A) \[\frac{A{{K}_{0}}}{bl}\,\left( \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right)\]
B) \[\frac{A{{K}_{0}}}{bl}\left( \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right)\]
C) \[\frac{A{{K}_{0}}}{bl}\,In\,\left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]
D) \[\frac{A{{K}_{0}}}{al}\,In\,\left[ \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right]\]
Correct Answer: C
Solution :
As, we know that, \[\Rightarrow \] \[\frac{dQ}{dt}=-KA\frac{dT}{dx}\] \[\Rightarrow \] \[\frac{dQ}{dt}=-\frac{{{K}_{0}}A}{a+bT}\frac{dT}{dx}\] On entegrating both sides within the proper limits. \[\frac{dQ}{dt}\,\int_{0}^{l}{dx=-{{K}_{0}}A}\,\int_{{{T}_{1}}}^{{{T}_{2}}}{\frac{dT}{a+bT}}\] This gives, \[\frac{dQ}{dt}=\frac{A{{K}_{0}}}{b\,l}\,In\,\,\left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]You need to login to perform this action.
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