If I wanted to calculate how hot (worse case scenerio) a piece of metal just outside the earth's atmosphere would get due only to the sun's radiation I used:

q = sigma * emiss * (Tsun^4 - Tmetal^4)

knowing q = 1367 W/m^2 Tsun - 6000K assuming emiss = 1 for black body

I am getting a really hot piece of metal - 4236K

did I miss something?

I think I am missing a view factor..

Is your piece of metal actually a black body? Surely some of the radiation will be reflected...

I'm doing worse case scenerio, so I am treating it as a black body. I am having trouble computing a view factor because of the distance from the sun..any suggestions?

Hmmm not sure on the mechanics of actually calculating the view factors (I let the CFD code I use do it for me!). How is the piece of metal cooled?

it is not cooled at all..i am just trying to figure out how hot it would get if it was in direct sunlight basically at the same distance the earth is. So, I know how much energy gets to the earth from the sun, the sun's temp, the area of the metal piece, and the diameter of the sun. But, the view factor is throwing me off.

j = sigma T^4, using T = 6000K is the radiative power per square metre on the surface of the Sun. Bear in mind that there is some 1.4x10^11m between the surface of the Sun and that piece of metal...

The 1367W/m^2 is the heat energy supplied by the Sun at the Earth's surface. You need an equation that defines the temperature rise that will be obtained by the metal conducting 1367J of energy (assuming complete absorption of the Sun's Energy).

using q*A=sigma*emiss*T^4 and knowing the area (A) of the plate I can solve for T, which gives me around 350K, which is reasonable. I know the equilibrium temp is of the earth is roughly 270K, so this sounds better.

So it receives heat from the sun but has no mechanism to pass that heat on thus achieving continuity, ummmm....

The piece of metal emits black body radiation. The view issue seems simple enough. Assume that the sun emits radiation isotropically.

R : distance from sun to object

P_s : total power emitted by sun

P_a : power absorbed by object

P_e : power emitted by object

A : cross-sectional area of object facing the sun

S : total surface area of object

so we have

P_a = P_s A /(4 \pi R^2)

P_e = S \sigma T^4

The steady state condition is P_a = P_r. Suppose the object is spherical with radius r. Then A = \pi r^2 and S = 4 \pi r^2 so

\sigma T^4 = P_s / (16 \pi R^2)

putting in numbers gives about 280 K.

let sun temperature be Ts and Radius of earth orbit be d and radius of earth be R ... then assuming thermal equilibrium betweem earth and sun u will get

{sigma*4*pi*Rsun^2*Ts^4/(4*pi*d^2)}*pi*R^2=sigma*4*pi*R^2*Te^4

This would give you the temperature of earth... now your steel metal piece is in thermal equilibrium with the sun and earth... so u need to find thermal resistance between metal and sun and metal and earth... all r in sries.. then u need to find the potential(temperature) using series law of resistances... to figure out the resistances see any book on heat transfer(radiation) ... it will depnd on the shape and orientation of ur metal piece.