Thermal Conduction, Thermal Conductivity – Problems and Solutions – Laman 4

04. The thermal conductivity of a rod is 210 W m^-1 K^-1. Find the temperature gradient of the rod if the cross-sectional area of the rod is 3.0 cm² and the rate of heat flow is 0.126 J s^-1.

Quick Note!

The rate of heat flow $\frac {dQ}{dt}$ perpendicular to the cross-sectional area A depends on :

the cross-sectional A.

the temperature gradient $-\frac {d\theta}{dt}$ .

the material of the solid.

It is found that

$\frac { dQ }{ dt } \propto \quad A\\ \\ \frac { dQ }{ dt } \propto -\frac { d\theta }{ dx } \\ \\$
Therefore

$\frac { dQ }{ dt } \propto -A\frac { d\theta }{ dx } \\ \\ \\ \\$
Where k = thermal conductivity of material.
Therefore, thermal conductivity k is defined as the negative rate of heat flow per unit area perpendicular to the flow per unit temperature gradient. Metals, being good conductors, have higher value of k compare to non-metals. For same metals, k decreases slightly as the temperature increases. However, the decrease is mostly negligible.

Solution

$\frac { dQ }{ dt } =-kA\frac { d\theta }{ dx } \\ \\ \frac { d\theta }{ dx } =\frac { -\frac { dQ }{ dt } }{ kA } \\ \\ \\ \frac { d\theta }{ dx } =\frac { -0.126 }{ (210)(3x{ 10 }^{ -4 }) } K{ m }^{ -1 }\\ \\ \\ \frac { d\theta }{ dx } =\quad -2.0\quad K{ m }^{ -1 }\\ \\$