Chapter 19, Problem 16SP

A certain double-pane window consists of two glass sheets, each $80\text{ cm}\times 80\text{ cm}\times 0.30\text{ cm}$, separated by a 0.30-cm stagnant air space. The indoor surface temperature is 20 °C, while the outdoor surface temperature is exactly 0 °C. How much heat passes through the window each second? $k_T=0.84\text{ W/k}\cdot \text{m}$ for glass and about 0.080 $\text{W/k}\cdot \text{m}$ for air.

To determine

The heat that passes through the double pan window in each second if it consists of two glass sheets, each $80\text{ cm}\times \text{80 cm}\times 0.3\text{ cm}$, separated by a $0.3\text{cm}$ stagnant of air space and the indoor space temperature is $20°\text{C}$ while the outdoor surface temperature is $0°\text{C}$. Thermal conductivity of the glass plate is $0.84\text{W/m}\cdot \text{K}$ and thermal conductivity of air is $0.08\text{ W/m}\cdot \text{K}$.

Answer to Problem 16SP

Solution:

$72.5\text{cal/s}$

Explanation of Solution

Given data:

The indoor surface temperature is $20°\text{C}$.

The outdoor surface temperature is $0°\text{C}$.

The dimension of the glass window is $80\text{ cm}\times \text{80 cm}\times 0.\text{30 cm}$.

The separation between the double pane window is $0.30\text{ cm}$.

The thermal conductivity of the glass is $0.84\text{W/m}\cdot \text{K}$.

The thermal conductivity of the air is $0.080\text{W/m}\cdot \text{K}$.

Formula used:

Consider the expression for the equilibrium condition for heat flow from one plate to another:

$k_{T_1}{A}_1\frac{T_1-T}{L_1}=k_{T_2}{A}_2\frac{T-T_2}{L_2}$

Here, $k_{T_1}$ is the thermal conductivity of first plate, $k_{T_2}$ is the thermal conductivity of second plate, $A_1$ is the area of first plate, $A_2$ is the area of second plate, $T_1$ is the temperature of first plate, $T_2$ is the temperature of second plate, $T$ is the temperature of the interface, $L_1$ thickness of the first plate, and $L_2$ thickness of the second plate.

Consider the expression for the heat passes through surface:

$\frac{\Delta Q}{\Delta t}=k_{T_1}{A}_1\frac{T_1-T}{L_1}$

Write the expression for the area of the rectangular ice sheet:

$A=lb$

Here, $l$ is the length of the glass sheet and $b$ is the width of the glass sheet.

Explanation:

Shape of the window is rectangular. Therefore, recall expression for the area of the window:

$A=lb$

Substitute $80\text{cm}$ for $l$ and $80\text{cm}$ for $b$

$\begin{array}{c}A=\left(80\text{ cm}\right)\left(80\text{cm}\right)\\ =6400{\text{ cm}}^2\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\\ =0.64{\text{ m}}^2\end{array}$

Consider the expression for the equilibrium condition for heat flow from air to glass plate:

$k_{T_1}{A}_1\frac{T_1-T}{L_1}=k_{T_2}{A}_2\frac{T-T_2}{L_2}$

Substitute $0.84\text{W/m}\cdot \text{K}$ for $k_{T_1}$, $0.08\text{W/m}\cdot \text{K}$ for $k_{T_2}$, $20°\text{C}$ for $T_1$, $0°\text{C}$ for $T_2$, $0.64{\text{m}}^2$ for $A_1$, $0.64{\text{ m}}^2$ for $A_2$, $0.3\text{cm}$ for $L_2$, and $0.3\text{cm}$ for $L_1$

$\begin{array}{c}\left(0.84\text{ W/m}\cdot \text{K}\right)\left(0.64{\text{m}}^2\right)\left(\frac{20°\text{C}-T}{0.30\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)=0.08\text{ W/m}\cdot \text{K}\left(0.64{\text{m}}^2\right)\left(\frac{T-0°\text{C}}{0.3\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)}\right)\\ 179.2\left(20-T\right)=17.067T\\ -179.2T-17.067T=-3584\\ 196.267T=3584\end{array}$

Further solve for $T$

$\begin{array}{c}T=\frac{3584}{196.267}\\ =18.26°\text{C}\\ =18.3°\text{C}\end{array}$

Temperature of the glass air interface is $18.3°\text{C}$.

Consider the process of the heat transfer is steady. Therefore, the heat transfer rate will be constant.

Again, consider the expression of heat passes the window:

$\frac{\Delta Q}{\Delta t}=k_{T_1}{A}_1\frac{T_1-T}{L_1}$

Substitute $0.84\text{W/m}\cdot \text{K}$ for $k_{T_1}$, $20°\text{C}$ for $T_1$, $0.64{\text{ m}}^3$ for $A_1$, $0.3\text{cm}$ for $L_1$, and $18.3°\text{C}$ for $T$ .

$\begin{array}{c}\frac{\Delta Q}{\Delta t}=\left(0.84\text{ W/m}\cdot \text{K}\left(\frac{1\text{cal/s}\cdot \text{cm}°\text{C}}{418.4\text{ W/m}\cdot \text{K}}\right)\right)\left(6400{\text{ cm}}^2\right)\left(\frac{20°\text{C}-18.3°\text{C}}{0.3\text{ cm}}\right)\\ =72.5\text{ cal/s}\end{array}$

Conclusion:

Therefore, the amount of heat passes through the window is $72.5\text{cal/s}$.