Calculating Earth's Absorbed Solar Power Using Stefan's Law

[Alt+F4]

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.)



(b) If the Earth is in thermal equilibrium, how much power must the Earth absorb from the Sun?

I know that this means that net flux is zero, that heat in equals heat out.

So how can i use Stefan's law to help me solve this?

 

[mezarashi]

The question is just asking you to calculate the power emitted by the Earth, or how much power is the Earth radiating into space.

Now with that answer, the additional question is, if the Earth is emitting so much power and it is in total not losing any power (due to the thermal equilibrium stated), then how much power must it be getting (absorbing from the Sun)?

Hint: A = B ;)

 

[Alt+F4]

(c) 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 = K

the total energy radiated by the sun and then use Stephan's law and solve for T_sun. For black body, emissivity is 1.0. Rsun = 7×10^8 m.)


So i tried to be smart about this one and googled surface area of sun in kelvin but none of the answers will work

So i got the Power emited by Earth which was 1.233 *10^17

So Power emited by sun : (1/10^10) * 1.23*10^17 = 12300000 W

So I = eAsigma T^4

1230000 = (1)(5.67*10^-8)(4*pi * (7*10^8)^2))* X^4


After all of that i get .077 Kelvin which makes no sense

I know from google it is around 4800-6000 K

 

[Alt+F4]

(c) 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 = K

the total energy radiated by the sun and then use Stephan's law and solve for T_sun. For black body, emissivity is 1.0. Rsun = 7×10^8 m.)


So i tried to be smart about this one and googled surface area of sun in kelvin but none of the answers will work

So i got the Power emited by Earth which was 1.233 *10^17

So Power emited by sun : (1/10^10) * 1.23*10^17 = 12300000 W

So I = eAsigma T^4

1230000 = (1)(5.67*10^-8)(4*pi * (7*10^8)^2))* X^4


After all of that i get .077 Kelvin which makes no sense

I know from google it is around 4800-6000 K

Am i even using the right Forumula?

 

[Alt+F4]

I don't Need help anymore, i Found out the Answer