The average rate at which heat flows out through the surface of the Earth in North America is 54 mW/m2 and the average thermal conductivity of he near surface rocks is2.5 W/m ‘K, Assuming a surface temperatureof 10ºC, what should be the temperature at a depth of 33 km (near the base of the crust)? Ignore the heat generated by radioactive elements; the curvature of the Earth can also be ignored.

First we have to find out the temperature difference.
To obtain the temperature difference ΔT, substitute 54 mW/m² for H/A, 33 km for Δx and 2.5 W/m.K for k in the equation ΔT=(H/A) (Δx/k),
ΔT=(H/A) (Δx/k)

Rounding off to two significant figure, the temperature difference would be 710 K.
The heat flow is out. Thus at the base of the crust, the temperature is higher. Therefore the temperature T at a depth will be,
T = (ΔT) + (10° C)
= (710 K) + (10° C)
= (710 K) + ((10+273) K)
= 710 K + 283 K
= 993 K
Rounding off to two significant figures, the temperature at a depth of 33 km would be 990 K.