are then carried
upward by thermal currents and turbulence), such temperature variations in the horizontal and
vertical are important when considering the energetics of the atmosphere. This is especially true
in considerations of the “boundary layer” , i.e., the lowest few thousand feet of the troposphere.
The related energy may be in the form of sensible heat, (approximately the heat we feel and can
measure with a thermometer; related to internal energy), or latent heat, (the heat energy released
or absorbed in a change of phase or change of state); and the energy fluxes (process of
movement of energy) is either by conduction, convection, or electromagnetic radiation, each of
which is dependent on, or associated with, temperature of the ground surface as well as of the
atmosphere. Conduction is the transfer of energy by molecular motion within a fluid resulting
in transport and mixing of the properties (e.g., energy) of that fluid. Radiation is the transport of
energy by virtue of joint undulatory variations in the electric and magnetic fields of space.
Depending on weather conditions that prevail at the time of the experiment, we will
attempt to show how temperatures of surfaces vary as a function of: material, color, degree of
wetness, and degree of sunshine, i.e., in direct sunlight or in the shade. Since we have little or no
control of conditions in our “outdoor laboratory,” (e.g., clouds may interfere with our efforts),
we may meet with limited success.
52
Each student will have the opportunity to use the hand-held IRT to determine the
temperatures of the various surfaces under different conditions, as indicated on the table on the
first and second pages of the answer sheet. Copy the temperatures measured on your answer
sheet.
Complete the following and answer the questions.
Problem 9.
On the answer sheet, rank the top ten warmest objects (only those without the *)
according to temperature with the highest temperature first, number 1, and the
last, number 10).
Problem 10.
Next, on the answer sheet, rank the top ten coldest objects (only those without a
*) according to temperature with the coldest temperature as number 1 and the
last number 10.
Answer the following questions about the data you collected. Place your answers on the
answer sheet.
Problem 11.
In the warmest objects ranking, were most objects in the sun or in the shade?
Problem 12.
In the coldest objects ranking, were most objects in the sun or in the shade?
Problem 13.
In the warmest objects ranking, were most objects light colored or dark colored?
Problem 14.
In the coldest objects ranking, were most objects light colored or dark colored?
Problem 15.
Considering your answers to problems 13 and 14, which objects absorb energy
more efficiently, light colored or dark colored?
All of the radiation which strikes an object will either be absorbed, reflected or
scattered. Objects which absorb radiation efficiently, and thus reflect or scatter only a small
amount, will warm more quickly to higher temperatures. Objects which absorb radiation less
efficiently, and thus reflect or scatter a larger amount of radiation, will tend to warm more slowly
and have a cooler temperature; assuming the same amount of radiation is falling on the two
objects. The amount of radiation which is either absorbed, reflected or scattered depends on the
characteristics of the material. You have just seen that the color of an object plays a role in its
ability to absorb, reflect or scatter radiation.
Problem 16.
In the warmest objects ranking, what type of material was warmest: metal,
water, grass, wood (tree), glass, dirt, concrete, rubber, asphalt?
Problem 17.
In the coldest objects ranking, what type of material was coldest: metal, water,
grass, wood (tree), glass, dirt, concrete, rubber, asphalt?
Problem 18.
Which object was warmest: the dry dirt patch in the sun or the green grass in the
sun?
53
Considering the answers to problems 16, 17 and 18, you can see that the ability of an
object to absorb, reflect or scatter radiation also depends on the type of material of which the
object is composed. Even though some materials may have nearly the same color, the amount of
radiation they absorb, and thus the temperature to which they warm, depends on the material of
which they are composed.
Problem 19.
Considering the articles of clothing, which color was warmest when in the sun?
Problem 20.
Considering the articles of clothing, which color was coolest when in the sun?
Problem 21.
Besides energy from the sun through radiation, from what other source are the
articles of clothing receiving energy?
Considering the answers to problems 19, 20 and 21, you can see that solar radiation is
only one source for the energy that an object receives which will determine its temperature. In
the above problems, the second source of energy was the human body; however, for the
atmosphere, that second source may be energy being conducted from the ground into the air
which is in contact with the ground. Objects also emit energy. The ground or ocean surface and
objects on the ground, such as trees, cars, buildings, trees, etc., emit energy. In the atmosphere,
clouds, particles, and the molecules of the air all emit energy. As objects or air molecules emit
energy, other objects and air molecules will absorb some of this energy. So, the amount of
energy (and thus the temperature) an object or molecule has at any moment is dependent of the
amount received by absorption of radiation and by conduction into the object or molecule (a
plus) and that lost due to emission or conduction away from the object (a minus).
Problem 22.
The temperature reading from the “clear sky” should have a low temperature
value as was obtained in January 1994. Why does the clear sky show such a
low temperature reading? (Consider the location from which the radiation is
emanating.)
Part III. Measuring Temperature with Mercury Thermometers
During a hot summer, one often hears statements such as “the temperature is 95o in the
shade” — implying that the temperature “in the sun” is higher than 95o. Certainly, a person in
the sun will feel more uncomfortable than in the shade; but, is the air really at a higher
temperature in the sun than in the shade? Why do we seek the shade on a hot, sunny day? Are
you warmer in the sun because the air is warmer, or are there for other reasons? Why are lightcolored clothes apt to be cooler than dark ones? You should be able to answer these questions
after working this exercise.
Exercise:
Before making any readings, place the two wicks, the aluminum foil, and the black cloth
on the ground in the shade. It is important that each of these comes into thermal equilibrium
with its surroundings.
A.
Using an ordinary mercury-in-glass thermometer: (Be certain to allow the thermometer
time to come to thermal equilibrium each time — it will cease changing.)
54
a.
Read the temperature in the shade. Shade the entire thermometer from direct
sunlight with a sheet of paper, book, or place it in your shadow.
Record the temperature here and on the answer sheet. __________
b.
Read the temperature with just the bulb exposed to the sun. Shade just the stem
of the thermometer.
Record the temperature here and on the answer sheet. __________
c.
Read the temperature in the sun, but shade just the bulb, leaving the stem exposed
to the sun.
Record the temperature here and on the answer sheet. __________
Answer the following questions. Place your answers on the answer sheet.
Problem 23.
Considering the above data Part III A, which reading should have given you a
more accurate reading of the true air temperature: a, b, or c?
Problem 24.
Why would the other two readings not give you a more accurate reading of the
true air temperature?
B.
Now shade all of your thermometer again and allow it to reach equilibrium.
a.
Place the white wick over the bulb and expose the entire thermometer to the sun.
Allow it to reach equilibrium.
Record the reading here and on the answer sheet. __________
b.
Place the black wick over the bulb and expose the entire thermometer to the sun.
Allow it to reach equilibrium.
Record the reading here and on the answer sheet. __________
Answer the following questions. Place your answer on the answer sheet.
Problem 25.
Which of these different colors in Part III B was more efficient in absorbing
energy from the sun?
Considering the answers to the previous questions concerning radiation and the ability of
objects of different color to absorb radiation, answer the following questions about the proper
placement of thermometers for the purpose of accurately measuring air temperature.
Problem 26.
When mounting instruments to measure the temperature of the atmosphere for
meteorological purposes:
a. Should the instrument be shaded or exposed to the direct rays of the sun to
obtain an accurate measure of the temperature of the air?
55
b. If the instrument is to be shaded, what color should the cover be which
shades the instruments?
c. From the data you have collected to this point, explain why a person will
“feel” hotter when in the sunlight than when in the shade.
C.
Record the temperature at a height of about 2 - 3 inches over the following materials.
Shield the entire thermometer from direct sunlight from above, but not from below.
(Allow sunlight to reflect off the material onto the thermometer from below, but shield
the thermometer from receiving any direct sunlight from above.) Record your answers
below and on the answer sheet.
Reading over the black cloth.
__________
Reading over the aluminum foil.
__________
Answer the following questions. Put your answers on the answer sheet.
The data collected in part C concerns reflection of radiation from an object.
Problem 27.
One thermometer should indicate a higher temperature than the other. Explain
the particular characteristic the material (either the black cloth or aluminum
over which the thermometer was located) had in causing this thermometer to
indicate a higher temperature.
The amount of energy reflected from a surface is termed the albedo of the surface. The
greater the albedo of a surface, the more energy is reflected from the surface and thus, the less
energy is absorbed. Under the same conditions of sunlight intensity and angle at which the sun's
rays strike the surface, the surface with the higher albedo will reflect more energy, and absorb
less, and thus will be cooler; while the surface that has a lower albedo will reflect less and absorb
more energy and thus have a higher temperature.
Problem 28.
Part IV.
Which surface, black cloth or aluminum foil, has a higher albedo?
Record your answer on the answer sheet.
Equivalent Temperatures
It is important in meteorology that one be familiar with the different scales used to
represent temperature. There are three scales that are primarily used in meteorology. In the
United States, the Fahrenheit scale is most commonly used, and if someone were to say that the
temperature is 75 degrees outside, you would know that it is a comfortable temperature and short
sleeves would be appropriate to wear. But, if you were told it was 24 degrees, you would want
to put on a warm coat before going out. However, if you were in France, 24 degrees would
sound quite comfortable, since France uses the Celsius scale. The third scale is the Kelvin scale
and is used primarily for energy determinations. The temperature scale used in Wien's
Displacement equation is the Kelvin scale. Appendix A shows a method to convert between
these three scales.
56
Problem 29.
Part V.
Complete the table on the answer sheet by calculating the equivalent
temperature value for the scales shown. Indicate the temperatures to the nearest
hundredth of a degree.
Heating and Cooling Degree Units
Energy consumption, especially in the United States, has become a major concern due to
the gradual depletion of oil reserves and the increased demand for energy. The use of indicators
for determining the projected energy usage has been very useful. Two of these indicators are
heating-degree units and cooling degree units. When the outdoor air temperature falls below a
certain base value, indoor heating is required to bring room temperatures to a comfortable level.
When the outdoor air temperature rises above a certain base value, indoor cooling is required to
bring the indoor air temperature down to a comfortable level. Use of such data is especially
useful to the construction industry in planning insulation requirements and to energy companies
in planning fuel and electrical usage.
The base value for both heating and cooling degree units is 65oF. Since the outdoor
temperature changes through the day, the mean daily temperature is used. The mean daily
temperature is the average of the daily maximum and daily minimum temperatures. For long
term considerations, the monthly mean temperature or annual mean temperature may also be
used in comparison with the base value of 65oF. The degree units are then the difference
between the mean daily (or monthly mean or annual mean) temperature and 65oF. For example,
if the daily maximum temperature were 72oF, and the daily minimum temperature were 51oF,
then the daily mean temperature would be 61.5oF. The difference between 61.5oF and 65oF is
3.5 Fahrenheit degrees. Since the daily mean temperature is lower than the base of 65oF, then
people will turn on heating units to raise the temperature. Thus, the 3.5 Fahrenheit degrees are
called heating degree units. If the daily mean temperature was greater than 65oF, then the degree
units would be called cooling degree units because people would want to turn their air
conditioners on to cool their homes and businesses. If the daily mean temperature is equal to
65oF, then there are zero heating degree units and zero cooling degree units.
Problem 30.
The table on Part IV of the answer sheet gives the maximum and minimum
temperatures for several days. Determine the heating degree units or cooling
degree units for each day. Only one blank on each line will have a value greater
than zero. Place an X on the other blank.
July and August are normally the warmest months in Texas and thus the months when the
most amount of energy must be expended to cool homes, businesses, etc. January and February
are normally the coolest months and thus the months when the most amount of energy must be
expended to warm homes and businesses.
Problem 31.
The monthly mean temperature for Bryan/College Station for August is 84oF.
Determine the total cooling degree days for August based on this mean
temperature value.
57
Problem 32.
The monthly mean temperature for Bryan/College Station for January is 49oF.
Calculate the total heating degree days for January based on this mean
temperature value.
Below are figures showing annual heating and cooling degree days for Texas, derived
from nearly thirty years of temperature observations in Texas.
Fig. 1. ANNUAL. Heating degree-days. In the TransPecos area, values will exceed 3000 above 5000 ft. and
6000 above 6000 ft
Problem 33.
Fig. 2. ANNUAL. Cooling degree-days. In Trans-Pecos
area, values will generally be less than 1800; above 5000 ft.
values will be less than 1400; and above 6000 ft. less than
1000
Using the Annual Heating Degree Days figure and the Annual Cooling Degree
Days figure, estimate the expected heating degree days and the expected cooling
degree days for Brazos County, shown by being shaded. Write these values on
the answer sheet in the blanks provided for College Station.
Below, and on the answer sheet, are the Annual Heating Degree Days and
Annual Cooling Degree Days for several additional cities in the United States.
Assuming that in these cities, the cost of raising the temperature of a home by
1oF is the same as lowering the temperature of a home by 1oF. Upon
graduating from college, you decide that you would like to live in a location in
which you will spend the least amount of money on energy to heat or cool your
home. Based on the annual values, in which city would residents spend the
least amount of money on energy?
City
Orlando, FL
Denver, CO
New York, NY
Austin, TX
Dallas, TX
Sacramento, CA
College Station, TX
Annual Heating Degree Days
580
6128
4754
1648
2370
2666
________
58
Annual Cooling Degree Days
3428
696
1151
2974
2568
1248
________
Part VI.
Growing Degree Units
The concept of heating or cooling degree units can be used in another beneficial manner;
that is in the production of crops. Each species of plant has certain environmental requirements
for growth, from necessary nutrients, the right intensity of sunlight, the right wavelength of
sunlight, the proper amount of moisture occurring at the proper stage of maturing, to the proper
temperature. Plant growth toward maturity (Net Available Production - NAP) can occur only
when the gross energy production by the plant (GP) exceeds that energy required for respiration
(Rsa) and energy losses due to organisms which feed on the plant: insects, diseases, etc. and
losses due to breakage as when leaves, stems are broken due to strong winds, etc. (Rsh). We can
then write the following expression to show these relationships.
NAP = GP - (Rsa + Rsh).
Ideally, the agricultural production would result in maximum energy being utilized
toward production of that part of the plant which is the desired crop product: grain seeds, leaves,
fruit, or cotton, whichever is the desired part of the plant, not just making a bigger plant. Energy
that does not go toward making more crop product; such as, Rsh would be minimized or
eliminated and Rsh would, ideally, be as low as possible. Current computer models for various
types of agricultural crops include all the elements necessary to monitor and evaluate plant
growth; such as radiation, moisture, temperature, wind speed, etc.. By monitoring these
elements in the field, predictions concerning the date of maturity, date for spraying insecticides
and fungicides, when to irrigate, and when to fertilize can be made.
In the case of temperature, each species of plant has a certain base temperature below
which plant growth ceases. Photosynthesis may continue and energy is produced, but all the
energy goes into maintaining the plant. None is available for growth. Additionally, each species
of plant also has a maximum temperature above which growth is not increased, and above which
harm may occur to the plant. Within this boundary, the plant produces energy and grows,
assuming moisture, radiation, and plant food are available, and insects and diseases are minimal.
Each species of plant requires a certain period of time with these favorable conditions to reach
maturity, that stage when the crop is ready for harvest. The concept of Growing-Degree Units
aids in determining if, and when the plant has reached the stage of maturity. A good example is
cotton which is grown extensively in this area. Let us assume that we are growing a species of
cotton which requires 2000 GDUs to reach maturity. The base temperature for this species is
60oF and the maximum is 95oF. Below 60oF the cotton can only maintain itself and if the
temperature drops too low, the destruction to the cotton is greater than plant material production.
At 95oF, the plant is producing plant material as fast as it can. Any temperatures above 95oF do
not help the plant produce any more material.
Normally, to determine the GDUs for this plant the daily mean temperature for each day
is calculated. If the daily mean temperature is below 60oF, the GDUs for that day are set at zero.
If the mean is between 60oF and 95oF, then 60 is subtracted from the mean and the remainder is
the number of GDUs for that day. If the daily maximum temperature is over 95oF, the
maximum daily temperature used to calculate the daily mean temperature for that day is set at
59
95oF and the daily mean temperature is then the average of 95oF and whatever the daily
minimum temperature is for that day. The GDUs are summed cumulatively (i.e., a running
summation).
For the problems below, we are just going to use mean temperatures for the month, or
part of the month, rather than calculating the GDUs for each day. Thus, we will have one GUD
representing the total GDUs for that month or part of the month.
When the cumulative sum of GDUs reaches 2000, the cotton should be at maturity and
ready for harvest. Other elements play a role in this process also, such as moisture availability,
but a rough estimate of the maturity date can be made.
Problem 34.
The table on part V of the answer sheet provides the mean temperature for the
last 15 days of April, the mean temperature for the month of May, (31 days),
and the mean temperature for the first 15 days of June. Note: not all of the
growing season is included. For this exercise, we will use a cumulative level of
1000 GDUs for the time at which, for example, spraying for insects should be
done, and 2000 GDUs for the time of harvesting.
a. Determine the total GDUs for April, (15 days worth), for all of May and for 15
days in June.
What are the total GDUs for each month period; April, May, June?
b. Plot the cumulative GDUs for each day on the graph provided. Note: the plot
for each period should be at the end of each period.
Using the last 15 days of the graph as a guide, draw a straight line extending the
graph until a value of 2000 GDUs is reached. This is your estimation of how
the GDUs will accumulate for the remaining period until the plants reach
maturity.
Problem 35.
What is the date, month, and day on which spraying for insects should be done?
Problem 36.
For a normal season, would the rate of accumulation of GDUs per day remain
the same, increase, or decrease during the remainder of the growing season?
Consider what the daily maximum and minimum temperatures are doing during
this time of year.
Problem 37.
Estimate a date; month and day, on which cotton maturity should be complete
and harvesting could begin.
60
ANSWER SHEET
Name __________________________
Object
IRT DATA FOR LAB #3
Temperature from IRT (oC)
Data collected
Student
3 Jan. 1994
Data
Green portion of sign in sun
____28____
__________
Green portion of sign in shade
____2_____
__________
*Clear sky (not pointed at sun)
Too cold to register
__________
____29____
__________
White shirt/sweater
*Both in sun
Red shirt/sweater
____27____
__________
____30____
__________
White shirt/sweater
*Both in shade
Red shirt/sweater
____25____
__________
____26____
__________
*Pants leg, brown pants in sun
____26____
__________
*Pants leg, brown pants in shade
____24____
__________
Puddle in shade
____4____
__________
Puddle in sun
____9____
__________
Green grass in sun
____5____
__________
Green grass in shade
____-2____
__________
Dry grass in sun
____10____
__________
Tree - shaded side
____2____
__________
Tree - sunny side
____20____
__________
Large O&M window in shade
____5____
__________
Large O&M window in sun
____33____
__________
Dry dirt patch in sun
____10____
__________
*Hand
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Dry dirt patch in shade
____-2____
__________
Side of O&M building in sun
____18____
__________
Side of O&M building in shade
____8____
__________
Black mat outside O&M bldg. in sun
____19____
__________
Black mat outside O&M bldg. in shade
____6_____
__________
Brown car/truck metal in sun
____23____
__________
Black car/truck metal in sun
____35____
__________
White car/truck metal in sun
____13____
__________
Red car/truck metal in sun
____20____
__________
Black rubber tire on car/truck in sun
____30____
__________
____41____
__________
____31____
__________
Chrome bumper on car in sun
____-5____
__________
“Asphalt” in sun
____11____
__________
“Asphalt” in shade
____-1____
__________
Tinted glass on car in sun
Same car
Normal glass on car in sun
Part I. Problem 1. Energy emission by Mars. __________________J/(m2 s) or (W/ m2)
Problem 2. Wavelength of maximum amount of radiation from Mars. ______________
Problem 3. Energy emission by Venus.
___________________J/(m2 s) or (W/ m2 )
Problem 4. Wavelength of maximum amount of radiation from Venus. ______________
Problem 5. Estimated solar constant 2.3 billion years ago. ______________________
Problem 6. Estimated solar constant 700 million years ago. ______________________
Problem 7. Current solar constant for Mars.
_______________________
Problem 8. Years before Mar’s solar constant equals Earth’s solar constant. _________
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Part II. Problem 9.
Warmest Object
Temperature
Rank
__________________________________
______
___1___
__________________________________
______
___2___
__________________________________
______
___3___
__________________________________
______
___4___
__________________________________
______
___5___
__________________________________
______
___6___
__________________________________
______
___7___
__________________________________
______
___8___
__________________________________
______
___9___
__________________________________
______
___10___
Coldest Object
Temperature
Rank
__________________________________
______
___1___
__________________________________
______
___2___
__________________________________
______
___3___
__________________________________
______
___4___
__________________________________
______
___5___
__________________________________
______
___6___
__________________________________
______
___7___
__________________________________
______
___8___
__________________________________
______
___9___
__________________________________
______
___10___
Problem 10.
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Problem 11.
_____________________________
Problem 12.
_____________________________
Problem 13.
_____________________________
Problem 14.
______________________________
Problem 15.
______________________________
Problem 16.
______________________________
Problem 17.
______________________________
Problem 18.
______________________________
Problem 19.
______________________________
Problem 20.
______________________________
Problem 21.
______________________________
Problem 22. Explain why the clear sky shows a low temperature reading.
Part III. A.
Data collected from thermometers
a. Read the temperature in the shade.
Record the temperature.
__________
b. Read the temperature with just the bulb exposed to the sun.
Record the temperature here.
__________
c. Read the temperature in the sun, but shade just the bulb, leaving the stem exposed to
the sun.
Record the temperature here.
64
__________
Problem 23.
_______________________________
Problem 24. Explain why the other two readings would not give a more accurate reading of the
true air temperature.
Part III. B
Data collected from thermometers
a.
Place the white wick over the bulb and expose the entire thermometer to the sun.
Record the reading here.
b.
__________
Place the black wick over the bulb and expose the entire thermometer to the sun.
Record the reading here.
Problem 25.
Problem 26.
__________
______________________________
a.
______________________________
b.
_____________________________
c.
Explain why a person will “feel” hotter.
65
Part III. C.
Data collected from thermometers.
Reading over the black cloth.
Reading over the aluminum foil.
__________
__________
Problem 27. One of the thermometers should show a higher temperature value. Explain the
particular characteristic the material (either the black cloth or aluminum over
which the thermometer was located) had in causing this thermometer to indicate a
higher temperature than the other thermometer.
Problem 28.
______________________________
Part IV. Problem 29
Equivalent Temperature
Fahrenheit
Celsius
Kelvin
76.8
______
______
______
14.0
______
-12.0
______
______
______
______
222.0
______
-05.0
______
______
______
300.0
66
Part V.
Problem 30.
Heating/Cooling Degree Units
Maximum
Temperature
oF
Minimum
Temperature
oF
Heating
Degree
Units
Cooling
Degree
Units
92
68
_______
_______
78
54
_______
_______
48
29
_______
_______
86
75
_______
_______
29
06
_______
_______
Problem 31.
Total Cooling Degree Days for August.
______________________
Problem 32.
Total Heating Degree Days for January.
______________________
Problem 33.
Annual Heating Degree Days for Brazos County. _________________
Annual Cooling Degree Days for Brazos County. _________________
City with the least expenditure on energy. ______________________
Part VI. Problem 34a.
Period
16 April - 30 April
(15 days)
1 May - 31 May
(31 days)
1 June - 15 June
(15 days)
Growing-Degree Units for Cottom
Mean Temp.
Total GDUs
Cumulative
for Period
for Period
GDUs
o
67.3 F
72.5oF
81.6oF
Problem 34b. Graph of Growing Degree Units.
Problem 35. Date for spraying insects:
______________________________
Problem 36. Will the rate remain same, increase, decrease? _______________
Problem 37. Date of maturity to begin harvesting: __________________
67