Surface Pressure
Distance from the Sun
Temperature
Albedo
radius
atmospheric Mass
CO2
N2
O2
Argon
H20
Earth
101,325 Pa
149,598,261 km
287.2 K
0.37
6371
1
0.00039
0.78080
0.20950
0.00930
0.01000
Venus
9,200,000 Pa
108,208,000 km
735 K
0.67
6051.8
93
0.97
0.04
-
Force, pressure, energy, power, and intensity all mean the different
things. You may need a pretty firm grasp of the distinctions.
Pressure
Momentum
Force
Energy
Power
Intensity
SI Unit
Pascal
Newton
Joule
Watt
Another description
Newton per square meter
kilogram meter per second
kilogram meter per second per second
Newton · meter (or Pascal · cubic-meter)
Joule per second
Watts per square meter
My conjecture would be that Venus high temperature is due to recent
formation; due perhaps to a colliding moon, or tectonic plate motion,
(or both.)
Equations to use: Stefan-Boltzmann’s Law PA = σεT 4 P = Power.
Measured in Watts, or Joules per second, it is the energy produced
in a second. A = Area. In general, A = 4πR2 is the surface area of
a sphere of radius R. σ is the Greek letter ”sigma” used to represent
the Stefan Boltzmann Constant. For our purposes here, we’ll use
5.67×10−8 Watts per square meter per Kelvin to the fourth.
Disclaimer: I’ve made up this problem set myself, and I’m going to
work out the answers to the best of my current ability. I strongly
suspect that there are large conceptual errors. I’ll try to highlight
any claims that I find somewhat dubious. I hope to encourage people
with better ideas of how to approach these questions to reply.
1. The sun is considered to be an almost perfect “blackbody.” It
reflects no light off its surface, so its albedo = 0, and emissivity
ε = 1, Surface temperature T = 5778 Kelvin, Radius R = 696 ×
106 m. Use Stefan Boltzmann’s Law PA = σεT 4 to find the power
generated by the sun in Watts.
Answer: The sun has a total power output of 3.85×1026 Watts
1
2. (a) The earth is not a blackbody, but reflects much of the light
incident on its surface. The earth’s average temperature is 287.2
Kelvin, albedo is 0.37, radius = 6371 km. Use Stefan Boltzmann’s Law to find the thermal energy radiated from the earth
in Watts. (b) Venus’ average temperature is 735 K, with albedo
0.67, and radius 6051.8 km. Use Stefan Boltzmann’s Law to find
the thermal energy radiated from Venus in Watts.
3. The sun generates a power output of 3.85×1026 Watts. A portion
of this energy lands on Earth’s surface, with r = 6371 km, 149.6
million km away, and a portion lands on Venus at 108.2 million
km away. Find the amount of radiant energy falling on the Earth
and on Venus from the sun in Watts.
4. Compare the amount of radiant energy incident vs thermal energy emanated, for Venus and Earth.
5. Another application for the Stefan Boltzmann equation I often
see used is to find out the “radiant temperature” at long distances from the sun. (a) Find this radiant temperature based on
the intensity of the light from the sun at 108.2 million km and
149.6 million km, and compare these to the actual temperatures
of Venus and Earth. (b) Should the Second Law of Thermodynamics, “Heat will not spontaneously flow from a cooler body
to a hotter body” apply in these cases? Is the vacuum of space
filled with radiant energy a “body?” (c) What are possible explanations to explain the results? Are Carbon Dioxide and “the
greenhouse effect” the major explanation for Venus extreme temperature?
6. Discuss the differences between emission, absorption, and specular reflection, and diffuse reflection. (b) If you see an emission
spectrum coming off the corona of the sun during a solar eclipse,
are you really seeing an emission spectrum, or are you seeing
specular reflection? (c) In a greenhouse gas, does it make a difference whether a gas actually absorbs the light, or reflects it?
Can our empirical observations of the behavior of the gas tell
the difference between reflection vs. absorption and re-emission?
(d) How is infrared absorption energy converted into thermal
energy?
7. (Red Herring) Find the intensity of the light at the sun’s surface.1
8. Given the total power coming out of the sun, and Mercury’s
distance of 108.2 billion meters, find the intensity of the sun’s
radiation that reaches mercury 2
9. Given the total power coming out of the sun, and Earth’s distance of 149.6 billion meters, find the intensity of the sun’s radiation that reaches Earth.
1
2
P/A = σT 4 = 63.2 Mega Watts per square meter
P
P/A = 4πR
2
2
10. Discuss the second law of thermodynamics as it relates to radiation. According to the second law of thermodynamics heat will
not flow spontaneously from a region of lower temperature into
a region of higher temperature. Is “temperature” a well-defined
property of the vacuum of space, and does it vary according to
the Stefan Boltzmann’s Law?
11. Assuming the albedo of earth is 0.37, what is it’s emissivity
3
12. Should the thermal radiation of earth be calculated using the
temperature at the some outer surface of the atmosphere, or by
the temperature at the ground.
3
I’m pretty sure, these are related by the simple relationship ε = 1 − albedo
3