Two different thermal conductivity constants

[a_lawson_2k]

Homework Statement

Solid cylindrical copper rod 0.2m long has one end maintained at temperature 20K, other end blackened and exposed to thermal radiation from surrounding walls at 500K. As the rod is insulated, no energy is lost or gained except at the ends of the rod. When equilibrium is reached, what is the temperature of the blackened end? hint: at 20K, copper's thermal conductivity constant is 1670 W/(mK), so the blackened end will only be slightly over 20K

Homework Equations

[tex]H=\frac{dQ}{dt}=kA\frac{T_h-T_c}{L}[/tex]

The Attempt at a Solution

Thus far, I'm not sure how to approach it given the presence of two different thermal conductivity constants; the one specified in the book is 385, but the other specified is only for one specific circumstance...

 

[Astronuc]

The key is 'equilibrium', which means the temperature at the boundary does not change, which implies a constant heat flux, i.e. the radiation heat flux = conductive heat flux.

q''(radiation) = q''(conduction)

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/stefan.html#c2

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/heatra.html#c2


The thermal conductivity of copper is 385.0 W/mK, but that is likely room temperature (~20-25°C), so perhaps this implies K is a function of temperature.

 

[a_lawson_2k]

I don't see how to do that, as I was not given a constant of emissivity. All I have is an equation which, frankly, I'm not sure how to apply here. The only example in the book with two heat currents was one with two rods between each heat/cooling source...how would you set it up?