Question

Two rods of the same length and cross-section are joined along the length. The thermal conductivities of the first and second rod are \[{K_1}\]​ and \[{K_2}\]. The temperature of the free ends of the first and second rods are maintained at \[{\theta _1}\]​ and \[{\theta _2}\]​ respectively. Find the temperature of the common junction.
A. \[\dfrac{{{\theta _1} + {\theta _2}}}{2}\]
B. \[\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}\left( {{\theta _1} + {\theta _2}} \right)\]
C. \[\dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
D. \[\dfrac{{{K_2}{\theta _1} + {K_1}{\theta _2}}}{{{K_1} + {K_2}}}\]

Answer 1

Hint: In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time. Here, using the formula for heat flow we are going to find the solution. When the two rods are kept in contact with each other then, the temperature at the junction of 2 rods is known as junction temperature.

Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta \theta }}{L}\]
Where,
A is cross-sectional area of metal rod
\[\Delta \theta \] is temperature difference between two ends of the metal rod
L is length of the metal rod
K is thermal conductivity

Complete step by step solution:
Consider two rods of the same length and cross-section are joined along the length. The thermal conductivities of the first and second rod are given as \[{K_1}\]​ and \[{K_2}\]. If the temperature of the free ends of the first and second rods are maintained at \[{\theta _1}\]​ and \[{\theta _2}\]​. We need to find the temperature of the common junction.

The rate of flow of heat for both blocks is,
\[{\left( {\dfrac{Q}{t}} \right)_1} = {K_1}A\dfrac{{\Delta \theta }}{{{L_1}}}\]
And, \[{\left( {\dfrac{Q}{t}} \right)_2} = {K_2}A\dfrac{{\Delta \theta }}{{{L_2}}}\]
As we know that, at steady-state, the rate of flow of heat for both blocks will be the same. That is,
\[{\left( {\dfrac{Q}{t}} \right)_1} = {\left( {\dfrac{Q}{t}} \right)_2} \\ \]
\[\Rightarrow {K_1}A\dfrac{{\Delta \theta }}{{{L_1}}} = {K_2}A\dfrac{{\Delta \theta }}{{{L_2}}} \\ \]
\[\Rightarrow {K_1}A\dfrac{{\left( {{\theta _1} - \theta } \right)}}{{{L_1}}} = {K_2}A\dfrac{{\left( {\theta - {\theta _2}} \right)}}{{{L_2}}} \\ \]

Given, \[{L_1} = {L_2}\]
Then, above equation will become,
\[{K_1}A\left( {{\theta _1} - \theta } \right) = {K_2}A\left( {\theta - {\theta _2}} \right) \\ \]
\[\Rightarrow {K_1}A{\theta _1} - {K_1}A\theta = {K_2}A\theta - {K_2}A{\theta _2} \\ \]
\[\Rightarrow {K_1}A{\theta _1} + {K_2}A{\theta _2} = {K_1}A\theta + {K_2}A\theta \\ \]
\[\Rightarrow A\left( {{K_1}{\theta _1} + {K_2}{\theta _2}} \right) = A\theta \left( {{K_1} + {K_2}} \right) \\ \]
\[\Rightarrow \left( {{K_1}{\theta _1} + {K_2}{\theta _2}} \right) = \theta \left( {{K_1} + {K_2}} \right) \\ \]
\[\therefore \theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Therefore, the temperature of the common junction is, \[\dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\].

Hence, option C is the correct answer.

Note: Here, we have considered two rods, in such cases, two rods have different thermal conductivities and are maintained at different temperatures. We found the value of temperature of the common junction by using the heat flow formula.