Conduction of Heat and Radiation problem

In summary, the question asks to find the temperature of the outer surface of a spherical asbestos shell with a small sphere at its center, given the temperatures of the inner surface of the shell and the small sphere. The thickness of the shell is negligible compared to its inner and outer radii. Using the equations Q=(kA deltaT)t/L and Q= e sigma T^4^At, and setting them equal to each other due to conservation of energy, we can solve for the temperature of the outer surface of the shell. The given ratio of r2/r1 = 10 does not provide enough information to solve the problem.
  • #1
Lexxian
1
0

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.
 
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  • #2
Hi Lexxian, welcome to PF. Yes, it would be a good idea to apply conservation of energy to this problem. At steady state, all the energy leaving the inner sphere as radiation will be conducted through the outer sphere. Does this help you get started?

(Also, check your equations - it looks like there are a few typos.)
 

Related to Conduction of Heat and Radiation problem

What is conduction of heat?

Conduction of heat is the transfer of thermal energy through a material or between two materials that are in contact with each other. This transfer of energy occurs due to the vibration of atoms and molecules in the material.

What factors affect the rate of conduction of heat?

The rate of conduction of heat is affected by several factors such as the type of material, its thickness, temperature difference between the two materials, and the thermal conductivity of the material.

How does conduction differ from radiation?

Conduction and radiation are two different methods of heat transfer. In conduction, heat is transferred through direct contact between materials, while in radiation, heat is transferred through electromagnetic waves without the need for any medium.

What are some examples of conduction of heat?

Some examples of conduction of heat include touching a hot stove, using a metal spoon in hot soup, and placing a cold hand on a warm forehead. In all these cases, heat is transferred from a warmer object to a cooler one through direct contact.

How can conduction of heat be reduced or increased?

Conduction of heat can be reduced by using materials with low thermal conductivity, increasing the thickness of the material, or creating a barrier between two materials. It can be increased by using materials with high thermal conductivity or decreasing the distance between two materials.

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