crysix
Apr10-11, 12:55 AM
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 thats correct at all
would any of you offer me in assistance on how to approach this question?
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 thats correct at all
would any of you offer me in assistance on how to approach this question?