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$A$ wall is made up of two layers $A$ and $B$ . The thickness of the two layers is the same, but materials are different. The thermal conductivity of $A$ is double than that of $B$ . In thermal equilibrium the temperature difference between the two ends is ${36^o}C$. Then the difference of temperature at the two surfaces of $A$ will be ....... $^oC$
$6$
$12$
$18$
$24$
Solution
(b) Suppose thickness of each wall is $x$ then ${\left( {\frac{Q}{t}} \right)_{combination}} = {\left( {\frac{Q}{t}} \right)_A}$
==> $\frac{{{K_S}A({\theta _1} – {\theta _2})}}{{2x}} = \frac{{2KA({\theta _1} – \theta )}}{x}$
$\because {K_S} = \frac{{2 \times 2K \times K}}{{(2K + K)}} = \frac{4}{3}K$ and $({\theta _1} – {\theta _2}) = 36^\circ $
==> $\frac{{\frac{4}{3}KA \times 36}}{{2x}} = \frac{{2KA({\theta _1} – \theta )}}{x}$
Hence temperature difference across wall $A$ is $({\theta _1} – \theta ) = {12^o}C$
