Posted by tim on Thursday, April 14, 2011 at 8:27pm.
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 nearsurface 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.

physics  Jai, Thursday, April 14, 2011 at 8:51pm
recall that for an energy transfer,
q = kA * dT / dx
where
q = rate of heat tranfer (in Watts)
k = thermal conductivity (in W/mK)
dT = differential change in Temperature (in Celsius or in Kelvin)
dx = differential change in position (xdirection)
*note: this is negative since (T2T1) 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 (T2T1)/(x2x1) ; 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~ :)

physics  Jai, Thursday, April 14, 2011 at 9:19pm
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~ :)
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