Question

A wall has two layers A and B, each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is ${36^ \circ }C$. The temperature difference across the layer A is:
$\left( A \right){6^ \circ }C$
$\left( B \right){12^ \circ }C$
$\left( C \right){18^ \circ }C$
$\left( D \right){24^ \circ }C$

Answer 1

Hint: The ability of a given material to conduct or transfer heat is called thermal conductivity. Here the thickness and thermal conductivity of the two walls A and B is given. Heat flow will be constant since it is a steady state. Hence equate the heat flow on both the walls. Also find the temperature difference between the walls. Combine all the equations to obtain the temperature difference at layer A.

Formula used:
$Q = KA\dfrac{{dt}}{{dx}}$ where, $Q$ is the heat flow, $K$ is the thermal conductivity, $A$ is the area.

Complete step by step answer:
The ability of a given material to conduct or transfer heat is called thermal conductivity. It is denoted generally by the symbol $k$. Thermal resistivity is the reciprocal to thermal conductivity. Thermal conductivity can be expressed in terms of temperature, length, mass and time. Watt per meter kelvin is the unit of thermal conductivity of a material.
High value thermal conductivity materials are used in heat sinks and with low value of thermal conductivity are used in thermal insulators. Transferred heat is proportional to the negative of the temperature gradient and is proportional to the area through the heat flows.
Let us consider two walls of thickness $t$ each and thermal conductivity of $2k$ and $k$ respectively.
Temperature at wall A = ${T_1}$
Temperature at wall B = ${T_2}$
Temperature at the junction = $T$
Heat flow is constant since it is in steady state
$\dfrac{{2KA\vartriangle {T_1}}}{l} = \dfrac{{KA\vartriangle {T_2}}}{l}$
$ \Rightarrow \dfrac{{\vartriangle {T_1}}}{{\vartriangle {T_2}}} = \dfrac{1}{2}$
$ \Rightarrow \dfrac{{\vartriangle {T_1}}}{{36 - \vartriangle {T_2}}} = \dfrac{1}{2}$
$ \therefore \vartriangle {T_1} = {12^ \circ }C$

Hence the correct option is $\left( B \right)$.

Note: Thermal conductivity is directly proportional to the area of cross-section and inversely proportional to the distance. Thermal conductivity can be expressed in terms of temperature, length, mass and time. Coefficient of thermal conductivity depends on the nature of the material. Here the temperature decreases with increase in length in the direction of heat transfer.