Chapter 26, Problem 43AP

A close analogy exists between the flow of energy by heat because of a temperature difference (see Section 19.6) and the flow of electric charge because of a potential difference. In a metal, energy dQ and electrical charge dq are both transported by free electrons. Consequently, a good electrical conductor is usually a good thermal conductor as well. Consider a thin conducting slab of thickness dx, area A, and electrical conductivity σ, with a potential difference dV between opposite faces. (a) Show that the current I = dq/dt is given by the equation on the left:

$\begin{array}{cc}\text{Charge}\text{<x-custom-btb-me data-me-id="2204" class="microExplainerHighlight">conduction</x-custom-btb-me>}& \text{Thermal}\text{conduction}\\ \frac{dq}{dt}=\sigma A\left|\frac{dV}{dx}\right|& \frac{dQ}{dt}=kA\left|\frac{dT}{dx}\right|\end{array}$

In the analogous thermal conduction equation on the right (Eq. 19.17), the rate dQ/dt of energy flow by heat (in SI units of joules per second) is due to a temperature gradient dT/dx in a material of thermal conductivity k. (b) State analogous rules relating the direction of the electric current to the change in potential and relating the direction of energy flow to the change in temperature.