Calculating Heat Energy in Conduction: A Mars Case Study

[stevemclaren]

Homework Statement

the problem is one of understanding. if heat is incident on a surface raising the temperature to some number then conducts through the solid to the other side how much heat energy is actually at the other side?? i understand the temperatures will be different and that heat and temperature aren't the same thing but how could one calculate heat energy at the other side?

Homework Equations

i suppose the conduction equation Q = ((lambda*c.s.a)/length)*delta T)

q in equals a fraction of q out

The Attempt at a Solution

if i use the solar constant on the side a spacecraft around Mars the temperatuere is around 30 degrees C i understand the temperature on the other side will be less and i know the incoming radiation so can work it out, but how much heat energy will be on the other side?

cheers all, this is really playing on my mind

 

[stevemclaren]

anyone got any ideas?

 

[Hootenanny]

What is 'heat energy'?

 

[stevemclaren]

http://en.wikipedia.org/wiki/Heat

to quote "energy transferred from one body or system to another due to a difference in temperature"

 

[Hootenanny]

http://en.wikipedia.org/wiki/Heat

to quote "energy transferred from one body or system to another due to a difference in temperature"

If heat is the transfer of energy, how can one calculate the 'heat energy' at the other side? Surely, once the energy is at the other side it is no longer being transfered, and therefore the 'heat energy' is zero?

 

[stevemclaren]

radiation conduction convection, all that jazz

 

[Hootenanny]

I think that you mean you want to calculate the 'energy transferred' (or 'heat') across the boundary. There is no such quantity as 'heat energy' and something cannot 'have heat', however, energy can be transferred across a boundary via 'heat'.

From what I understand of your question, you have light incident on some surface, you want to calculate the energy transferred through the surface?