How fast does heat travel via conduction?

Question

I have read this question which seems to ask an identical question, but I'm not sure - it had far too many words I don't understand, let alone the equations. Perhaps someone can answer with a heat-for-dummies answer.

I understand that given thermal conductivity, and thermal mass, it takes a while for heat applied to one end of a material to make the other side rise in temperature. Like for a heat sink for electronics, with the component dissipating a constant power, and the ambient temperature staying constant, a thermal gradient will develop until it reaches an equilibrium, where the power dissipated by the electronic component equals the power dissipated by the heat sink to the ambient environment.

But my question is: say the heat sink is exactly at ambient temperature. And say the electronic component instantly starts dissipating to the heatsink a given power. How long would it take for the heatsink to start dissipating even the smallest amount of power to the environment? My guess is it would be equal to the speed of the molecules' vibrations that is otherwise known as heat.

Or perhaps at the speed of light, since thermal radiation would penetrate the material, even if it is very, very little.

I hope this makes sense?

Answer 1

The problem is quite complex to solve quantitatively and requires a differential calculus of multi-variable functions, but I'll try to simplify it.

Imagine that the object consist of many thin slices across the temperature gradient. Every second slice is a heat container with heat capacity $c \left[ \frac{J}{K}\right]$ and the remaining are heat conductors of a heat conductivity $h \left[ \frac{W}{K}\right]$ (just to separate two effects -- conductivity and capacity)

A the first instant all layers are at an ambient temperature. When you touch left side of the conductor, heat starts flowing, but the object is not at the equilibrium.

Heat flows into segment $A$ at the rate: $$P_{A\,in}=h(T_{left}-T_A)$$ And the temperature of $A$ starts increasing. $$\frac{dT_A}{dt} = \frac{P_{A\,in}-P_{A\,out}}{c}$$ When $T_A$ is just slightly over the ambient temperature it starts giving heat out to segment $B$ $$P_{A\,out}=P_{B\,in}=h(T_A-T_B)$$ And the temperature of $B$ starts rising which affects segment $C$, and so on. This process continues until temperature gradient is established and heat absorbed and dissipated are equal.

The speed ot this process depends on density of the material, heat capacity, thickness, surface area, conductivity and much more. Although the process starts immediately, it takes some time until it becomes observable. In this case radiation does not contribute much to energy transfer since most conductors are opaque. Heat is mainly transferred through the collisions of the molecules.

I can show you derivation of temperature function for a single segment $A$ to illustrate the whole process. $$P_{in}=h(T_{left}-T)$$ $$P_{out}=h(T-T_{amb})$$ $$\frac{dT}{dt} = \frac{P_{in}-P_{out}}{c} = \frac{h}{c}(T_{left}-2T+T_{amb})$$ $$\frac{dT}{T_{left}-2T+T_{amb}} = \frac{h}{c}dt$$ Solving this differential equation we obtain: $$\frac{\ln(-2T+T_{left}+T_{amb})}{-2} = \frac{h}{c}t + C$$ Which, after some transformations, gives: $$T = A \exp\left(\frac{-2h}{c}t\right)+\frac{1}{2}(T_{left}+T_{amb})$$ Where $A$ depends on initial conditions. In this case it is: $$A=\frac{1}{2}(T_{amb}-T_{left})$$

You may notice, that for $t=0$, $T=T_{amb}$; and, after very long time, when equilibrium is established, $T$ is just between $T_{left}$ and $T_{amb}$.

Answer 2

I think the average velocity can be estimated from the average time required to move distance L: $$t = L^2/2\alpha$$ where $\alpha$ is the thermal conductivity (k) divided by heat capacity (C).

Since velocity (u) is distance/time:

$$ u = L/(L^2/2\alpha) = 2\alpha/L $$

Answer 3

At near the speed of light. The same maxwells equations that describe the conduction of electric discribe the conduction of thermal energy.