Question

A composite slab consists of two slabs $A$ and $B$ of different materials but of the same thickness placed one on top of the other. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. A steady temperature difference of $12{}^{\circ }C$ is maintained across the composite slab. If $k_1=\frac{k_2}{2},$ the temperature difference across the slab $A$ will be

• A. $4^{\circ }C$
• B. $8^{\circ }C$
• C. $12^{\circ }C$
• D. $16^{\circ }C$

Answer 1

Answer: (B)

For slab $A$ ,
$\frac{\Delta Q}{\Delta t}=\frac{k_{1}A}{\frac{l}{2}}\Delta T_1$
For slab $B$ ,
$\frac{\Delta Q}{\Delta t}=\frac{k_{2}A}{\frac{l}{2}}\Delta T_2$
$\therefore \frac{k_{1}A}{\frac{l}{2}}\Delta T_1=\frac{k_{2}A}{\frac{l}{2}}\Delta T_2$
$\frac{k_2}{2}\Delta T_1=k_2\left(12-\Delta T_2\right)$
$\Delta T_1=24-2\Delta T_1$
$\Delta T_1=8^{\circ }C$