Two materials having coefficients of thermal conductivity ′3K′ and ′K′ and thickness 'd' and ′3d′, respectively, are joined to form a slab as shown in the figure. The temperature of the outer surfaces are $ \prime {\theta _2}\prime $ and $ \prime {\theta _1}\prime \; $ respectively, $ \left( {{\theta _2} {\theta _1}} \right). $ . The temperature at the interface is?

A) $ \dfrac{{{\theta _1} + {\theta _2}}}{2} $ B) $ \dfrac{{{\theta _1}}}{{10}} + \dfrac{{9{\theta _2}}}{{10}} $ C) $ \dfrac{{{\theta _1}}}{3} + \dfrac{{2{\theta _2}}}{3} $ D) $ \dfrac{{{\theta _1}}}{6} + \dfrac{{5{\theta _2}}}{6} $

Question

Two materials having coefficients of thermal conductivity ′3K′ and ′K′ and thickness 'd' and ′3d′, respectively, are joined to form a slab as shown in the figure. The temperature of the outer surfaces are $' θ_{2}'$ and $' θ_{1}'$ respectively, $(θ_{2} > θ_{1}) .$ . The temperature at the interface is?

A) $\frac{θ_{1} + θ_{2}}{2}$
B) $\frac{θ_{1}}{10} + \frac{9 θ_{2}}{10}$
C) $\frac{θ_{1}}{3} + \frac{2 θ_{2}}{3}$
D) $\frac{θ_{1}}{6} + \frac{5 θ_{2}}{6}$

A)  $ \dfrac{{{\theta _1} + {\theta _2}}}{2} $ B)  $ \dfrac{{{\theta _1}}}{{10}} + \dfrac{{9{\theta _2}}}{{10}} $ C)   $ \dfrac{{{\theta _1}}}{3} + \dfrac{{2{\theta _2}}}{3} $ D)  $ \dfrac{{{\theta _1}}}{6} + \dfrac{{5{\theta _2}}}{6} $

Vedantu Content Team · Accepted Answer

Hint :  In this solution, we will assume that the system has been in a steady-state for a long time. Then both the slabs will have the same rate of heat transfer and we will use this to determine the solution of the interface. Formula used: In this solution, we will use the following formula:Rate of heat transfer in a slab with different temperatures:  $ Q = \dfrac{{kA\Delta T}}{L} $  where  $ A $  is the area of the surface,  $ \Delta T $  is the temperature difference, and  $ L $  is the length of the slab. Complete step by step answer In the system given to us, we will assume that the system has been left in this situation for a long time. This implies that the heat transfer will have occurred over a long time and the system can be said to have achieved a steady-state. In this steady-state situation, we can say that the heat transfer in both the slabs will be the same. Let us assume that the temperature of the surface is  $ \theta  $ . Then since the heat flow rate is constant for both the surfaces will be the same and we can write $ \dfrac{{3kA({\theta _2} - \theta )}}{d} = \dfrac{{kA(\theta  - {\theta _1})}}{{3d}} $ Dividing both sides by  $ kA/d $  and cross multiplying the denominators, we get $ 9{\theta _2} - 9\theta  = \theta  - {\theta _1} $ Solving for  $ \theta  $ , we get $\Rightarrow \theta  = \dfrac{{9{\theta _2}}}{{10}} + \dfrac{{{\theta _1}}}{{10}} $  which corresponds to option (B).NoteUnless mentioned otherwise, we must take such systems to be in a steady-state i.e. the system has been in this state for a long time. Only with this assumption can we assume that both the slabs will have the same rate of heat transfer.

Repeaters Course for NEET 2022 - 23