91Ó°ÊÓ

1. Thickness$L_2=0.700L_1$
2. Thickness$L_3=0.350L_1$
3. Thermal conductivity$k_2=0.900k_1$
4. Thermal conductivity$k_3=0.800k_1$
5. The temperature oftheleft side of wall$T_H=30.0\text{°°äor}303\text{K}$
6. The temperature of the left side of wall$T_c=−15\text{°°äor}258\text{‿é}$

According to Fourier's Law, the amount of time it takes for heat to go through a material is proportional to the negative gradient of the temperature as well as the cross-sectional area that is perpendicular to the gradient. We will use the equation for the conduction rate through the slab under the assumption that the transfer is a steady-state process, i.e., the temperature everywhere in the slab and the rate of energy transfer do not change with time. In a steady-state process, the conduction rate through the two materials must be constant.

Formula:

The rate at which the sphere emits thermal radiation,$P_{cond}=\frac{Q}{t}=\frac{kA(T_H−T_C)}{L}$ …(¾±)

Considering the rate of heat transfer through each layer to be the same, the rate of heat transfer across the entire wallis equal to the rate across layer$2\left(P_2\right)$. Using equation (i) and cancelling out the common factor of area, we obtain

$\begin{array}{c}\frac{T_H−T_C}{\frac{L_1}{k_1}+\frac{L_2}{k_2}+\frac{L_3}{k_3}}=\frac{\text{Δ}T}{\frac{L_2}{k_2}}\\ \frac{30\text{°°ä}−(−15\text{°°ä})}{2.21}=\frac{\text{Δ}{T}_2}{\frac{7}{9}}\\ \frac{45\text{°°ä}}{2.21}=\frac{\text{Δ}{T}_2}{\frac{7}{9}}\\ \text{Δ}{T}_2=15.8\text{°°ä}\end{array}$

Hence, the value of temperature difference is$15.8\text{°°ä}$

If $k_2=1.1k_1$, then the rate at which energy is conducted through the wall is greater because we are multiplying it by conductivity $k_{2.}$Hence the conductivity of $k_{2.}$is going to increase.

Repeat the above calculation of part (a) with new value for$k_2$.We get the temperature difference as:

$\begin{array}{c}\frac{45}{1+\frac{7}{11}+\frac{35}{80}}=\frac{\text{Δ}{T}_2}{\frac{7}{11}}\\ \text{Δ}{T}_2=13.8\text{°°ä}\end{array}$

This is less than the result calculated in (a), which shows that the temperature gradient across 1 and 3 is greater than in part (a).The larger temperature gradient leads to large conductive heat currents.

Hence, the required temperature difference is$13.8\text{°°ä}$