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°.You need to login to perform this action.
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