Q.SI unit of thermal conductivity is

You have probably touched a metal spoon left in a hot pan and yelped, then touched the wooden handle a second later and felt nothing. That difference — the speed at which heat travels through the material — is what thermal conductivity measures.

The intuition: heat as a "flow"

Think of heat like water. A thick copper rod is like a wide, open pipe — heat rushes through it easily. A piece of wood is like a narrow, clogged pipe — heat barely trickles through. Thermal conductivity is the number that tells you how "open" that pipe is for heat flow.

When one end of a rod is hot and the other is cold, heat energy moves from the hot end to the cold end. Some materials let this happen fast (metals), others are stubborn (wood, air, plastic). That property is thermal conductivity, usually denoted by $k$ (or $K$).

The precise statement

For a slab of material of thickness $L$ and cross-sectional area $A$, with a temperature difference $\Delta T$ across its faces, the rate at which heat flows through it is:

Here:

• $Q/t$ is the heat transfer rate (in joules per second, i.e. watts)
• $k$ is the thermal conductivity (units: $\text{W} / (\text{m} \cdot \text{K})$)
• $A$ is the area (larger area = more heat flow)
• $\Delta T$ is the temperature difference (bigger difference = faster flow)
• $L$ is the thickness (thicker slab = slower flow)

This is Fourier's law of heat conduction in its simplest form. It says: heat flows at a rate proportional to the area and the temperature gradient, and inversely proportional to the thickness.

What the number $k$ actually means

A high $k$ means the material is a good conductor — heat moves through it quickly. Copper has $k \approx 400 \, \text{W/(m·K)}$. A low $k$ means the material is an insulator — heat moves slowly. Wood has $k \approx 0.1 \, \text{W/(m·K)}$, and still air is even lower, around $0.025$.

Why the formula makes sense

If you double the area $A$, twice as many atoms are passing heat along, so the rate doubles. If you double the thickness $L$, the heat has to travel twice as far, so the rate halves. If you double the temperature difference $\Delta T$, the "push" for heat is twice as strong, so the rate doubles. The constant $k$ just scales everything to match the material.

A common exam point …