No idea of how to really start this question

In summary, the problem involves finding the temperature at a depth of 35.0 km, given the surface temperature of 8.70°C and the average thermal conductivity of near-surface rocks. Using the formula P = KA x [(T(h) - T(c)) / L], where P is the rate of energy conducted outward, K is the thermal conductivity, A is the surface area, T(h) is the surface temperature, T(c) is the temperature at the given depth, and L is the depth, the correct temperature at a depth of 35.0 km is 579.65°C. This calculation takes into account the heat generated by the presence of radioactive elements.
  • #1
crysix
2
0
The average rate at which energy is conducted outward through the ground surface in a certain region is 63.7 mW/m2, and the average thermal conductivity of the near-surface rocks is 3.96 W/m·K. Assuming a surface temperature of 8.70°C, find the temperature (in Celsius) at a depth of 35.0 km (near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

So i understand that the surface is radiating energy at 63.7 mW/m^2
and that k for the surface rocks is 3.96
whilst the surface temperature is 8.70°C


I thought of using P = KA x [(T(h) - T(c)) / L]
L = 35500 M
K = 3.96
P = 0.0637 W
A = 1 m ^ 2
T(h) - T(c) = 8.60 - T(c)

Solving for T(c) = -562.45 <---- I don't think that's correct at all

would any of you offer me in assistance on how to approach this question?
 
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  • #2
How is 8.6 T(h)? It's hotter deeper down.
 
  • #3
haha silly me,
hence if i use T(c) as 8.6
then i get T(H) as 579.65 which I can understand because it gets quite hot down there, but would that be correct?
 

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