Question

We focus on the geothermal effect in a slightly different manner than in Example 11.3. We know that the average rate at which heat is conducted through the surface of the ground in North America is 55 mJ/(s m 2 ). Assuming a surface temperature in summer of 25°C, what is the temperature at a depth of 30 km (near the base of Earth’s crust, which together with the brittle upper portion of the mantle forms the 100-km-deep lithosphere)? Hint: Ignore the heat generated by the presence of radioactive elements and use 1.7 J/(m s K) for the average thermal conductivity of the near-surface rocks. Start with Fourier’s law.

Answer 1

Using a formula, we have

(Q / t) = A (T_hot - T) / d

Q / (A t) = (T_hot - T) / d

where, = average thermal conductivity of the near-surface rocks = 1.7 J/m.s.K

d = depth = 30 km = 30000 m

T_hot = surface temperature in summer = 25 ⁰C = 298.15 K

then, we get

(55 x 10^-3 J/m².s) = (1.7 J/m.s.K) [(298.15 K) - T] / (30000 m)

[(55 x 10^-3 J/m².s) (30000 m)] = (1.7 J/m.s.K) [(298.15 K) - T]

[(1650 J/m.s) / (1.7 J/m.s.K)] = [(298.15 K) - T]

(970.58 K) = [(298.15 K) - T]

T = [(298.15 K) - (970.58 K)]

T = - 672.43 K