Need help with a-c. Thanks. During the day, heat from the Sun is absorbed by the

Question

Need help with a-c. Thanks.

During the day, heat from the Sun is absorbed by the Earth's surface. It is then radiated into space at night. Calculate the rate, in watts, at which this heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 m^2 to the atmosphere at night. Assume the emissivity is 0.90, the temperature of the surface of the Earth is 15degree C, and that of outer spacc is 2.7 K. The intensity of the Sun's radiation at the Earth's distance is about 800 W/m^2, 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? What is the maximum magnetic field strength, in microteslas, of the outgoing radiation, assuming it is a continuous wave?

Explanation / Answer

Here ,

A = 1 m^2

emmisivity , e = 0.90

T = 15 degree C = 288 K

To = 2.7 K

part a)

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

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

rate of energy output = 351.1 W/m^2

the rate of energy output is 351.1 W/m^2

part b)

fraction of radiation absorption by the earth = energy output in night/energy intake during day

fraction of radiation absorption by the earth = 351.1/(400)

fraction of radiation absorption by the earth = 0.878

part c)

let the maxium magentic field is B

as Intensity = 0.5 * Bm^2 * c/u0

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

solving for Bm

Bm = 1.715 *10^-6 T

the maximum magnetic field is 1.715 *10^-6 T

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