Question

A partition wall has two layers of different materials $A$ and $B$ in contact with each other. They have the same thickness but the thermal conductivity of layer $A$ is twice that of layer $B$. At steady state, if the temperature difference across the layer $B$ is $50K$, then the corresponding difference across the layer $A$ is
$\begin{align}
  & A)50K \\
 & B)12.5K \\
 & C)25K \\
 & D)60K \\
\end{align}$

Answer 1

Hint: When two different materials are in contact with each other, heat transfer in both these materials will be the same. Heat transfer in a material is dependent on the difference in temperature across the material. It is also dependent on the area of the cross section of the material. These concepts are given by Fourier’s law or the law of heat conduction.

Formula used:
$\Delta Q=-\dfrac{KA\Delta T}{l}$

Complete step-by-step answer:
Fourier’s law or the law of heat conduction states that heat transfer in a material is proportional to the temperature difference across the material as well as the area of cross-section of the material. Fourier’s law is mathematically represented as follows.
$\Delta Q=-\dfrac{KA\Delta T}{l}$
where
$\Delta Q$ is heat transfer or change in heat during a particular time interval, in a material
$K$ is the thermal conductivity of the material
$\Delta T$ is the difference in temperature across the material
$l$ is the thickness of the material
Let this be equation 1.
Coming to the question, we are provided with a partition wall made of two different layers of materials $A$ and $B$, in contact with each other. It is given that thermal conductivity of material $A$ is twice that of material $B$. It is also given that the thickness of both these materials are the same. At steady state, if the temperature difference across material $B$ is $50K$, then, we are required to find the corresponding temperature difference in material $A$.
Applying the law of heat induction to material $A$, we have
$\Delta {{Q}_{A}}=-\dfrac{{{K}_{A}}{{A}_{A}}\Delta {{T}_{A}}}{{{l}_{A}}}$
where
$\Delta {{Q}_{A}}$ is the change in heat or heat transfer in material $A$
${{K}_{A}}$ is the thermal conductivity of material $A$
${{A}_{A}}$ is the area of cross-section of material $A$
${{l}_{A}}$ is the thickness of material $A$
$\Delta {{T}_{A}}$ is the change in temperature across material $A$
Let this be equation 2.
Similarly, if we apply the law of heat conduction to material $B$, we have
$\Delta {{Q}_{B}}=-\dfrac{{{K}_{B}}{{A}_{B}}\Delta {{T}_{B}}}{{{l}_{B}}}$
where
$\Delta {{Q}_{B}}$ is the change in heat or heat transfer in material $B$
${{K}_{B}}$ is the thermal conductivity of material $B$
${{A}_{B}}$ is the area of cross-section of material $B$
${{l}_{B}}$ is the thickness of material $B$
$\Delta {{T}_{B}}$ is the change in temperature across material $B$
Let this be equation 3.
Now, since both these materials are in contact, heat transfer in both these materials will be equal. Therefore, equation 2 and equation 3 can be equated as follows:
$\Delta {{Q}_{A}}=\Delta {{Q}_{B}}\Rightarrow -\dfrac{{{K}_{A}}{{A}_{A}}\Delta {{T}_{A}}}{{{l}_{A}}}=-\dfrac{{{K}_{B}}{{A}_{B}}\Delta {{T}_{B}}}{{{l}_{B}}}$
Here, we know that
$\begin{align}
  & {{K}_{A}}=2{{K}_{B}} \\
 & {{l}_{A}}={{l}_{B}} \\
 & {{A}_{A}}={{A}_{B}} \\
\end{align}$
and
$\Delta {{Q}_{B}}=50K$, as provided in the question
Substituting these values in the above expression, we have
$-\dfrac{{{K}_{A}}{{A}_{A}}\Delta {{T}_{A}}}{{{l}_{A}}}=-\dfrac{{{K}_{B}}{{A}_{B}}\Delta {{T}_{B}}}{{{l}_{B}}}\Rightarrow 2{{K}_{B}}\Delta {{T}_{A}}={{K}_{B}}\Delta {{T}_{B}}\Rightarrow \Delta {{T}_{A}}=\dfrac{\Delta {{T}_{B}}}{2}=\dfrac{50K}{2}=25K$
Therefore, at steady state, if the temperature difference across material $B$ is $50K$, then, the corresponding temperature difference in material $A$ is equal to $25K$.

So, the correct answer is “Option C”.

Note: Students need not get worried on the negative sign in the expression for heat transfer in a material, as given by Fourier’s law. Fourier’s law or law of heat conduction states that change in heat in a material during a particular time interval is proportional to the negative temperature gradient across that material. This suggests that direction of heat transfer in a material will be in a direction, opposite to the direction of temperature change.