Question

The figure shows the face and interface temperature of a composite slab consisting of four materials, of identical thickness, through which the heat transfer is steady. Which material has minimum thermal conductivity?

• A. $\text{a}$
• B. $\text{b}$
• C. $\text{c}$
• D. $\text{d}$

Answer 1

The correct option is C $\text{c}$

Rate of heat transfer i.e thermal current is given by

$i_{th}=-kA\frac{\Delta T}{\Delta x}$

$i_{th}==kA\frac{T_{big}-T_{small}}{\Delta x}$

$\Rightarrow k=\left(\frac{i_{th}\times \Delta x}{A}\right)\left(\frac{1}{T_{big}-T_{small}}\right)$

In steady state, $i_{th}$ is the same for all slabs. Since all the slabs have the same thickness and cross-sectional area, $\Delta x$ and $A$ are constant.

i.e $k=\frac{C}{T_{big}-T_{small}}$ for constant $C$.

Therefore, $k_a=\frac{C}{25-15}=\frac{C}{10}$
$k_b=\frac{C}{15-10}=\frac{C}{5}$
$k_c=\frac{C}{10-(-5)}=\frac{C}{15}$
$k_d=\frac{C}{-5-(-10)}=\frac{C}{5}$

So. the minimum thermal conductivity is $k_c$.