Q. 1.56 - An Introduction to Thermal Physics

Question

Calculate the rate of heat conduction through a layer of still air that is $1 mm$ thick, with an area of $1 m^2$, for a temperature difference of ${20}^{\circ }C$.

Step-by-Step Solution

Answer

The rate of heat conduction through a layer of still air is $\frac{Q}{\mathrm{\Delta }t}=-520\text{ watts .}$

Step: 1 Heat conduction definition:

Heat conduction is the transfer of thermal energy from one part of a substance to another via atomic or molecular interactions. Conduction is one of three primary ways of heat transport, and including convection and radiation.

Step: 2 Finding the heat conduction rate value:

The thermal conductivity of air is $k_t=0.026\mathrm{J}\times {\mathrm{s}}^{-1}\times {\mathrm{m}}^{-1}\times {\mathrm{K}}^{-1}$.

The still layer air with surface area of $1 m^2$and thickness of $1 mm$ and $∆T={20}^{\circ }K$,

$\begin{array}{l}\frac{Q}{\mathrm{\Delta }t}=-k_{t}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }x}\\ \frac{Q}{\mathrm{\Delta }t}=-0.026\times 1\times \frac{20}{0.001}\\ \frac{Q}{\mathrm{\Delta }t}=-520\text{ watts .}\end{array}$

The negative sign indicates the heat flows down the temperature gradient.