Question

Assume heat producing elements are uniformly distributed within a 16 km thick layer in the crust in a heat flow province. Given that the surface heat flow and reduced heat flow are 54 mW/m$^2$ and 22 mW/m$^2$, respectively, the radiogenic heat production in the given crustal layer in $\mu$W/m$^3$ is ________ (in integer).

Hint:

When thickness is in km and heat flows are in mW/m$^2$, the shortcut is: \[ A(\mu\text{W m}^{-3}) = \frac{q_0 - q_r \; (\text{mW m}^{-2})}{H \; (\text{km})}. \]

Answer 1

Step 1: Use the surface heat–production relation.
For a uniformly radiogenic layer of thickness \(H\), \[ q_0 \;=\; q_r \;+\; A\,H \] where \(q_0\) = surface heat flow, \(q_r\) = reduced heat flow, and \(A\) = volumetric heat production. Step 2: Rearrange and plug in numbers.
\[ A \;=\; \frac{q_0 - q_r}{H} = \frac{54 - 22}{16} = \frac{32}{16} = 2 \] \textit{Unit check:} \(A\) in \(\mu\)W/m\(^3\) because \((\mu\text{W}/\text{m}^3)\times (\text{km}) = \text{mW}/\text{m}^2\). Final Answer:
\[ \boxed{2 \ \mu\text{W m}^{-3}} \]