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Properties of Water

Two walls of ...

Question

Two walls of thickness $d_{1}$ and $d_{2}$ , thermal conductivities $K_{1}$ and $K_{2}$ are in contact. In the steady state if the temperature at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature are the common wall will be

A

\frac{K_{1} T_{1} + K_{2} T_{2}}{d_{1} d_{2}}

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B

\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}

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C

\frac{K_{1} d_{1} T_{1} + K_{2} d_{2} T_{2}}{K_{1} d_{1} + K_{2} d_{2}}

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D

\frac{(K_{1} d_{1} + K_{2} d_{2}) T_{1} T_{2}}{T_{1} + T_{2}}

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Solution

The correct option is B $\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$
$\begin{matrix} k_{1} \frac{T_{1} - T_{c}}{d_{1}} = k_{2} \frac{T_{c} - T_{2}}{d_{2}} w h e r e T_{c} i s c o m m o n w a l l t e m p e r a t u r e, s o l v i n g f o r T_{c} w e w i l l g e t T_{C} = \frac{T_{1} + α T_{2}}{α + 1} u sin g α = \frac{d_{1}}{d_{2}} \frac{k_{2}}{k_{1}} T_{c} = \frac{T_{1} + \frac{d_{1}}{d_{2}} \frac{k_{2}}{k_{1}} T_{2}}{\frac{d_{1}}{d_{2}} \frac{k_{2}}{k_{1}} + 1} = \frac{T_{1} (d_{2} k_{1}) + T_{2} d_{1} k_{2}}{d_{1} k_{2} + d_{2} k_{1}} H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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