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