# 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.