Question

Geologists measure conductive heat flow out of the earth by drilling holes (a few hundred meters deep) and measuring the temperature as a function of depth. Suppose that in a certain location the temperature increases by${20}^{\circ }\mathrm{C}$per kilometer of depth and the thermal conductivity of the rock is $2.5\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$. What is the rate of heat conduction per square meter in this location? Assuming that this value is typical of other locations over all of the earth's surface, at approximately what rate is the earth losing heat via conduction? (The radius of the earth is $6400\mathrm{km}$.)

Short answer

Rate of heat conduction and the rate at which the earth loses heat via conduction

$\frac{Q}{\mathrm{\Delta }t}=0.05\mathrm{W}$

$\frac{Q_{\text{total}}}{\mathrm{\Delta }t}=2.573\times {10}^{13}$

Step-by-step solution

Calculation of ratio

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:$\mathrm{\Delta }T={20}^{\circ }\mathrm{K}$per $\mathrm{\Delta }x=1000\mathrm{m}$and $k_t=2.5{\mathrm{W}}^{\circ }{\mathrm{K}}^{-1}$, the rate of heat conduction in an area of $1m^2$is, therefore:

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

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

Calculation of heat

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:

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

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

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