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Conduction

Two slabs A...

Question

Two slabs $A$ and $B$ of equal surface areas are placed one over the other such that their surfaces are completely in contact. The thickness of slab $A$ is twice that of $B$ . The coefficient of thermal conductivity of slab A is twice that of $B$ . The first surface of slab $A$ is maintained at $100$ $^{o}$ C , while the second surface of slab B is maintained at $25$ $^{o}$ C . The temperature at the contact of their surface is

A

15^{o} C

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B

{62.5}^{o} C

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C

55^{o} C

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D

85^{o} C

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Solution

The correct option is B ${62.5}^{o} C$
We know that for a plane wall heat dissipation is give as:
$Q = \frac{K A (T_{1} - T_{2})}{l}$
So, equating the heat dissipating from both slab we get a relation:
$\frac{T_{1} - T_{2}}{L} = \frac{T_{c} - T_{2}}{l} = \frac{100 - 25}{2 l} = \frac{T - 25}{l}$

$T = {62.5}^{0} C$

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