# why does a larger thermal conductivity provide a smaller temperature gradient?

I was thinking about Fourier's Law in heat transfer today and for some reason I am just not understanding the relationships it gives us. Fourier's tells us that if the heat transfer rate is kept constant, then a larger thermal conductivity provides a smaller temperature gradient.

I am confused about the physical reason behind this or I am just misunderstanding the definition of thermal conductivity.

I thought that since thermal conductivity is ease of heat transfer through a material then a high thermal conductivity would mean heat is easily transferred so one side of the material is at a much temperature than the other and a large temperature gradient is created. In fact, this is actually the opposite

-

It's exactly the analog of electric resistance. A low resistance junction has a small voltage difference, a high resistance means you get a high potential difference for the same current

-

Think of it this way: In the middle of a cold winter, when you want to keep a large gradient between the outside world and the inside of your home, would you go for a high or a low thermal conductivity material for the walls?

-

The heat flux $Q [W/m^2]$ is determined via the temperature gradient and heat conductivity coefficient $\lambda$ like this:$$Q=\lambda\cdot(T_2-T_1)/L.$$ If you keep the temperature difference constant, it is the flux that increases with increasing $\lambda$. It is possible in case two liquids with high velocities are separated with a solid heat-conducting layer.

If you keep the flux constant, then there is no need in large temperature difference to provide it because high conductivity decreases the heat resistance of the material. The heat resistance is proportional to $1/\lambda$. Such a situation is possible when a heat source of a given power is cooled from exterior. The higher conductivity of its insulator, the colder is the source with respect to the exterior temperature.

-

Let's look at Fourier's law:

$$\vec q = -k \nabla T$$

So from basic dimensional analysis assuming we know the heat flux and the distance over which the gradient occurs, if the thermal conductivity is large, then the temperature difference should be small.

-