Problem — From graph to conductivity

Problem

A slab of material with area $A = 0.5$ m² and thickness $d = 0.02$ m separates two environments. The heat power transmitted is recorded as the imposed temperature difference is varied:

$\Delta T$ (K)	5	10	20	30
$\dot Q$ (W)	5	10	20	30

Determine the material’s thermal conductivity $k$.

Solution

Reading the data. The power is proportional to the temperature gap: the ratio is constant, $\frac{\dot Q}{\Delta T} = 1\ \mathrm{W/K}$ This is the slope of the line $\dot Q$ against $\Delta T$. Fourier’s law. Since $\dot Q = k \, A \, \dfrac{\Delta T}{d}$, the slope is $\dfrac{\dot Q}{\Delta T} = \dfrac{k \, A}{d}$. We obtain $k$: $k = \frac{\dot Q}{\Delta T} \cdot \frac{d}{A} = 1 \cdot \frac{0.02}{0.5} = 0.04\ \mathrm{W/(m \cdot K)}$ The very low value indicates a good thermal insulator. $\ev{k = 0.04\ \mathrm{W/(m \cdot K)}}$

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Topics: Termologia Concepts: Trasmissione del calore