The Earth has a surface temperature around 270 K and an emissivity of 0.8, while

Question

The Earth has a surface temperature around 270 K and an emissivity of 0.8, while space has a temperature of around 2 K. Use Stefan's law: P(net) = 5.67×10-8 W/K4 m2 *A*e*(T4 - To4) to solve the following questions. (Radii of the Earth and the Sun are Re = 6.38×106 m, Rs = 7×108 m.)

The net power radiated from our Earth into free space in Watt W is 1.234E17W and the power must the Earth absorb from the Sun P_abs =1.234E17W.

1.)If the energy absorbed by the Earth from the Sun constitutes only 1/(1010) of the Suns total radiant energy to the space, and consider the Sun is a black body, what is the surface temperature of the Sun in Kelvin K?

T_sun =

Explanation / Answer

P(net) = [5.67×10-8*A*e*(T^4 - To^4)][W/(K^4)*(m^2)]

For a black body, the emissivity "e" has the ideal value of
simply 1. So for the sun radiating to space
As = 4*pi*Rs^2,
e = 1,
To = 2 K.

1) I think the area A here should be the "circular face area" of the earth as presented to the sun at any time (pi.Re^2), not the entire surface area of the earth (4.pi.Re^2).

A = pi*(6.38*10^6)^2 = 1.28*10^14 m^2

So P = 5.67×10^-8 W/K^4 m^2 *1.28*10^14 * 0.8 * (270^4 - 2^4)
P = 3.08*10^16 W


For the last question, solve for T with P = 3.08*10^16*10^10 W = 3.08*10^26 W
And for A it should be the entire surface area of the sun (4.pi*Rs^2) since the sun radiates in all directions

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