Animated Solution for Physics - Properties of Solids and Liquids: Two materials having coefficients of thermal conductivity '$3K$' and '$K$' and thickness '$d$' and '$3d$' respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are '${\theta }_2$' and '${\theta }_1$' respectively, (${\theta }_2>{\theta }_1$). The temperature at the interface is

A Tale of Two Slabs

Imagine heat flowing like a river through a channel. In our problem, this channel is made of two distinct materials joined end-to-end, forming a composite slab. The first slab is highly conductive with a thermal conductivity of $3K$ and a relatively thin profile of $d$. The second slab is less conductive, with a thermal conductivity of $K$, and is three times as thick, measuring $3d$.

The outer surfaces are maintained at temperatures ${\theta }_2$ and ${\theta }_1$, where ${\theta }_2>{\theta }_1$. Because of this temperature difference, heat naturally flows from the hotter end (${\theta }_2$) to the cooler end (${\theta }_1$). Our mission is to find the exact temperature at the junction where these two slabs meet. Let's call this unknown interface temperature $\theta$.

Steady State

The magic phrase in this problem is steady state. But what does it actually mean? In a steady state, the temperature at any given point within the slabs no longer changes with time. Consequently, the rate at which heat enters the first slab is exactly equal to the rate at which it leaves the second slab. There is no accumulation or loss of heat energy anywhere in between.

Mathematically, this means the heat current $H_1$ through the first slab is perfectly equal to the heat current $H_2$ through the second slab:

Fourier's Law in Action

To express these heat currents, we rely on Fourier's Law of Heat Conduction. It states that the rate of heat flow $H$ is directly proportional to the cross-sectional area $A$, the temperature difference $\Delta \theta$, and the thermal conductivity $K$, while being inversely proportional to the thickness $L$:

Let's apply this law to each slab individually.

For the first slab, the heat flows from ${\theta }_2$ to $\theta$. So, the temperature drop is $({\theta }_2-\theta )$. Plugging in its specific properties:

For the second slab, the heat continues its journey from $\theta$ to ${\theta }_1$. The temperature drop here is $(\theta -{\theta }_1)$:

Now, we equate the two expressions based on our steady-state principle:

Algebra to the Rescue

At first glance, the equation might look a bit cluttered. However, notice that the area $A$, the base thermal conductivity $K$, and the base thickness $d$ are common to both sides. We can elegantly cancel them out, leaving us with a much cleaner relationship:

To eliminate the fraction, we cross-multiply by moving the $3$ from the denominator on the right to the numerator on the left:

Expanding the bracket gives:

Our goal is to isolate $\theta$. Let's group all the terms containing $\theta$ on the right side and the rest on the left:

Finally, dividing both sides by $10$ reveals the interface temperature:

Which can be rewritten to match the options perfectly:

Notice how the interface temperature is heavily weighted towards ${\theta }_2$ (with a factor of $9/10$). This makes perfect physical sense! The first slab is highly conductive and very thin, meaning it offers very little thermal resistance. Therefore, the temperature doesn't drop much across it, keeping the interface temperature quite close to the hot end.