The average rate at which energy is conducted outward through the ground surface in a certain region is 66.8 mW/m2, and the average thermal conductivity of the near-surface rocks is 2.13 W/m·K. Assuming a surface temperature of 8.01°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.
To find the temperature at a depth of 35.0 km, we can use the formula for heat conduction:
q = k * A * (dT/dx)
where:
q is the rate of heat transfer per unit area (in watts per square meter),
k is the thermal conductivity (in watts per meter-kelvin),
A is the cross-sectional area (in square meters), and
(dT/dx) is the temperature gradient (in kelvin per meter).
Given:
q = 66.8 mW/m2 = 66.8 * 10^-3 W/m2
k = 2.13 W/m·K
dT/dx = ? (to be calculated at a depth of 35.0 km)
We can determine the temperature gradient (dT/dx) using the formula:
(dT/dx) = q / (k * A)
First, let's find the cross-sectional area (A) at a depth of 35.0 km:
A = π * r^2
where:
r is the radius of the earth at a depth of 35.0 km,
which can be calculated using the formula:
r = R - d
where:
R is the radius of the earth (approximately 6,371 km) and
d is the depth (35.0 km).
Substituting the values, we get:
r = 6,371 km - 35.0 km = 6,336 km = 6,336,000 m
Now, we can find the cross-sectional area:
A = π * (6,336,000 m)^2
Next, we can calculate the temperature gradient (dT/dx):
(dT/dx) = 66.8 * 10^-3 W/m2 / (2.13 W/m·K * π * (6,336,000 m)^2)
Finally, we can find the temperature at a depth of 35.0 km:
T = T_surface + (dT/dx) * depth
where:
T_surface = 8.01°C
Substituting the values, we get:
T = 8.01°C + (dT/dx) * 35,000 m
Calculate the value of (dT/dx) and substitute it into the equation to find T.
To find the temperature at a depth of 35.0 km, we can use the concept of heat conduction and the formula:
Q = kA (T2 - T1) / d
where:
Q = rate of heat conduction
k = thermal conductivity
A = surface area
T2 = temperature at depth
T1 = surface temperature
d = depth
In this case, we are given the rate of heat conduction Q = 66.8 mW/m^2, the thermal conductivity k = 2.13 W/m·K, the surface temperature T1 = 8.01°C, and the depth d = 35.0 km.
First, we need to convert the given values to their proper units.
1 mW = 0.001 W, so the rate of heat conduction Q is 66.8 * 0.001 = 0.0668 W/m^2.
35.0 km = 35000 m.
Next, we need to calculate the surface area A of a square meter at a depth of 35.0 km. Since the area remains constant, A is equal to 1 m^2.
Now, we rearrange the formula to solve for T2:
T2 = (Q * d / k) + T1
Substituting the values:
T2 = (0.0668 * 35000) / 2.13 + 8.01
Calculating, T2 ≈ 1080.23°C
Therefore, the temperature at a depth of 35.0 km is approximately 1080.23°C.