The average rate at which energy is conducted outward through the ground surface in North America is$\mathbf{54}\mathbf{.0}\mathbf{}{\mathbf{\text{mW/m}}}^{\mathbf{\text{2}}}$, and the average thermal conductivity of the near-surface rocks is$\mathbf{2}\mathbf{.50}\mathbf{}\mathbf{\text{W/mK}}$. Assuming a surface temperature of$\mathbf{10}\mathbf{.0}\mathbf{°}\mathbf{\text{C}}$, find the temperature at a depth of$\mathbf{35}\mathbf{.0}\mathbf{}\mathbf{\text{km}}$(near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

1. The average rate at which energy is conducted outward through the ground surface$\frac{P_{cond}}{A}=54.0\frac{\text{mW}}{{\text{m}}^{\text{2}}}\text{or}54.0\times {10}^{-3}{\text{W/m}}^{\text{2}}\text{.}$
2. The average thermal conductivity of the near-surface rocks is.$k=2.50\text{W/}\left(\text{m.K}\right)$
3. Surface temperature is,$T_C=10.0°\text{C}.$
4. Depth is,$L=35.0\text{kmor}35\times {10}^3\text{m}.$

Conduction heat transfer is the transfer of heat through matter without the bulk motion of the matter. This process involves the transfer of the high-energy particle to the end of a low energetic particle, when two bodies with different temperatures are brought in contact inside a system. Thus, here in the given problem, when the ground surface of North America is brought in contact with the near-surface, the flow rate is from the ground to the near-surface end.

Formula:

The rate of conduction of the energy by the body,$P_{cond}=\frac{kA(T_H-T_C)}{L}$ …(i)

whereis the thermal conductivity of the material, A is the surface area of radiation,$T_H$ is the temperature at the hotter end,$T_L$ is the temperature at the colder end, and L is the length of the conduction.

The temperature at the depth 35 km using equation (i) is

$\begin{array}{c}{T}_H=\frac{PL}{Ak}+T_C\\ =\frac{(54.0\times {10}^{-3}{\text{W/m}}^2)(35\times {10}^3\text{m})}{\left(2.50\text{W/m.K}\right)}+10.0°\text{C}\\ =756\text{K}+10.0°\text{C}\\ =483°\text{C}+10.0°\text{C}\\ =493°\text{C}\end{array}$

Therefore,the temperature at a depth of $35.0\text{km}$ is$493°\text{C}$