Question

The average rate at which energy is conducted outward through the ground surface in a certain region is 60.9 mW/m2, and the average thermal conductivity of the near-surface rocks is 1.99 W/m·K. Assuming a surface temperature of 5.65°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.

Answer 1

Power = Q/t = k*A*dT/L

dT = P*L/(k*A)

given values are

P/A = 60.9 mW/m^2 = 60.9*10^-3 W/m^2

L = 35*10^3 m

k = 1.99 W/m-K

dT = 60.9*10^-3*35*10^3/1.99

dT = 1071.10

Th - Tc = 1071.10

Tc = 5.65 C

Th = 1071.10 + 5.65

Th = 1076.75 C