Animated Solution for Physics - Properties of Solids and Liquids: A heat source at $T=10^3$ K is connected to another heat reservoir at $T=10^2$ K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is $0.1{\text{ WK}}^{-1}{\text{m}}^{-1}$, the energy flux through it in the steady state is

Understanding Energy Flux

Imagine you are standing in a freezing room, holding a piping hot cup of coffee. You can feel the heat radiating into your hands. This transfer of thermal energy is one of the most fundamental processes in physics. But how do we measure exactly how much heat is flowing?

In this problem, we are tasked with finding the energy flux through a copper slab connecting two heat reservoirs. Let's break down the physics and the math behind this steady-state heat transfer.

We have two massive thermal bodies, known as heat reservoirs. A reservoir is so large that its temperature doesn't change even if it gains or loses heat.

Our hot reservoir is sitting at a scorching $T_1=10^3\text{ K}$ (which is $1000\text{ K}$). Our cold reservoir is at $T_2=10^2\text{ K}$ (or $100\text{ K}$). Connecting these two extremes is a copper slab with a thickness of $l=1\text{ m}$.

Because nature hates imbalances, heat will spontaneously flow from the hot reservoir to the cold one through the copper slab. The question states that the system is in a steady state. This is a crucial detail! It means the temperature profile inside the copper slab is no longer changing with time. The heat entering one side exactly equals the heat leaving the other side.

Heat Current and Flux

To find out how fast the heat is flowing, we use Fourier's Law of Heat Conduction. The rate of heat flow, often called the heat current ($H$), is given by:

Here, $k$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $T_1-T_2$ is the temperature difference, and $l$ is the thickness.

However, the question doesn't give us the area $A$, and it asks for the energy flux. What is energy flux? It is simply the heat current per unit area. Think of it as the intensity of the heat flow.

By dividing our heat current equation by the area $A$, we get the formula for energy flux:

Notice how beautifully the area $A$ cancels out. We don't need to know how wide the slab is; the flux is an intensive property that depends only on the material and the temperature gradient.

Now, we just need to substitute our known values into the flux equation. We are given the thermal conductivity of copper as $k=0.1{\text{ W m}}^{-1}{\text{K}}^{-1}$.

First, we calculate the temperature difference driving the heat flow:

Now, multiply this gradient by the thermal conductivity:

The final answer is $90{\text{ W m}}^{-2}$. This means that for every square meter of the copper slab, 90 Joules of heat energy pass through it every single second.

If you were to double the thickness of the slab to 2 meters, the flux would drop to $45{\text{ W m}}^{-2}$ because the temperature gradient would be less steep. Conversely, if you replaced the copper with a material that had a higher thermal conductivity $k$, the flux would proportionally increase. Always remember to look beyond the numbers and see the physical story they tell!