Earth absorbs solar energy and radiates in the infrared The solar energy inciden

Question

Earth absorbs solar energy and radiates in the infrared The solar energy incident on earth is pi R^2 J where R is earth's radius and J = 1350 W m^-2 is the solar constant. Also, earth radiates from its entire surface area 4 pi R^2. In the following questions, assume earth's surface temperature to be uniform at T_1. Find the steady state value of the earth's surface temperature T_1 assuming the earth absorbs ALL the incident solar energy, and assuming that the earth is black (emissivity = 1) for infrared (Numerical answer required). Find the steady state value of T_2 taking into account that earth absorbs only 65% of the incident solar energy (The remaining 35% is reflected from clouds and ice). You'll discover that at this temperature earth would not support life. Now find the steady state value of T_3 taking into account the "greenhouse effect'' of atmospheric infrared absorption and emission The diagram below shows the ground at the surface temperature T_1 and the atmosphere, represented as a thin black layer, at temperature T_a. The atmosphere absorbs 95% of the infrared radiation emitted by the earth, and can therefore be approximated as a single black layer You should note that there is infrared radiation emitted upward by the earth, and both upward and downward radiation emitted by the atmosphere Assume that the ground absorbs 47.5% of the incident solar energy and the atmosphere absorbs 17 5% of the incident solar energy (for a total of 65%) Write the heat energy balance (heat gained = heal lost) for the ground AND tor the atmospheric layer to obtain two equations one for T_1 and one for T_a: Solve tor these temperatures and evaluate them numerically Then look up on WolframAlpha the average surface temperature for the earth and compare this value with your answer obtained for the ground temperature using the above greenhouse effect model.

Explanation / Answer

a) pi * R^2 * 1350 = 5.67* 10^-8 * 4* pi* R^2 * T^4

So 1350 * 10^8 / (4*5.67) = T^4

T = 277.762 K

= 4.7 degrees

b) If only 65% is absorbed then

0.65* 1350 * 10^8 / (4*5.67) = T^4

T = 249.5 K

So Temperature = -23.5 degrees Celsius

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