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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let ${\theta }_{1}and{\theta }_2$ be the temperatures at the two faces of the composite slab and let $\theta$ be the temperature at the common face of the slab. If is the length of each slab and A the area of their face, then, in the steady state, the rate of flow of heat across A = rate of flow of heat across B, i.e

$\frac{k_1({\theta }_1-\theta )}{1}=\frac{k_{2}A(\theta -{\theta }_2)}{1}Or{k}_1({\theta }_1-\theta )=k_2(\theta -{\theta }_2)$ Now $k_2=2k_1$.

Therefore $({\theta }_1-\theta )=2(\theta -{\theta }_2)$ (i)

Also, ${\theta }_1-{\theta }_2={12}^{\circ }Cor{\theta }_2-{\theta }_1=12$ (ii)

Using (ii) in (i) we have $({\theta }_1-\theta )=2\{\theta -({\theta }_1-12\left\}Or3\right({\theta }_1-\theta )=24Or{\theta }_1-\theta =8^{\circ }C$

Hence the correct choice is (b).