91Ó°ÊÓ

Because the rock that makes up the Earth's crust has a thermal conductivity and there is a temperature differential between a point underground and the Earth's surface, the Earth loses energy through heat conduction. We've calculated the following using Schroeder's values:$\mathrm{Δ}T={20}^{∘}\mathrm{K}$per $\mathrm{Δ}x=1000\mathrm{m}$and $k_t=2.5{\mathrm{W}}^{∘}{\mathrm{K}}^{−1}$, the rate of heat conduction in an area of $1m^2$is, therefore:

localid="1651746825640" $\frac{Q}{\mathrm{Δ}t}=k_{t}A\frac{\mathrm{Δ}T}{\mathrm{Δ}x}=2.5×1×\frac{20}{1000}=0.05\mathrm{W}/{\mathrm{m}}^2$

localid="1651746839506" $\frac{Q}{\mathrm{Δ}t}=0.05\mathrm{W}/{\mathrm{m}}^2$

Even if the rate of heat loss for a square metre is fairly low, if we assume that this figure applies to the entire Earth, the total heat loss is:

$\frac{Q_{\text{total}}}{\mathrm{Δ}t}=\text{Loss per meter square}×\text{Total area of earth}$

$\frac{Q_{\text{total}}}{\mathrm{Δ}t}=0.05×\left[4\mathrm{π}{r}^2\right]=0.05×\left[4\mathrm{π}{\left(6400×{10}^3\right)}^2\right]$

$\frac{Q_{\text{total}}}{\mathrm{Δ}t}=2.573×{10}^{13}W$