First slide

Conduction

Question

Two slabs A and B of different materials but of the same thickness are joined end- to-end to form a composite slab. The thermal conductivities of A and B are K₁ and K₂, respectively. A steady temperature difference of 12⁰C is maintained across the composite slab. If $K_{1} = \frac{K_{2}}{2}$ the temperature difference across slab A is

Moderate

A

4⁰C
B

6⁰C
C

8⁰C
D

10⁰C

Solution

The given situation can be shown as below.

Rate of flow of heat will be equal in both the slabs,
$(12 - x) K_{1} = K_{2} (x - 0)$
$12 - x = 2 x$
$\Rightarrow x = 4^{0} C$ $(∵ K_{1} = \frac{K_{2}}{2})$
The temperature difference across slab,
$A = (12 - x) = (12 - 4) = 8^{\circ} C$

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