Geothermal energy loss due to heat conduction

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.61.

The temperature of the Earth increases as we go further underground so, given that the rock that makes up the Earth’s crust has a thermal conductivity and there is a temperature difference between a point underground and the Earth’s surface, the Earth is actually losing energy by heat conduction. Using the values given by Schroeder, we have {\Delta T=0.02\mbox{ K m}^{-1}} and {k_{t}=2.5\mbox{ W m}^{-1}\mbox{K}^{-1}}. The rate of heat conduction in an area of {1\mbox{ m}^{2}} is therefore

\displaystyle \frac{Q}{\Delta t} \displaystyle = \displaystyle -k_{t}A\frac{T_{2}-T_{1}}{\Delta x}\ \ \ \ \ (1)
\displaystyle \displaystyle = \displaystyle \left(2.5\right)\left(1\right)\frac{0.02}{1}\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle 0.05\mbox{ W} \ \ \ \ \ (3)

Although the rate of heat loss is quite small for a square metre, if we assume this value applies over the entire Earth (radius 6400 km), the total heat loss is

\displaystyle \frac{Q_{tot}}{\Delta t}=4\pi\left(6.4\times10^{6}\right)^{2}\left(0.05\right)=2.57\times10^{13}\mbox{ W} \ \ \ \ \ (4)

Although this sounds like a lot, to put it in perspective, the sun’s luminosity is {3.839\times10^{26}\mbox{ W}}, and the peak energy received on Earth from the Sun is around {1365\mbox{ W m}^{-2}}, which is around 27,000 times the rate at which the geothermal energy is being lost.

Leave a Reply

Your email address will not be published. Required fields are marked *