A 1.5-m^2 black surface at 120 degrees C is losing heat to the surrounding air at 30 degrees C by convection with a convection heat transfer coefficient of 18 W/(m^2 degree C), and by radiation to the surrounding surfaces at 10 degrees C. The total rate of heat loss from the surface is : ?
Q (convection) = h * A *( delta T)
Q (radiation) = [itex]\epsilon[/itex] * [itex]\sigma[/itex] * A * (T^4 (surface) - T^4 (surroundings))
The Attempt at a Solution
I calculate the heat lost by convection to be 2430 Watts. However, I need to find the total heat loss from the surface, which includes the loss from radiation.
I have the surface Area, the sigma value (Stefan-Boltzmann constant) , and the temperatures (393 K for the emitting surface and 283 K for the surrounding surface) but no epsilon value is given. (emissivity). All it says is "black surface" and not the material. Should I assume the surface is ideal (blackbody radiation) and put a value of 1? Should I disregard the term altogether?
I wanted to post this here before I emailed my instructor, because I unless I'm doing something wrong I feel I lack enough information to complete the problem.