A) \[{{6}^{o}}C\]
B) \[{{12}^{o}}C\]
C) \[{{18}^{o}}C\]
D) \[{{24}^{o}}C\]
Correct Answer: B
Solution :
Suppose thickness of each wall is x then \[{{\left( \frac{Q}{t} \right)}_{combination}}={{\left( \frac{Q}{t} \right)}_{A}}\]Þ \[\frac{{{K}_{S}}A({{\theta }_{1}}-{{\theta }_{2}})}{2x}=\frac{2KA({{\theta }_{1}}-\theta )}{x}\] \[\because \] \[{{K}_{S}}=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}K\] and \[({{\theta }_{1}}-{{\theta }_{2}})=36{}^\circ \] Þ \[\frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA({{\theta }_{1}}-\theta )}{x}\] Hence temperature difference across wall A is \[({{\theta }_{1}}-\theta )={{12}^{o}}C\]You need to login to perform this action.
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