(5 %) Problem 17: During the day, heat from the Sun is absorbed by the Earth\'s

Question

(5 %) Problem 17: During the day, heat from the Sun is absorbed by the Earth's surface. 33% Part (a) Calculate the rate in watts, at which this heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 m2 to the atmosphere at night. Assume the emissivity is 0.90, the temperature of the surface of the Earth is 15 C, and that of outer space is 2.7 K. 33% Part (b) The intensity of the Sun's radiation at the Earth's distance is about 800 W/m2 but only half of the incoming radiation is actually absorbed by the Earth's surface. What fraction of the rate of radiation absorption by the Earth's surface during the day is reemitted at night? Grade Summary l 0.25 - 0251 Dedactions 16% Potential 84% l coso I tan()-1 acoso 1 (E718191 sinO | cos() | tan() | | (11) 7| 8| 9|HOME sino Submissions cotan) atan acotan) sinh0 cosh t cotanhO (4% per attempt) detailed view 4% 4% 4% 4% END Degrees ORadians VOI BACKSPACE "" CLEAR Submit I give up Hints: for a 0%deduction. Hints remaining: Feedback: deduction per feedback. Form a ratio between your result from part (a) and the amount of radiation absorbed from the Sun during a day. 33% Part (c) what is the maximum magnetic field strength in microtesla, of the outgoing radiation, assuming it is a continuous wave?

Explanation / Answer

Here,

part a)

e = 0.90

T = 15 degree c = 288 K

To = 2.7 K

area , A = 1 m^2

rate of energy lost = e * A * sigma * (T^4 - To^4)

rate of energy lost = 0.90 * 1 * 5.6703 *10^-8 * (288^4 - 2.7^4)

rate of energy lost = 351.1 W

the rate of energy lost at night is 351.1 W

part b)

for the fraction = power emitted at night/power absorbed during day

fraction = 351.1/(800/2)

fraction = 0.89

the fraction is 0.89

part B)

let the maximum magnetic field is B

output intensity = 0.5 * Bm^2*c^2/(c * u0)

351.1 = 0.5 * (Bm^2 * 3 *10^8)/(4pi *10^-7)

solving for Bm

Bm = 1.72 *10^-6 T

the maximum magnetic field is 1.72 uT

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