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Lexxian
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Homework Statement
The question - A small sphere (emissivity = 0.90, radius r1) is located at the center of a spherical asbestos shell (thickness=1.0cm, outer radius r2). The thickness of the shell is small compared to the inner and outer radii of the shell. The temperature of the small sphere is 800.0 degree C, while the temperature of the inner surface of the shell is 600.0 degree C, both temps remain constant. Assuming that r2/r1 = 10.0 and ignoring any air inside the shell, find the temperature of the outer surface of the shell.
k(asbestos)=0.090 J/s x m x degrees Celsius
Homework Equations
Q=(kA deltaT)t/L (minus the t)
Q= e sigma T^4^At (minus the t)
A= 4 pi r^2^
The Attempt at a Solution
I know that I will be using both equations. I am unsure if I set them equal to each other (similar to conservation of energy equations) which I feel I was lead to that thought process by the statement of ignoring any air inside the shell. Also, in order to find the radius thus giving me the Area I need to solve the equations listed. Unless...the r2/r1 = 10 will actually provide me with...nah...because the Area of the smaller sphere is smaller than the area of the whole sphere. Please help.