CHEMISTRY 253

Spring, 2015 - Dixon

Group Assignment #9 - Solutions

One feedback associated with climate change is where increasing temperatures melt snow or sea ice, resulting in a decreased albedo (e.g. open water is much less reflective than ice or snow covered ice).  The decreased albedo, then can result in more solar radiation used for warming the Earth.  This is particularly important in polar regions in early summer (even though the greatest ice loss occurs in late summer). The Stephan-Boltzmann constant is 5.6 x 10^-8 W m^-2 K^-4. and 0°C = 273.15 K.

1.  To examine the affect of the change in albedo, a) Calculate the expected temperature (in K) on the surface of the Earth for an albedo of 0.30 and for an albedo of 0.27 in which there are no greenhouse gases and the incoming solar radiation is 1370 W m^-2.  Do this with the assumption that the average Earth’s temperature is an appropriate quantity to use and that the Earth acts as a blackbody.

F_S(1 – A)/4 = F_E (the ¼ is a consequence of the ratio between intercepted solar radiation – given by cross-sectional area and out-going Earth radiation – given by surface area).

Normal conditions:

F_E = (1370 W m^-2)(1 – 0.3)/4 = 239.75 W m^-2 = sT⁴

or T = (239.75 W m^-2/5.6 x 10^-8 W m^-2 K^-4)^0.25 = 255.8 K

For A reduced to 0.27, T = (250.0 W m^-2/5.6 x 10^-8 W m^-2 K^-4)^0.25 = 258.5 K

b) What is the change in temperature (including sign) associated with the decrease in albedo?

DT = 258.5 – 255.8 = 2.7 K ( = +2.7°C).

c) Is this a positive or negative feedback?

This is a positive feedback where warming leads to additional warming.

d) Explain why the calculated temperature appears too cold (both under standard and under reduced albedo changes).

We are ignoring the greenhouse effect where the amount of radiation leaving the Earth is reduced by absorption and re-radiation at lower temperatures in the atmosphere.

2.  Although this feedback is in one direction, an opposite type of feedback can occur in fall.  In fall, there is little to no sun in polar regions (at the poles, the sun sets for six months on the autumnal equinox).  Thus, there can be a large difference in energy loss between energy loss from open ocean at the sea water freezing point (~ -2°C) and what is typical above ice covered ocean (~ -20°C) in fall.

a) Calculate the ratio of energy loss above open ocean and above polar ice at the two temperatures mentioned above.  Assume greenhouse gases don’t affect this calculation.

E₂/E₁ = (T₂/T₁)⁴ = [(273.15 – 2)/(273.15 – 20)]⁴ = (1.071)⁴ = 1.32

b)  Is this feedback positive or negative?

Negative feedback.  Warming induced loss of ice results in greater heat loss in fall.