SPU 25 Energy and Climate: Vision for the Future, Spring 2018
Homework Set #1 Climate Basics, DUE Monday 2/12
(1:00 pm in lecture)
Name: _____________________________________ Section: __________ TF: _____________________ Score: ______/_______
Collaborators: _________________________________________________________________________________________________
Earth’s climate is a complex reflection of the distribution and movement of energy around the planet. The
primary source of Earth’s energy is radiation from the Sun. In this problem set, you will see how this energy is
received, lost, and stored, and the role of greenhouse gases in altering the climate. You will also understand
how the distribution of energy influences climate and get some practice with basic quantitative problems and
unit conversions. (
40 points total)
You will only receive credit if you show all your work and write legibly.
I. Earth’s Energy Balance
Planet Earth (including the solid earth, oceans, and atmosphere) is best described as an isolated system:
energy, but not mass, can pass from Earth to space and vice versa. In this type of system, the energy balance
may be described by:
Energy gained = Energy lost – change in storage.
1.
Greenhouse gases in the atmosphere prevent the radiation of energy to space, so they impact which
term in the above equation?
(1 point)
Energy lost
2.
Changing surface and ocean temperatures reflect which term in the above equation?
(1 point)
Change in storage
II. Energy from the Sun
3.
True or
False
:
Summer is warmer than winter because the Earth is closer to the Sun.
(1 point)
4.
Currently, the energy from the Sun arriving at Earth’s orbital position amounts to 1,379 W m
-2
,
measured perpendicular to the direction of the sunlight (see figure below).
a.
Given that
the Earth is
a circular
target for
this energy
and has a
radius of
6.38 x 10
3
km, calculate the total rate at which
solar energy is intercepted by the Earth. Report your answer in Watts. Recall that the area of a
circle is
r
π
2
.
(2 points)
1
Figure 1. The shadow formed by an
illuminated sphere has a circular
shape with a radius R equal to the
radius of the sphere.

A = 1.28*10
14
m
2
I = 1.76*10
17
W
We calculate the cross-sectional area of the earth the multiply by the rate at which the sun is radiating energy
to earth.
(1 pt) Cross-sectional area of earth
=
r
π
2
=
π
* (6.38 * 10
3
* 10
3
m)
2
=
1.27877 * 10
14
m
2
(1 pt) Solar energy intercepted
=
Cross-sectional area * rate at which the sun is radiating energy
to earth
=
1.27877 * 10
14
m
2
* 1379 W/m
2
=
1.76342 * 10
17
W
The total rate at which solar energy is intercepted by earth is
1.76 * 10
17
W.
b.
Assuming that the energy from the Sun is eventually distributed equally over the entire surface
area of the Earth, use your answer from part (a) to calculate the current average rate of
incoming solar radiation. Express your answer in Watts per meter squared. Recall that the
surface area of a sphere is 4
r
π
2
.
(2 points)
I
0
= 1379 W.m
-2
Divide by 4 to get
I
S
= 344.8 W.m
-2
We take the answer from part a and divide it by the surface area of earth (a sphere)
(1 pt) Surface area of earth
=
4
r
π
2
=
4
π
(6.38 * 10
3
* 10
3
m)
2
=
5.11507 * 10
14
m
2

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