question_answer

A wall has two layers A and B, made of two different materials. The thermal conductivity of material A is twice that of B. If two layers have the same thickness and under thermal equilibrium, the temperature difference across the wall is \[48{}^\circ C\], the temperature difference across layer B is:

A) \[40{}^\circ C\]

B) \[32{}^\circ C\]

C) \[16{}^\circ C\]

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

Correct Answer: B

Solution :

                 Let \[\Delta {{\theta }_{1}}\]and \[\Delta {{\theta }_{2}}\]be temperature differences across layers A and B. In series rate of flow is same. \[{{H}_{1}}={{H}_{2}}\] \[\frac{{{K}_{1}}\Delta {{\theta }_{1}}}{l}=\frac{{{K}_{2}}\Delta {{\theta }_{2}}}{l}\]  (Given: \[{{K}_{1}}=2{{K}_{2}}\] temperature difference across wall \[={{48}^{o}}\]) \[\frac{\Delta {{\theta }_{1}}}{\Delta {{\theta }_{2}}}=\frac{{{K}_{2}}}{{{K}_{1}}}\] \[\frac{\Delta {{\theta }_{1}}}{\Delta {{\theta }_{2}}}=\frac{{{K}_{2}}}{2{{K}_{2}}}=\frac{1}{2}\] \[\Delta {{\theta }_{2}}=2\Delta {{\theta }_{1}}\]                                              ?(i) Also, in series \[\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}}=48\]                                         ?(ii) Using Eqs.(i) and (ii) \[\Delta {{\theta }_{1}}+2\Delta {{\theta }_{1}}={{48}^{o}}\] \[3{{\theta }_{1}}={{48}^{o}}\] \[{{\theta }_{1}}={{16}^{o}}\]                 Also, \[\Delta {{\theta }_{2}}=48-\Delta {{\theta }_{1}}={{48}^{o}}-{{16}^{o}}={{32}^{o}}\] Hence, the temperature difference across wall B is 32°.