Earth absorbs solar energy and radiates in the infrared. The solar energy incide

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 in the following questions, assume earth's surface temperature to be uniform at T. Find the steady state value of the earth's surface temperature T assuming the earth absorbs ALL the incident solar energy, and assuming that the earth is block (emissivity = 1) for infrared. (Numerical answer required) Find the steady slate value of T. 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, taking into account the "greenhouse effect" of atmospheric infrared absorption and emission. The diagram below shows the ground at the surface temperature T. 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 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 gamed - heat lost) for the ground AND for the atmospheric layer to obtain two equations: one for T, and one for T.. Solve for these temperatures and evaluate them numerically. Then look up on Wolfram Alpha 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) At steady state
Incident energy = radiated energy
INcident energy = I = pi*R^2*J {J = 1350 W/m^2]
Radiant energy, R = sigma*T^4*(pi*Re^2) [ from stefan-Boltzmann's law, Re is radius of earth]
here sigma = boltzmann's constant = 5.67*10^-8
T is temperature ( steady state)
R = Radius of earth = 6371 km
so, from energy conservation in steady state
1350 = 5.67*10^-8*T^4
T = 392.814 K
b) But earth absorbs only 65 pc of the radiation, and emits back as a blackbody
0.65*1350 = 5.67*10^-8*T^4
T = 352.708 K
c) Consider ground
Absorbed power = 0.475*1350 + 0.65*Pr' [ because heat emitted by the atmosphee is absorbed by earth]
Radiated power Pr = sigma*Ts^4
for steady state conditions
0.475*1350 + 0.65*sigma*Ta^4 = sigma*Ts^4

Consider Atmosphere
Absorbed power = 0.175*1350 + 0.95*Pr
Radiated power, Pr' = sigma*Ta^4
for steady state condition
0.175*1350 + 0.95*(sigma*Ts^4) = sigma*Ta^4

so, solving for Ts
0.475*1350 + 0.65*(0.175*1350 + 0.95*(sigma*Ts^4))= sigma*Ts^4
Ts = 437.534 K
Ta = 444.341 K

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