91Ó°ÊÓ

$T_4=−10{}^{\text{0}}\text{C}$a) The given thermal conductivities of the three layers$k_1=0.060\text{W/m.K}$, $k_2=\text{unknown}$, $k_3=0.040\text{W/m.K}$,$k_4=0.12\text{W/m.K}$,

b) Thickness of the layer,$L_1=1.5\text{cm}$,$L_2=\text{unknown}$,$L_3=2.8\text{cm}$, $L_4=3.5\text{cm}$

c) The known temperatures, $T_1=30{}^{\text{0}}\text{C}$,localid="1661847835464" $T_{12}=25{}^{\text{0}}\text{C}$,

The thermal conductivity of a substance is the measurement of the substance's ability to conduct heat during a heat transfer process. Here, we are given four layers that can conduct heat by coming in contact with each other. Heat transfer involves the process in which the temperature from a hotter body is transferred to a cooler one. Thus, the transfer takes place from the left layer to the last right layer due to a temperature difference. Again, considering that the rate of conduction per unit area is uniform in every layer, the required value of the temperature at the given interface can be calculated.

Formulae:

The rate of conduction of energy by the body, $P_{rad}=\frac{kA(T_H−T_C)}{L}$ …(¾±)

where,$k$is the thermal conductivity of the material,$A$is the surface area of radiation,$T_H$is the temperature at the hotter end,data-custom-editor="chemistry" $T_L$is the temperature at the colder end,$L$is the length of the conduction.

As, its is given that the energy transfer through the wall is steady, it can be considered that the rate of conduction of heat per unit area is uniform throughout the layers. Thus, the rate of transfer per unit area from layer 1 to layer 2 will the equal to the heat transfer by layer 3 to layer 4 considering their cross sectional view along the horizontal direction. Thus, it can be given using equation (i) as follows: (From the given data, layer 1 is at higher temperature, thus heat transfers from layer 1 to layer 4)

$\begin{array}{c}\frac{k_1(T_1−T_{12})}{L_1}=\frac{k_4(T_{34}−T_4)}{L_4}(âˆ\mu \frac{P_{rad}}{A}=\text{³Ü²Ô¾±´Ú´Ç°ù³¾â€‰o°ù³¦´Ç²Ô²õ³Ù²¹²Ô³Ù})\\ \frac{\left(0.060\text{W/m.K}\right)({30}^0\text{C}−{25}^0\text{C})}{1.5\text{cm}}=\frac{\left(0.12\text{W/m.K}\right)\left(T_{34}−(−{10}^0\text{C)}\right)}{3.5\text{cm}}\\ \left(T_{34}−(−{10}^0\text{C)}\right)=\frac{\left(0.060\text{W/m.K}\right)\left(5^0\text{C}\right)\left(3.5\text{cm}\right)}{\left(1.5\text{cm}\right)\left(0.12\text{W/m.K}\right)}\\ ={5.83}^0\text{C}\\ T_{34}={5.83}^0\text{C}−{10.0}^0\text{C}\\ =−{4.17}^0\text{C}\\ ≈−{4.2}^0\text{C}\end{array}$

Hence, the required value of the interface temperature is$−{4.2}^0\text{C}$ .