Rate of Heat Flow

1. Thermal Conductivity

Consider a slab (or cube) of a material of thickness $x$ and cross-sectional area $A$, with its two opposite faces kept at temperatures $T_1$ and $T_2$ where $T_1 > T_2$. Heat then flows steadily from the hot face to the cold face.

Experiment shows that the quantity of heat $Q$ conducted across in time $t$ is:

• directly proportional to the area $A$ of each face, $Q \propto A$;
• directly proportional to the temperature difference, $Q \propto (T_1 - T_2)$;
• directly proportional to the time of flow, $Q \propto t$;
• inversely proportional to the thickness $x$, $Q \propto \dfrac{1}{x}$.

Combining these:

$Q = \frac{k A (T_1 - T_2)\,t}{x}$

where $k$ is the constant of proportionality called the thermal conductivity of the material. The rate of heat flow is:

$\frac{Q}{t} = \frac{k A (T_1 - T_2)}{x}$

If $A = 1\ \text{m}^2$, $T_1 - T_2 = 1\ \text{K}$ and $x = 1\ \text{m}$, then $k = \dfrac{Q}{t}$. Hence the thermal conductivity of a material is numerically equal to the rate of heat flow between two faces, each of area $1\ \text{m}^2$, separated by a distance of $1\ \text{m}$ and maintained at a temperature difference of $1\ \text{K}$.

• SI unit of $k$: $\text{W m}^{-1}\text{K}^{-1}$.
• The quantity $\dfrac{T_1 - T_2}{x}$ is the temperature gradient, so $\dfrac{Q}{t} = k A \times (\text{temperature gradient})$.