b) Would the average albedo of the Earth increase or decrease? Why? (2 points) The average albedo of the Earth would decrease.

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1 EPS 5 Spring 2010 Problem Set 2 Due Wed, Feb. 17, 2010 (at the beginning of lecture) 1. Climate change and albedo feedbacks (16 points) a) Although 70% of the Earth's surface is covered by ocean, Earth's average albedo is 0.33 (Approximately 33% of the incoming solar energy is reflected back to space). Using the table below, explain why the average albedo of the Earth is so high. (2 points) Surface cover Albedo Fresh snow Clouds Ice 0.35 Desert Grassland Ocean 0.05 Forest 0.1 While 70% of the Earth's surface is covered by ocean (with a low albedo of 0.05), 30% of the Earth's surface is covered by land surfaces that all have a higher albedo. Given the known distribution of land and ocean, the Earth's albedo would never reach 0.33 without the influence of clouds. Clouds are very reflective, and they prevent a significant portion of solar radiation from ever reaching the surface of the Earth. Now consider the effect of snow cover on the Earth s albedo. Suppose that in the next twenty years, snow cover retreats by 5%. b) Would the average albedo of the Earth increase or decrease? Why? (2 points) The average albedo of the Earth would decrease. Land covered with snow has a high albedo. Replacing snow cover with land surfaces that have a lower albedo would decrease the average albedo of the Earth. c) If all other factors were held constant, what would happen to the average temperature? Why? (2 points) The temperature of the Earth would increase. Less of the energy from the sun is reflected, more is absorbed and reemitted by the ground. d) Based on your answer in c), do you think the snowmelt will continue? Explain why. Is this a positive or a negative feedback, or neither? (2 points) The snowmelt would continue this is an example of a positive feedback. Earth warms -> snow cover retreats -> albedo decreases -> Earth warms some more -> snow cover retreats some more -> albedo continues to decrease -> etc. e) In 1980, the area of sea ice in the Arctic at the annual minimum was around 7 million square kilometers. In 2007, the minimum area was around 4 million square kilometers. Consider the Arctic

2 Ocean to be a 10 million square kilometer box composed of sea ice and ocean. What was the albedo at the minimum sea ice extent in 1980? What was it in 2007? (2 points) 1980: A = 0.7* *0.05 = : A = 0.4* *0.05 = 0.17 f) Following up to part e), consider the one-layer greenhouse model from the previous class, just for the Arctic Ocean box. Do not consider clouds. Use a solar constant appropriate for Arctic summer, F s = 2000 W/m 2 and number of atmospheric layers n = 1. How much does the ground temperature in this simple energy balance change between 1980 and 2007? How much does the effective temperature change over the same period? (5 points) Energy balances (2 points) (1) Whole system: Fs(1-A)/4 = σt1 4 (2) Ground: Fs(1-A)/4 + σt1 4 = σtg 4 (3) Atmosphere: σtg 4 = 2σT : T1 = [Fs(1-A)/4σ] 1/4 = [2000(1-0.26)/(4*5.67x10-8 )] 1/4 = K Tg = (2T1 4 ) 1/4 = K 2007: T1 = [Fs(1-A)/4σ] 1/4 = [2000(1-0.17)/(4*5.67x10-8 )] 1/4 = K Tg = (2T1 4 ) 1/4 = K Tg = 9.8 K Te = 8.3 K g) The ground temperatures you calculated for the simplified model in part f) are much higher than actual ground temperatures in the Arctic. What are some reasons why this model may be unrealistic in this scenario? The global average for n is around 0.7, and it is likely lower in the Arctic. We only considered ocean and ice, while land is also present at high latitudes and would have a higher albedo than ocean. There is considerable heat capacity in the Arctic Ocean. There is significant transport of heat away. 2. Short questions on feedback (3 points) a) As the concentration of CO 2 increases in the atmosphere, some plants may be able to grow more and take up more CO 2 for photosynthesis and storage as organic matter in their tissues. Is this a positive or negative feedback for atmospheric CO 2 concentrations? Negative feedback. Higher CO2 More uptake Lower CO2. b) As the concentration of CO 2 increases in the atmosphere, the ocean will become more acidic and be able to take up a smaller fraction of atmospheric CO 2 in the future. (We will learn more about this when we study the carbon cycle) Is this a positive or a negative feedback for atmospheric CO 2 concentrations?

3 Positive feedback. Higher CO2 Less uptake Higher CO2 Less uptake. c) As the climate warms, there may be more forest fires, which emit black soot to the atmosphere. Is this a positive or a negative feedback for climate change? Positive feedback. Warmer More fires More black soot Warmer More fires. 3. Scale heights (11 points) a) On Mars, the atmosphere is 95.72% carbon dioxide (you may assume all CO 2 for the purposes of this question). The temperature is 220K, and the acceleration of gravity is 3.7 m s -2. What the scale height of the Martian atmosphere? How is the scale height of the Earth s atmosphere different from the scale height of Mars? Why? (Remember, the scale height of the Earth s atmospheres = 7.4 km.) (3 points) Scale height H = RT/Mag R = J mol -1 K -1 T = 220 K M = kg mol -1 g = 3.7 m s -2 H = (8.314*220)/(0.044*3.7) = 11.2 km This is larger than the scale height of the Earth s atmosphere. The atmosphere of Mars extends deeper because of smaller size of Mars and the resulting weaker gravitational pull on its atmosphere. b) In case of Venus, the atmosphere is 96.5% carbon dioxide, so you may again assume ALL). The temperature is735k, and the acceleration of gravity is 8.9 ms -2. What the scale height of Venus atmosphere? Compare to the scale height of the Earth s and Mars atmosphere. (3 points) Scale height H = RT/Mag R = J mol -1 K -1 T = 735 K M = kg mol -1 g = 8.9 m s -2 H = (8.314*735)/(0.044*8.9) = 15.6 km Venus has a similar gravitational pull on the atmosphere compared to the Earth, but its much hotter temperature leads to a much larger scale height than Earth. c) You might have wondered if the scale height varies around the Earth. The temperature around the Earth affects the scale height. Calculate the scale height for the equator (280K) and the poles (250K). Which is larger? By what percentage? (3 points) Equator: H = (8.314*280)/(0.029*9.8) = 8.2 km Pole: H = (8.314*250)/(0.029*9.8) = 7.3 km Percent: ( )/7.3 = % greater scale height at the equator.

4 The scale height is larger at the equator, due to the higher atmospheric temperature. The equator-pole difference is 12%, the same as the difference in temperatures. For many purposes this difference is not important, so we usually treat scale height as constant. d) Describe in your own words the definition of scale height. Why is it a useful concept? (2 points) Scale height describes the relationship between pressure and altitude. For every increase in altitude of H, pressure falls off by a factor of e. A plot of altitude versus the natural log of pressure is a straight line with a slope of -H. Scale height is a measure of depth and can be thought of as the thickness of an atmospheric layer. (2 points) 4. Barometric law (8 points) a) Calculate the concentration, in moles per liter, of oxygen at sea level. Pressure at sea level is 1013 hpa. In this case, use a surface temperature of 280 K. Oxygen comprises 21% of the atmosphere. (2 points) PV = NkT air concentration = N/AvV = P/AvkT = (1.013x10 5 ) / (6.022x10 23 * 1.38x10-23 * 280) =43.5 mol m -3 oxygen concentration = 0.21 * 43.5 mol m -3 * (1 m 3 / 1000 L) = 9.14x10-3 mol L -1 b) Assume the altitude of Mt. Everest is 9km. Calculate the pressure at the top of Mt. Everest, given a 7.4km scale height. (2 points) P = P0e -z/h P = (1.013x10 5 )e (-9/7.4) = 300 hpa c) Calculate the concentration, in moles per liter, of oxygen at the top of Mt. Everest. Assume the temperature at the top of Mt. Everest is 215K. (2 points) PV = NkT air concentration = N/AvV = P/AvkT = (3.00x10 4 ) / (6.022x10 23 * 1.38x10-23 * 215) =16.8 mol m -3 oxygen concentration = 0.21 * 16.8 mol m -3 * (1 m 3 / 1000 L) = 3.53x10-3 mol L -1 d) Imagine that you've just hiked up Mount Everest and you're standing on top of the summit savoring the moment. You love atmospheric science, of course, so you've brought a very special, massless, incompressible Nalgene bottle to fill with the summit air to take back home with you. You open the bottle and let the air waft in, then close it. You accidentally drop the magic bottle. Does it sink, rise or stay in about the same place? Why? In 2005, a stunt pilot landed a helicopter on the summit of Mount Everest. Imagine the helicopter cabin is pressurized to sea level (not the case, but we're imagining) and is at the same temperature as the outside summit air. The pilot has also brought a magic bottle to bring home some souvenir mountain air, but he doesn't want to get out of the cabin. He lets the air into his bottle then closes it. He thinks better of it and decides to get out of the helicopter after all. He brings the bottle of air and drops it. Does it sink, rise, or stay in the same place? Why?

5 Your magic bottle of air stays in about the same place because it is the same density as the surrounding atmosphere. The bottle of air from the helicopter sinks because it is pressurized at sea level and is more dense than the surrounding atmosphere.