A) \[\frac{{{k}_{1}}{{T}_{1}}{{d}_{2}}+{{k}_{2}}{{T}_{2}}{{d}_{1}}}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}\]
B) \[\frac{{{k}_{1}}{{T}_{1}}+{{k}_{2}}{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}}\]
C) \[\left( \frac{{{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}}{{{T}_{1}}+{{T}_{2}}} \right){{T}_{1}}{{T}_{2}}\]
D) \[\frac{{{k}_{1}}{{d}_{1}}{{T}_{1}}+{{k}_{2}}{{d}_{2}}{{T}_{2}}}{{{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}}\]
Correct Answer: A
Solution :
In series both walls have same rate of heat flow. Therefore \[\frac{dQ}{dt}=\frac{{{K}_{1}}A({{T}_{1}}-\theta )}{{{d}_{1}}}=\frac{{{K}_{2}}A(\theta -{{T}_{2}})}{{{d}_{2}}}\] \[\Rightarrow {{K}_{1}}{{d}_{2}}({{T}_{1}}-\theta )={{K}_{2}}{{d}_{1}}(\theta -{{T}_{2}})\] \[\Rightarrow \theta =\frac{{{K}_{1}}{{d}_{2}}{{T}_{1}}+{{K}_{2}}{{d}_{1}}{{T}_{2}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}\]You need to login to perform this action.
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