Question

A carpenter builds an outer house wall with a layer of wood $2.0\,{\text{cm}}$ thick on the outside and a layer of an insulation $3.5\,{\text{cm}}$ thick as the inside wall surface. The wood has $k = 0.03\,{\text{W}}/{\text{m}} - {\text{K}}$ and the insulation has $k = 0.01\,{\text{W}}/{\text{m}} - {\text{K}}$ . The interior surface temperature is $19^\circ {\text{C}}$ and the exterior surface temperature is $- 10^\circ {\text{C}}$ . What is the temperature at the plane where the wood meets the insulation? (in ${\text{K}}$ )

Answer 1

First of all, we will find the expression for the heat current flow for the wood and the insulation. At steady state, the two heat currents will be the same. We will equate that and manipulate accordingly to obtain the answer.

Formula used:
The formula for the heat current is given by:
$Q = \dfrac{{KA}}{l}\left( {{T_1} - {T_2}} \right)$ …… (1)
Where,
$Q$ indicates the heat current.
$K$ indicates constant.
$A$ indicates the cross-sectional area.
${T_1}$ and ${T_2}$ are the two different temperatures.

Complete step by step answer:
In the given question we are supplied with the following data:
There are two layers on a wall, of a house.The outer layer is made of wood, which has thickness of $2.0\,{\text{cm}}$ .The inner wall is made of some kind of insulation which has thickness of $3.5\,{\text{cm}}$ .The interior surface temperature is $19^\circ {\text{C}}$ and the exterior surface temperature is $- 10^\circ {\text{C}}$ .We are asked to find out the temperature at the plane where the wood meets the insulation.

To begin with, we must take the heat current into account. We know at the steady state, the heat current will flow through the wall. Both the layers will have exactly the same area of the cross section. The heat flow through the two layers will definitely be the same.Let the outer temperature be ${T_{{\text{out}}}}$ .The inner temperature is ${T_{{\text{in}}}}$ .The temperature of the junction is $T$ .The heat current which flows from the outer layer to the junction can be written as:
${Q_1} = \dfrac{{{K_1}A}}{{{l_1}}}\left( {T - {T_{{\text{out}}}}} \right)$

The heat current which flows from the junction through the inner layer can be written as:
${Q_2} = \dfrac{{{K_2}A}}{{{l_2}}}\left( {{T_{{\text{in}}}} - T} \right)$
We know that at steady state, the heat flow through the two layers will definitely be the same.
So, we can write:
{Q_1} = {Q_2} \\\ \Rightarrow \dfrac{{{K_1}A}}{{{l_1}}}\left( {T - {T_{{\text{out}}}}} \right) = \dfrac{{{K_2}A}}{{{l_2}}}\left( {{T_{{\text{in}}}} - T} \right) \\\ \Rightarrow \dfrac{{{K_1}}}{{{l_1}}}\left( {T - {T_{{\text{out}}}}} \right) = \dfrac{{{K_2}}}{{{l_2}}}\left( {{T_{{\text{in}}}} - T} \right) \\\ \Rightarrow \dfrac{{0.03}}{2} \times \left( {T - \left( { - 10} \right)} \right) = \dfrac{{0.01}}{{3.5}} \times \left( {19 - T} \right) \\\ \Rightarrow 10.5 \times \left( {T + 10} \right) = 2 \times \left( {19 - T} \right) \\\ \Rightarrow 12.5T = - 67 \\\ \Rightarrow T = \dfrac{{ - 67}}{{12.5}} \\\ \Rightarrow T = - 5.36^\circ {\text{C}} \\\
Now, we convert the temperature in Kelvin scale:
$T = - 5.36 + 273 \\\ \therefore T = 267.64\,{\text{K}}$

Hence, the required temperature is $267.64\,{\text{K}}$.

Note: It is important to remember that steady-state heat flow means that temperatures are kept stable for a reasonable amount of time on both sides of a building envelope feature (while different) so that heat flow is steady on both sides of the assembly. The technique of steady-state heat flow is a simplification, since temperatures change continuously in the natural world. It is also important to note that even if we change the dimension of thickness into the S.I system of units, we would get the same answer as well.