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.
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~ :)
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~ :)
To solve this problem, we can use the concept of steady-state heat conduction. The rate of heat conducted through a material can be determined using the equation:
Q = (kAΔT) / d
where Q is the rate of heat conducted (in watts), k is the thermal conductivity of the material (in watts per meter-kelvin), A is the surface area (in square meters), ΔT is the temperature difference across the material (in kelvin), and d is the thickness of the material (in meters).
First, let's determine the temperature difference ΔT across the crust. We have the surface temperature (11.7°C), but we need to convert it to kelvin by adding 273.15:
ΔT = (11.7 + 273.15) K - T
Next, we can calculate the rate of heat conducted Q using the equation:
Q = kA(ΔT) / d
We are given the rate of heat conducted Q (51.6 mW/m², which is equivalent to 51.6 × 10⁻³ W/m²), the thermal conductivity k (3.31 W/m·K), and the thickness of the crust (35.0 km, which is equivalent to 35.0 × 10³ m).
Now, let's plug in the values and solve for ΔT:
51.6 × 10⁻³ = (3.31)A(ΔT) / (35.0 × 10³)
To eliminate the area A, we need to express it in terms of the crust's thickness. The area of a cylindrical crust with a radius r is given by:
A = 2πr²
The radius r is related to the thickness d by the equation:
r = d
Therefore, we can rewrite the equation for A as:
A = 2πd²
Now we can substitute this expression for A into the equation:
51.6 × 10⁻³ = (3.31)(2πd²)(ΔT) / (35.0 × 10³)
At this point, we have one equation with two unknowns (ΔT and d). To solve for ΔT, we need an additional equation. We can use the equation for the average temperature at a given depth:
T = Ts + (Tc - Ts) * (z / Z)
where T is the temperature at a given depth, Ts is the surface temperature, Tc is the temperature at the base of the crust, z is the depth, and Z is the total thickness of the crust.
In this case, we want to find the temperature at a depth of 35.0 km (near the base of the crust). We know the surface temperature (11.7°C) and the depth (35.0 km), but we need the temperature at the base of the crust (Tc).
Unfortunately, we don't have enough information to directly calculate Tc. We would need to know the temperature gradient or have additional data on the subsurface temperatures.
Therefore, without the necessary data, we cannot determine the temperature at a depth of 35.0 km near the base of the crust.