Question

Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities $K$ and $2K$ respectively. The equivalent thermal conductivity of the slab is

• A. $\sqrt{2}K$
• B. $3K$
• C. $\frac{4}{3}K$
• D. $\frac{2}{3}K$

Answer 1

The correct option is C

$\frac{4}{3}K$

Explanation for the correct option:

Step 1: Given

Thermal conductivity of first slab is $K$.
Thermal conductivity of second slab is $2K$

Since the length of the two slabs are equal, let it be denoted by $l$.
Also, as their areas are also same, let them be denoted be $A$.

Step 2: Formulas used

Thermal resistance of a material is given as,
$R=\frac{l}{kA}$
where $l$ is the length of the material, $A$is its area and $K$ is the thermal conductivity

Thermal resistance of two materials connected in series is,
$R_{eq}=R_1+R_2+\dots$

Step 3: Calculating thermal conductivity

Since the slabs are in series, their thermal resistances add up.

So,
$\frac{2l}{K_{eq}A}=\frac{l}{KA}+\frac{l}{2KA}$ {Since the slabs are connected one after the another, their lengths add up}
$\begin{array}{cc}& \frac{l}{A}·\frac{2}{K_{eq}}=\frac{l}{A}\left(\frac{1}{K}+\frac{1}{2K}\right)\\ \Rightarrow & \frac{2}{K_{eq}}=\frac{2K+K}{2K^2}\\ \Rightarrow & \frac{K_{eq}}{2}=\frac{2K}{3}\\ \Rightarrow & K_{eq}=\frac{4}{3}K\end{array}$

Therefore, the equivalent thermal conductivity is $\frac{4}{3}K$.

Hence, option C is correct.