question_answer

A wall has two layers A and B, each made of a different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal   equilibrium,   the   temperature difference across the wall is \[36{}^\circ C\]. The temperature difference across the layer A is

A) \[6{}^\circ C\]

B) \[12{}^\circ C\]

C) \[18{}^\circ C\]

D) \[24{}^\circ C\]

Correct Answer: B

Solution :

Heat flowing through the layers \[Q=\frac{KA\Delta \theta }{d}t\] Now, if Q, A, d, t are constant, then                 \[\Delta \,{{\theta }_{A}}{{K}_{A}}=\Delta \,{{\theta }_{B}}{{K}_{B}}\]                 \[({{\theta }_{1}}-\theta ).2K=(\theta -{{\theta }_{2}})K\] or            \[2({{\theta }_{1}}-\theta )=(\theta -{{\theta }_{2}})\]    ?.(i) \[\because \]     \[{{\theta }_{1}}-{{\theta }_{2}}={{36}^{o}}C\]                 \[{{\theta }_{2}}={{\theta }_{1}}-36\] From Eq. (i)                 \[2({{\theta }_{1}}-\theta )=\theta -{{\theta }_{1}}+36\] or            \[3({{\theta }_{1}}-\theta )=36\] or            \[{{\theta }_{1}}-\theta =12\] So, temperature difference across A is\[{{12}^{o}}C\].