
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} $

Answer
466.2k+ views
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).
Note
Unless 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.
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).
Note
Unless 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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How do I convert ms to kmh Give an example class 11 physics CBSE

Describe the effects of the Second World War class 11 social science CBSE

Which of the following methods is suitable for preventing class 11 chemistry CBSE
