1. Tardigrade
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  4. 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 12°C is maintained across the composite slab. If K1 = (K2/2) the temperature difference across slabs A is

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Q. 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 $K_1$ and $K_2$ respectively. A steady temperature difference of 12°C is maintained across the composite slab. If $K_1 = \frac{K_2}{2}$ the temperature difference across slabs $A$ is

AIIMSAIIMS 2012

Solution:

The given situation can be shown as
image
Rate of flow of heat will be equal in both the slabs
$\therefore (12 - x)K_1 = K_2 (x - 0)$
$ 12 - x = 2x \left(\because K_1 = \frac{K_2}{2}\right)$
$\Rightarrow x = 4^{\circ} C$
The temperature difference across slab
$A = (12 - x) = (12 - 4)$
$ = 8^{\circ}C$