The Solar Input: A Limit to Growth
Purpose
In spite of all our technical progress, we are still basically living creatures completely dependent
upon what the plant kingdom produces. Similarly, the plants are living materials and are
completely dependent upon solar radiation for their existence. The amount of solar radiation the
Earth receives is finite because the Earth has definite boundaries. Since the Earth can only
receive so much radiation (unless we intercept more via satellites, but that is very unlikely) there
is an upper limit to the amount of plant material that can efficiently be raised. Hence there is an
upper limit to our human population. The purpose of this lab is to explore this limit. Note: Clever
people can grow more food on a given plot of land than less clever people.
Determination of the Solar Input
A. Materials (per lab group)
1 - styrofoam tray approximately 7" x 9" and 1" deep (used at supermarkets to put
chicken parts in) -- should easily hold 600 ml. of water. The inside of the tray is
spray painted dull black.
1 - small thermometer (Celsius) that will fit in the tray.
1 - 1000 ml. graduated cylinder or container that has the 600 ml. level marked on it.
1 - sheet of glass to cover the tray. Carry the glass carefully to and from the lab site.
1 - watch or timing device.
2 - meter sticks to determine the sun's altitude (one to cast a shadow and one to measure it with).
B. Procedure for Gathering Data
NOTE: This experiment can only be done on a clear day and/or during a 20 min. period when no
clouds block out any of the sun's rays.
Gather the materials listed above. Measure the inside length and width of your styrofoam
tray in centimeters. Record all data. Place 600 ml. of water in a large graduated cylinder or other
container provided. The water should be 5 to 10 degrees Celsius colder than the ambient
(surrounding outside) air temperature. Place the styrofoam tray on the ground on a spot that is
level. It must receive direct sunlight and little or no wind. Do not place the tray next to a building
or other large object as it will reflect sunlight onto your tray. You want to receive only sunlight
that comes directly from the sun. After selecting your location, measure and record the
temperature of the ambient (surrounding) air. Your teacher may choose to do this for you. Next,
pour the water from its container into the styrofoam tray. Knowing the volume of the water,
determine its mass in grams (1 ml H2O = 1 gram). Place the small thermometer on the bottom of
the tray. Completely submerge it in the water. Position it so you can read it easily without
disturbing it. Cover the tray with a pane of clear glass. Record the temperature of the water once
a minute for twenty minutes. If the glass fogs up, you may shift the glass around a little to read
the thermometer, but don't remove the glass from the tray. If wind comes up as you are doing the
lab, protect your set-up as best you can. Do not however block out sunlight as you shield the setup from wind and as you read the thermometer. After about ten minutes have passed, have
someone mount a meter stick vertically (using a plumb line) on level ground and measure the
length of its shadow. Record this information on your data sheet. After taking your last
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temperature reading, pour out the water. Gather all your lab materials and return to your
classroom.
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C. Data
Location:__________________________
Date:_____________ Time:_______________
Weight of water = 600 ml = ____________ grams.
Area of pan =________________ cm x ________________ cm =________________ cm2.
Sun's Elevation Angle:
tan 



sin  


length of stick =__________cm = ___________
length of shadow
cm


Ambient (surrounding) air temperature = _________________C
Sample graph
Temp. vs. time
Time (min.)
Slope = 6 C/ 10 min. = 0.6 C/ min.
Note: Determine the slope of that segment of the graph that is near the ambient air temp. This is
necessary to eliminate the effects of heat gained by the water from its surroundings when it is
below ambient temperature; and the heat lost by the water to its surroundings when it is above
ambient temp. What we want to determine is only the heat gained by the water due to the sun's
energy.
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D. Calculating the Solar Input
Obtain a sheet of graph paper. Plot a graph of temperature vs. time (see sample provided).
Determine the slope of your graph near the ambient temperature. Find the slope in C/min. The
slope of the graph at the ambient temperature will give the rate of temperature rise nearly
independent of heat losses. You are now ready to solve for the value of the solar input.
The solar input is the amount of power a given cross-sectional area receives from the sun
when the sun is directly over that cross section. Hence, the method of this experiment is to
absorb a known cross section of sunlight and compute the heat absorbed from the rise in
temperature of the water. The glass and the water are transparent to the incoming sunlight.
However, the black paint absorbs essentially all of the sunlight that strikes it and then conducts
most of this heat energy to the water (conduction-convection). A very small fraction is radiated
back into space, but this fraction is so small it is negligible in this experiment. To get the crosssectional area of the styrofoam tray perpendicular to the sun's rays, multiply the area of the tray
by the sine of the sun's elevation angle (the sun's altitude angle). The sine corrects for the sun's
angle. See the Appendix at the end of this lesson for a more complete explanation of this.
Sun elevation
angle = 

In this experiment, we assume that the styrofoam tray is a perfect insulator, i.e., it gains
no heat from its surroundings nor loses any heat to them. This assumption is quite accurate.
Obviously, any heat added to or taken away from the water by its surroundings will give errors.
These errors are minimized by determining the rate of temperature rise when the water and the
surrounding air are at the same temperature.
Computation of the solar input consists of solving the equation which states that the heat
gained by the system is equal to the energy received from the sun. The key conversion factor is
that 1 watt equals 0.24 calories per second.
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Key Equation:
Energy RECEIVED from sun/sec. = Heat GAINED by water/sec.
The amount of energy received from the sun/sec, depends on three factors:
1. the area of the tray
2. the sun's elevation angle
3. the strength of the sun as a power source (the solar input).
The amount of heat gained by the water/sec, depends on two factors:
1. the mass of the water
2. the temperature rise/sec, (the slope of your graph).
Hence, our equation is:
Unit analysis:
Symbols of the Quantities
0.24 x A x sin  x s = m x R x 1/60 x 1.00
where
0.24
A

s
= calories per watt * second
= area of tray in square centimeters
= angle of elevation of the sun
= power density of the sun in watts per square centimeter perpendicular to
the sun's rays. The solar input.
m = mass of the water in grams
R = temperature rise rate in degrees Celsius per minute (slope of graph)
1.00 = the definition of the calorie. One calorie is the amount of heat required
to raise the temperature of one gram of water one Celsius degree.
1/60 = factor for changing minutes to seconds.
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Notes:
1. The cover glass is used to prevent large losses of heat from evaporation, but the glass also
causes some losses due to reflection. In most cases the reflective losses are small. See your
teacher for information on reflective losses.
2. This experiment must be run on a clear day. The effect of clouds on plant growth is taken into
account when the % of sunlight that goes into photosynthesis enters the experiment.
Your Calculation
Now that the equation has been determined, solve for the value of the solar input in watt/cm2 and
watt/ m2.
The accepted value of the solar input is approximately 1000 watt/m2 on a clear day at sea
level. Above the Earth's atmosphere (where there is no solar energy reflected away), the solar
input is 1353 watt/m2 2%. How much power is 1000 watt/m2? Well, 1000 watt/m2 equals
approximately 2,600,000 kilowatts per square mile or 2,600 megawatts per square mile. Today, a
1,000 megawatt electric power generating plant is considered a large plant. Hence, a square mile
receives more than enough solar power to operate a very large electric power generating plant
when the sun is directly overhead. This assumes that energy can be efficiently trapped and used.
A generating plant of this size could provide for the electrical needs of several hundred thousand
people.
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Calculating the Earth's Carrying Capacity
Note:
For all of the following determinations, use the sea level value of 1000 watt/m2 as the solar input.
Most of the world's food-growing regions are quite close to sea level.
A. Determination of the Total Solar Power Received by the Earth
PTotal
Given: Radius of the Earth = 6.38 x 106 meters
= Solar INPUT x Cross Sectional Area of the Earth.
PTotal = _____________________________________________________ watts.
B. Power Trapped in the Photosynthesis Process
In the photosynthesis process, plants receive the energy they need to carry on their functions and
store energy in their structures. It has been determined that approximately 0.0757% of the solar
energy that reaches the Earth is captured by the chlorophyll of plants and fixed in the
photosynthesis process. How many watts of potential plant power are fixed in the photosynthesis
process?
(Remember: 1% = 0.01, so 0.0757% = 0.000757)
Pfixed = _____________________________________________________ watts.
This power is involved in the production of all the organic plant matter of the Earth -- seaweed,
phytoplankton, weeds, grass, trees, crops, etc. (Roughly 32% in oceans, 68% on land.)
C. Total Amount of Food Available to Humans
George Borgstrom, professor of Food Science and Geography at Michigan State University and
author of the informative and comprehensive book. The Hungry Planet, states that close to 10%
of the total organic mass which through the sun's energy is mass produced in the cells of green
plants through photosynthesis is taken by humans as food. How many watts of power is this?
Pfood = _______________ watts.
Knowing that 1 watt = 0.24 cal/sec., how many calories of food are produced on the Earth/day?
Remember, the sun shines somewhere on Earth 24 hours/day.
Amount of food/day = _____________ cal/day.
D. A Ball-park (probably infield) Figure for the Maximum Population of the Earth
Fact #1:
Contemporary expert opinion says that the normal food requirement of a fullgrown person of 154 pounds (70 kilograms) weight is approximately 3,000 dietic Calories per
day -- recognizing that some is never used up by the body.
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Fact #2:
Scientists distinguish between several different kinds of calories when they are
dealing with energy and food value. The following should make clear all of these distinctions:
1. a gram calorie is the kind that raises 1 gram of water 1 C.
2. a kilocalorie (or Calorie) raises 1000 gm. of water 1 C.
3. primary Calories are a measure of the energy locked in plant material.
4. dietic Calories are a measure of the energy locked in food (plant and animal origin).
5. the mathematical relationship between these different types of calories is:
1 primary Calorie = 1 dietic Calorie = 1 kilocalorie
= 1000 gram calories.
Fact #3:
For every calorie of animal foodstuffs produced, five to eight calories are
required, represented by plant products and other feeding -- stuffs that the animals need for their
maintenance as well as for their food-producing activities. On such a simplified basis, the
number of calories required to feed the various categories of livestock, in terms of primary
Calories, can be computed. Since almost all humans eat animal products as well as plants, their
intake includes the primary Calories (plant Calories) needed to produce the animal products
consumed. In this light, each American has a daily food intake of approximately 3,240 dietic
Calories (plant and animal). However, this figure rises to 10,870 (approximately 11,000) primary
Calories/day when the conversions of plant to animal products is considered (Borgstrom, p. 32).
Fact #4:
Primary Calories, in the form of plant production, must be the type considered
when computing an upper limit for the world's population based on world food producing
capability.
Fact#5:
Recognizing only the fuel value of our food is an error and greatly misleading. In
our food, we need protein -- the living substrate of the cell's protoplasm -- and in addition our
protein intake has to satisfy very narrow specifications as to molecular structure. We,
furthermore, require a number of vitamins and minerals, special fats and, it would appear, certain
specified carbohydrates. The proteins, however, are key compounds. It is more than a
coincidence that, during recent decades, protein deficiency diseases have come to prevail in most
continents and must be regarded as the chief nutritional deficiency of the world (Borgstrom, p.
30).
If we use the American food intake as the standard for a well-fed world, how many people can
the Earth support?
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Questions to Think About
1. What are the possibilities of increasing our use of all photosynthesis products as food above
the 10% figure? Keep in mind that we need paper, lumber, forests and meadows and a stable
ecosystem. Also, it takes more than direct solar energy and compost to grow food as we do it in
the United States. In the U.S. we currently use the equivalent of 80 gallons of gasoline to
produce an acre of corn.
2. World population is now over 6 billion and has a doubling time of approximately 40 years.
What are the implications of this in light of what you found in this lab?
Reference
Borgstrom, George, The Hungry Planet, New York, Collier Books, 1972.
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