EPS 5 Spring 2010
Problem Set 1
Due: Wed, Feb 3, 2010 (at the beginning of lecture)
1. Pressure in the International Space Station (8 points)
In January 2004, NASA reported that the International Space Station was experiencing a leak
(http://www.clickorlando.com/technology/2742496/detail.html, units have been changed to SI).
The shuttle, which has a volume of roughly 7.0×105 m3 and is usually pressurized to
1.01352×105 Pa, had lost 1200 Pa over the course of 5 days.
(a) Assuming the temperature remained constant at 20°C, how many air molecules were lost
from the space station during the leak? (2 points)
We use the perfect gas law, PV = NkT. We can rearrange to have N = PV/kT. Since
temperature isn’t changing and the volume of the space station also cannot change, the
initial number of molecules can be written N1 = P1V/kT and the final number N2 = P2V/kT.
The change in the number of molecules is therefore ΔN = (V/kT)*ΔP. Plugging in the
numbers, and remembering to convert from °C to °K, we get ΔN = 2.1×1029 molecules.
(b) The “air” in the space station is a mix of 78% nitrogen (N2), 21% oxygen (O2), and 1% water
vapor (H2O). What is the average molar mass (g/mol) of space station air? (2 points)
First we determine the molar mass of each individual component: (14 g/mol)*2 = 28 g/mol
for nitrogen, (16 g/mol)*2 = 32 g/mol for oxygen, and (1 g/mol)*2 + (16 g/mol) = 18 g/mol
for water vapor. Then we add them according to their percent contributions:
m=0.78*28 + 0.21*32 + 0.01*18 = 28.7 g/mol.
(c) Using your answers from parts (a) and (b), calculate the mass of air molecules that was lost.
Give your final answer in kilograms of air. (2 points)
There are 6.02×1023 molecules in one mole, so using our answer from part (a), we find that
the number of moles lost was (2.1×1029 molecules)*(1 mole / 6.02×1023 molecules) = 3.4×105
moles. We can now use our calculate molar mass from part (b):
(3.4×105 moles)*(28.7 g/mol) = 9.9×106 g = 9.9×103 kg
(d) What percent mass was lost? Do you think the astronauts were worried? (2 points)
We can use the same calculations to figure out the mass of air originally present:
M1 = N1*m*(1/N0) = (P1*V*m) / (k*T*N0)
M1 = (1.01352×105 Pa)*(7.0×105 m3)*(28.7 g/mol) / ((1.38×1023 J/K)*(293.15 K)*( 6.02×1023
molecules)) = 8.4×108 g = 8.4×105 kg
So, the percent lost is 100*(ΔM/M) = 100*(9.9×103 kg / 8.4×105 kg) = 1.2% - not a lot to
worry about (but more than I expected)!
2. Solar Variability and Global Energy Balance (9 points)
Temporal variation in solar activity (and thus the total emission of solar radiation) is found to be
one of the major causes of pre-industrial climate change, causing the so-called Little Ice Age in
the latter half of the 17th century. Some scientists have proposed that the recent global warming
phenomenon could be a result of solar variability. Let us evaluate this claim based on our
understanding of global energy balance covered in class.
(a) Measurements since 1979 have indicated that the Sun’s surface “radiating” temperature
varies between 5765 K and 5768 K over an 11-year cycle, due to variations in sunspot activity.
Calculate the solar constants corresponding to the peak and trough of the solar cycle. Take the
Sun’s radius to be 6.955×105 km, and the average distance between the Earth and the Sun to be
1.490×108 km. The Stefan-Boltzmann constant σ is 5.670×10-8 W m-2 K-4. (3 points)
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Solar constant F is calculated by:
σ TS4 ⋅ 4 π RS4 σ TS4 RS4
(1 points)
F=
=
4 π dE4
dE4
where TS is the solar surface “radiating” temperature, RS = 6.955×108 m is the solar radius,
and dE = 1.490×1011 m is the average distance between the Earth and the Sun. Substituting
values, the solar constants are
For TS = 5765 K:
F = 1365 W m-2
(1 point)
-2
For TS = 5768 K:
F = 1367 W m
(1 point)
(b) Assume the Earth has a one-layer blackbody atmosphere, as we discussed in class. Find the
Earth’s mean ground temperatures corresponding to the peak and trough of the solar cycle.
Assume the Earth has an albedo of 0.33. (4 points)
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The Earth’s effective temperature is calculated by:
 F (1 − A) 1 4
(1 point)
Teff = 

 4σ 
For a one-layer blackbody atmosphere, the mean ground temperature is related to Teff by:
Tg = 21 4 Teff
(1 point)
Substituting values,
For F = 1365 W m-2:
Tg = 299.7 K
(1 point)
For F = 1367 W m-2:
Tg = 299.8 K
(1 point)
(c) The Earth’s global mean surface temperature has risen by > 0.5 K since 1980. Would solar
variability be responsible for the recent global warming phenomenon? Explain. (2 points)
No. (1 point) Solar variability can explain at most a 0.1 K change in the Earth’s surface
temperature over the past few decades, which is only a small fluctuation compared to the
actual temperature trend. The overall increase in recorded global surface temperature
must have been caused by other factors, including anthropogenic greenhouse gas emissions.
(1 point)
3. Simplified Feedback Loops (8 points)
Imagine a very simple planet that has no atmosphere but magically sustains one species of white
flower.1 Temperature is maintained by energy from the planet’s sun, and flower growth is
dependent only on temperature (higher temperatures mean more flowers can grow). Answer the
following questions about the planet.
(a) Would you expect the albedo of the bare ground to be higher or lower than the albedo of
flower-covered areas? How would that affect the local temperature? (2 points)
The albedo would be higher for the flowers (white) than for the bare ground. The flowercovered areas therefore reflect more solar radiation, leading to a decrease in local
temperature.
(b) If the sun suddenly brightened, leading to a sharp increase in temperature, how would the
flower population respond? How would the temperature respond to the change in flower
population? Is this a positive or negative feedback loop? (3 points)
The increased temperature would cause an increase in flower population, causing more
areas to be covered by flowers. As these areas change from bare ground to white flowers,
the albedo would increase. The result would be a decrease in temperature (and subsequent
decrease in flower population, etc.). This is therefore a negative feedback loop.
(c) How would your answers to parts (a) and (b) change if the flowers were black instead of
white? (3 points)
If the flowers were black, they would have a lower albedo than the bare ground, so their
growth would lead to increases in local temperature. In this case, a sudden increase in
temperature would lead to increased flower growth, decreased albedo, and subsequently
increased temperature. This cycle would continue unabated, making this a positive
feedback loop.
4. Greenhouse Gases and Feedback (10 points)
Listed below are the effective and ground temperatures for Mercury, Venus, Earth and Mars:
Planet
Teff (K)
Tg (K)
Mercury
442
442
Venus
227
750
Earth
253
288
Mars
216
240
(a) In class, we mentioned that the one-layer blackbody atmospheric model could be generalized
to any number n of layers, with the ground temperature related to the effective temperature by:
14
Tg = ( n +1) Teff
Find the effective number of blackbody layers n for the 4 planets. Contrast and comment on your
answers for Mercury and Venus. (3 points)
1
Idea
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courtesy of the Science Education Resource Center at Carleton College.
Substituting the corresponding values for Tg and Teff :
For Mercury:
n=0
(0.5 point)
For Venus:
n = 118
(0.5 point)
For Earth:
n = 0.679
(0.5 point)
For Mars:
n = 0.524
(0.5 point)
Mercury has no atmosphere (therefore n = 0), so Tg = Teff. In contrast, Venus has a very
thick atmosphere heavily composed of greenhouse gases (96.5% CO2!), so n is large,
representing the strong greenhouse effect and causing Tg to be much higher than Teff. (1
point)
(b) You may have noticed that n < 1 for the Earth. If n = 1 as assumed by the one-layer model,
the Earth’s surface will be warmer than it actually is. Explain why n is smaller than one for the
Earth. Hint: How valid is it to treat the Earth’s atmosphere as a blackbody? Is all radiation
emitted by the Earth’s surface absorbed by the atmosphere? (1 point)
In fact, the Earth’s atmosphere is not a perfect blackbody. Rather, it is a “leaky”
greenhouse, or a “greybody”, which means that only a fraction of the radiation emitted by
the Earth’s surface is absorbed by the atmosphere while the rest passes directly into outer
space. Also, the atmosphere emits only a fraction of its blackbody radiation. The overall
effect is that less radiation is reemitted back to the Earth’s surface, causing the mean
surface temperature to be lower than if the atmosphere were a perfect blackbody. (1 point)
(c) Suppose now we have a hypothetical planet at Earth's orbit with n = 0.75 and Teff = 250K.
What is Tg, the ground temperature? (1 point)
Tg = (0.75 + 1)1/4 * 250 K = 287.5 K
(1 point)
(d) If we add greenhouse gases sufficiently to increase n by 0.05 units, how much will Tg
increase? (1 point)
Now n = 0.75 + 0.05 = 0.80:
Tg = (0.80 + 1)1/4 * 250 K = 289.6 K, an increase of ~1 K.
(1 point)
(e) If this Tg increase leads to an additional increase of n by 1/2 as much as the initial addition of
greenhouse gases did (e.g. as a result of adding to the water vapor content of the atmosphere),
how much additionally would Tg change? Is this a positive or negative feedback? (2 points)
Now n = 0.75 + 0.05 + 0.025 = 0.825:
Tg = (0.825 + 1)1/4 * 250 K = 290.6 K, an additional 1K increase.
(1 point)
The initial increase in Tg will cause a further increase in Tg, and therefore this is a positive
feedback. (1 point)
(f) Can you guess the final temperature change if the atmosphere keeps adjusting in this way?
(Hint: 1 + 1/2 + 1/4 + 1/8 + … = 2) (2 points)
If the atmosphere keeps adjusting in this way,
n = 0.75 + 0.05*(1 + 1/2 + 1/4 + 1/8 + …) = 0.75 + 0.05*2 = 0.85 (1 point)
Tg = (0.85 + 1)1/4 * 250 K = 291.6 K
Feedback will induce a total temperature increase of (291.6 – 287.5) K ≈ 4 K. (1 point)