Question

A wall has two layers A and B made of different materials. The thickness of both layers is the same. The thermal conductivity of A and B are \[{K_A}\] and \[{K_B}\]​ such that \[{K_A} = 3{K_B}\]​. The temperature across the wall is \[{20^0}C\]. In thermal equilibrium
A. The temperature difference across \[A = {20^0}C\]
B. The temperature difference across \[A = {5^0}C\]
C. The temperature difference across \[A = {10^0}C\]
D. The rate of transfer of heat through A is more than that through B

Answer 1

Hint:In order to solve this problem we need to understand the rate of heat transfer and thermal conductivity. The rate of flow of heat is the amount of heat that is transferred per unit of time. Thermal conductivity is defined as the amount of heat transmitted through a material. Heat transfer occurs at a higher rate in materials of high thermal conductivity than in those of low thermal conductivity. Here, using the formula for heat flow we are going to find the solution.

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

Complete step by step solution:
Consider a wall that has two layers A and B made up of different materials. The thickness of both layers is the same. They have given the thermal conductivity of A and B are \[{K_A}\] and \[{K_B}\]​ such that \[{K_A} = 3{K_B}\]​. The temperature across the wall, that is,\[{\theta _1} - {\theta _2} = {20^0}C\]. The rate of flow of heat of A and B is,
\[{\left( {\dfrac{Q}{t}} \right)_A} = {K_A}A\dfrac{{\Delta \theta }}{L}\]
And, \[{\left( {\dfrac{Q}{t}} \right)_B} = {K_B}A\dfrac{{\Delta \theta }}{L}\]

In a series of connections, rate of flow of heat is same in A and B that is,
\[{\left( {\dfrac{Q}{t}} \right)_A} = {\left( {\dfrac{Q}{t}} \right)_B} \\ \]
\[\Rightarrow \dfrac{{{K_A}\left( {{\theta _1} - \theta } \right)}}{L} = \dfrac{{{K_B}\left( {\theta - {\theta _2}} \right)}}{L} \\ \]
\[{K_A}\left( {{\theta _1} - \theta } \right) = {K_B}\left( {\theta - {\theta _2}} \right)\]
Given that, \[{K_A} = 3{K_B}\] then,
\[3{K_B}\left( {{\theta _1} - \theta } \right) = {K_B}\left( {\theta - {\theta _2}} \right) \\ \]
\[\Rightarrow 3\left( {{\theta _1} - \theta } \right) = \left( {\theta - {\theta _2}} \right) \\ \]
\[\Rightarrow 3{\theta _1} - 3\theta = \theta - {\theta _2}\]……… (1)

Given that, \[{\theta _1} - {\theta _2} = {20^0}C\]
\[ \Rightarrow {\theta _2} = {\theta _1} - {20^0}C\]
Substitute the value in equation (1) we get,
\[3{\theta _1} - 3\theta = \theta - \left( {{\theta _1} - 20} \right) \\ \]
\[\Rightarrow 4{\theta _1} = 4\theta + 20 \\ \]
\[\Rightarrow 4{\theta _1} - 4\theta = 20 \\ \]
\[\Rightarrow 4\left( {{\theta _1} - \theta } \right) = 20 \\ \]
\[\Rightarrow \left( {{\theta _1} - \theta } \right) = \dfrac{{20}}{4} \\ \]
\[\therefore \left( {{\theta _1} - \theta } \right) = {5^0}C\]
Therefore, at thermal equilibrium, the temperature difference across the walls is \[{5^0}C\].

Hence, option B is the correct answer.

Note:Thermal conductivity is directly proportional to the cross-sectional area and inversely proportional to the distance. It can be expressed in terms of temperature, length, mass and time. The coefficient of thermal conductivity depends on the nature of the material.