# Total solar power received by Earth and The calculation

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```							ENV 312 Energy & the Environment                                               Fall 2008

Final Exam Problem Review Question

Total solar power received by Earth and
The calculation of the average temperature of the Earth
PART 1 (similar to quiz 3) – Calculate total solar power received by earth
First, recall the difference between energy and power. Power is the energy per time, or the rate
of energy striking the earth. Energy is measured in BTU, Joules, or kWhours, while Power is
measured in Watts, kilowatts, or BTU/hr.

The solar constant is defined as the total radiant power of the Sun, per unit of area, received at
the Earth's orbit (outside of the Earth's atmosphere).

The Earth' shadow is a circular disk.
Treat the illuminated face of the Earth as
if it were a disk, when calculating the total
area of sunlight falling on the Earth's
surface.

Using the radius of the Earth, and the
solar constant, calculate the TOTAL
radiant power incident on the Earth's
sunlight surface in Watts (W). You merely
multiply the solar constant times the area
that it illuminates.

Pin, total = Sc x Ashadow         Sc = Solar Constant = 1,370 W     Shadow area of a disk

R = Radius of the Earth           π = 3.14
= 6.37 x106 meters

Hint: this is a simple unit conversion and geometry problem. For partial credit points, please

solution for earth:                                    Please note that the units are in [brackets]

⎡W⎤
(               )
Powerin, total = 1,370 ⎢ 2 ⎥ ⋅ π 6.37 × 10 6 [m ] = 1.75 × 1017 [Watts ]
2

⎣m ⎦
ENV 312 Energy & the Environment                                                        Fall 2008

PART 2 (similar to quiz 7) – Calculate the average temperature of the earth
Using the answer from above, the total energy received by the Earth from the sun, find the
average temperature of the Earth with no atmosphere.

Use Pin, total from Part 1, above.

Pin,total x (1 - albedo) = Eout                   R = Radius of the Earth       Surface area of a sphere
= 6.37 x106 meters          (the radiating surface of
and                                                    earth)
Asurface = 4 π R2 (m2)
Radiative Power, Pout = σ T4 A

Average earth’s albedo, α = 31%                   σ = Stephan Boltzmann         π = 3.14
(or 31% is reflected away to space                constant = 5.67 x 10-8 W
and 69% is absorbed)                                                   m2 K4

Find T in the equation above
(and remember that you use the
absolute scale, or degrees Kelvin):

Pin ,total (1 − α )      Pin ,total (1 − α )
T=      4                           =
σ × Asurface             σ × Asurface

solution for earth:

1.75 × 1017 [W ] ⋅ (1 − 0.31)
T   [ K] =
O
= 254.2 O K        [ ]
-8 ⎡ W ⋅ K ⎤
(            )
4
× 4π 6.37 × 106 [m ]
2
5.67 × 10 ⎢       2   ⎥
⎣ m ⎦

[ ]
254.2 K − 273 = −18.8 o C
O
[ ]

```