Chapter 14, Problem 68P

To determine

Rank the walls according to the heat flow through the walls from greatest to smallest.

Answer to Problem 68P

The rank of walls according to the heat flow through the walls from greatest to smallest is $\left(\text{c}\right)\text{,}\left(\text{d}\right)\text{,}\left(\text{a}\right)=\left(\text{b}\right)\text{=}\left(\text{e}\right)$.

Explanation of Solution

Write an expression for the heat flow.

$P=\frac{\kappa A\Delta T}{d}$ (I)

Here, $P$ is the heat flow, $\kappa$ is the thermal conductivity, $A$ is the area, $\Delta T$ is the temperature difference and $d$ is the thickness.

The temperature difference is same since the heat flows till the equilibrium.

$P\propto \frac{\kappa A}{d}$ (II)

Conclusion:

For wall (a), substitute $120{\text{m}}^2$ for $A$, $10\text{cm}$ for $d$ and $0.030\text{W/}\left(\text{m}\cdot \text{K}\right)$ for $\kappa$ in equation (II).

$\begin{array}{c}P\propto \frac{\left(0.030\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ \propto \frac{\left(0.030\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(0.10\text{m}\right)}\\ \propto 36\text{W/K}\end{array}$

For wall (b), substitute $120{\text{m}}^2$ for $A$, $15\text{cm}$ for $d$ and $0.045\text{W/}\left(\text{m}\cdot \text{K}\right)$ for $\kappa$ in equation (II).

$\begin{array}{c}P\propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(\left(15\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ \propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(0.15\text{m}\right)}\\ \propto 36\text{W/K}\end{array}$

For wall (c), substitute $180{\text{m}}^2$ for $A$, $10\text{cm}$ for $d$ and $0.045\text{W/}\left(\text{m}\cdot \text{K}\right)$ for $\kappa$ in equation (II).

$\begin{array}{c}P\propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(180{\text{m}}^2\right)}{\left(\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ \propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(180{\text{m}}^2\right)}{\left(0.10\text{m}\right)}\\ \propto 81\text{W/K}\end{array}$

For wall (d), substitute $120{\text{m}}^2$ for $A$, $10\text{cm}$ for $d$ and $0.045\text{W/}\left(\text{m}\cdot \text{K}\right)$ for $\kappa$ in equation (II).

$\begin{array}{c}P\propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ \propto \frac{\left(0.045\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(120{\text{m}}^2\right)}{\left(0.10\text{m}\right)}\\ \propto 54\text{W/K}\end{array}$

For wall (e), substitute $180{\text{m}}^2$ for $A$, $15\text{cm}$ for $d$ and $0.030\text{W/}\left(\text{m}\cdot \text{K}\right)$ for $\kappa$ in equation (II).

$\begin{array}{c}P\propto \frac{\left(0.030\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(180{\text{m}}^2\right)}{\left(\left(15\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\\ \propto \frac{\left(0.030\text{W/}\left(\text{m}\cdot \text{K}\right)\right)\left(180{\text{m}}^2\right)}{\left(0.15\text{m}\right)}\\ \propto 36\text{W/K}\end{array}$

Thus, the rank of walls according to the heat flow through the walls from greatest to smallest is $\left(\text{c}\right)\text{,}\left(\text{d}\right)\text{,}\left(\text{a}\right)=\left(\text{b}\right)\text{=}\left(\text{e}\right)$.