Question

The average rate at which energy is conducted outward through the ground surface in a certain region is 51.6 mW/m2, and the average thermal conductivity of the near-surface rocks is 3.31 W/m·K. Assuming a surface temperature of 11.7°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

recall that for an energy transfer,
q = -kA * dT / dx
where
q = rate of heat tranfer (in Watts)
k = thermal conductivity (in W/m-K)
dT = differential change in Temperature (in Celsius or in Kelvin)
dx = differential change in position (x-direction)
*note: this is negative since (T2-T1) is always negative, for heat always flows from higher temp to lower temp.

assuming heat transfer is at steady state (does not change with time), this formula simplifies to
q = -kA (T2-T1)/(x2-x1) ; or
q/A = -k*(T2 - T2)/(x2 - x1)
where
T1 = heat source temp (at this problem it's the crust)
T2 = lower temp (at this problem, it's the surface)
first we convert 51.6 mW/m^2 into W by multiplying by 10^-3
substituting,
51.6 * 10^(-3) = -3.31*( 11.7 - T1 )/(35)
0.0516*35/(-3.31) = 11.7 - T1
-0.546 = 11.7 - T1
-12.246 = -T1
T1 = 12.246 deg C

hope this helps~ :)

Answer 2

oops, scrap my first answer,, i forgot to convert 35 km to meters:
35 km = 35000 m

substituting,
51.6 * 10^(-3) = -3.31*( 11.7 - T1 )/(35000)
0.0516*35000/(-3.31) = 11.7 - T1
-545.62 = 11.7 - T1
-557.32 = -T1
T1 = 557.32 deg C

sorry~ ^^;
hope this helps~ :)