A double-pane window used for insulating a room thermally from outside consists of two glass sheets

Q: Why do we assume the rate of heat flow is constant across all layers?

The problem states the system is in a 'steady state'. In steady state, no part of the system absorbs or stores heat; whatever heat enters a layer must exit it. Hence, the heat current $H$ is uniform throughout.

Q: Why is air used between the glass panes instead of just making a thicker glass?

Air has a significantly lower thermal conductivity ($0.08$) compared to glass ($0.8$). This means a thin layer of trapped, stagnant air provides much more thermal resistance than a thick block of glass, making it an excellent and lightweight insulator.

Q: Why did my answer for heat flow come out to $41.54\text{ W}$ instead of $41.6\text{ W}$?

If you calculate the equivalent resistance $R_{eq} = 0.65$ and use $H = 27 / 0.65$, you get exactly $41.538...\text{ W}$. The answer $41.6\text{ W}$ is obtained if you prematurely round the interface temperature $\theta_1$ to $26.48^\circ\text{C}$ and then calculate $H = (27 - 26.48) / 0.0125$. Both approaches are conceptually correct, but avoiding premature rounding is mathematically purer.

Q: Can I solve this without finding the equivalent resistance?

Absolutely! You can set up the continuous equality: $H = \frac{27 - \theta_1}{R_1} = \frac{\theta_1 - \theta_2}{R_2} = \frac{\theta_2 - 0}{R_3}$. By solving these simultaneous equations, you can find $\theta_1$ and $\theta_2$ first, and then calculate $H$. The equivalent resistance method is just a faster mathematical shortcut.

Q: What does 'stagnant' air mean, and why is it important?

'Stagnant' means the air is completely still. If the air were moving, heat would be transferred via convection, which is much faster than conduction. By trapping the air in a narrow space, convection currents are prevented, forcing heat to transfer only via poor conduction.

Question

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area $1	ext{ m}^2$ and thickness $0.01	ext{ m}$ separated by a $0.05	ext{ m}$ thick stagnant air space. In the steady state, the room glass interface and the glass-outdoor interface are at constant temperatures of $27^\circ	ext{C}$ and $0^\circ	ext{C}$ respectively. Calculate the rate of heat flow through the window pane. Also, find the temperatures of other interfaces. Given, thermal conductivities of glass and air as $0.8$ and $0.08	ext{ W m}^{-1}	ext{K}^{-1}$ respectively.

Sigma Percentile · Accepted Answer

The physics of heat transfer is not just a set of abstract equations; it is the very reason we can stay warm in our homes during a freezing winter. In this problem, we explore the elegant mechanics of a double-pane window, a marvel of everyday engineering.

### The Electrical Analogy of Heat

Imagine heat as a fluid trying to escape from a hot room to the cold outdoors. Just as electrical current faces resistance when flowing through a wire, heat current faces **thermal resistance** when flowing through a material.

This analogy is incredibly powerful. We can define the thermal resistance $R$ of any layer as:
$$R = \frac{l}{KA}$$
where $l$ is the thickness of the layer, $K$ is its thermal conductivity, and $A$ is the cross-sectional area. Because the heat must flow through the inner glass, then the trapped air, and finally the outer glass, these three layers act exactly like three electrical resistors connected in **series**.

### Calculating the Resistances

Let's break down the window into its three components.

For the two glass panes, the thickness is $0.01	ext{ m}$ and the thermal conductivity is $0.8	ext{ W m}^{-1}	ext{K}^{-1}$. Since the area is $1	ext{ m}^2$, the resistance for each glass pane is:
$$R_1 = R_3 = \frac{0.01}{0.8 	imes 1} = 0.0125	ext{ K/W}$$

Now, let's look at the trapped air layer. It is thicker ($0.05	ext{ m}$) and has a much lower thermal conductivity ($0.08	ext{ W m}^{-1}	ext{K}^{-1}$). Its resistance is:
$$R_2 = \frac{0.05}{0.08 	imes 1} = 0.625	ext{ K/W}$$

Notice how much larger the resistance of the air layer is compared to the glass! This is the secret behind the double-pane window.

### The Master Equation: Finding the Heat Flow

Because the layers are in series, the total equivalent thermal resistance $R_{eq}$ is simply the sum of the individual resistances:
$$R_{eq} = R_1 + R_2 + R_3 = 0.0125 + 0.625 + 0.0125 = 0.65	ext{ K/W}$$

The system is in a **steady state**, meaning the rate of heat flow $H$ is constant throughout. We can find this total heat current by dividing the total temperature difference by the total resistance:
$$H = \frac{\Delta T_{total}}{R_{eq}} = \frac{27 - 0}{0.65} \approx 41.54	ext{ W}$$
*(Note: If we use the rounded interface temperatures from the next step, the heat flow is often cited as $41.6	ext{ W}$ in textbooks).*

### Unveiling the Interface Temperatures

To find the temperatures between the layers, we apply our "Ohm's Law for Heat" to individual sections.

For the first glass pane, the temperature drops from $27^\circ	ext{C}$ to $	heta_1$. The temperature drop $\Delta T_1$ is the heat current multiplied by the resistance of the glass:
$$\Delta T_1 = H 	imes R_1 = 41.54 	imes 0.0125 \approx 0.52^\circ	ext{C}$$
Therefore, the temperature at the first interface is:
$$	heta_1 = 27 - 0.52 = 26.48^\circ	ext{C}$$

Similarly, for the outer glass pane, the temperature drops from $	heta_2$ to $0^\circ	ext{C}$:
$$\Delta T_3 = H 	imes R_3 = 41.54 	imes 0.0125 \approx 0.52^\circ	ext{C}$$
This means the temperature at the second interface is:
$$	heta_2 = 0 + 0.52 = 0.52^\circ	ext{C}$$

### The Real Hero: Stagnant Air

Look closely at the temperatures we just calculated. The temperature drops by a mere $0.52^\circ	ext{C}$ across each glass pane. However, it plummets by a massive $25.96^\circ	ext{C}$ across the trapped air layer!

This beautifully illustrates why we don't just use thicker glass to insulate our homes. Stagnant air is a phenomenal insulator, and by trapping it between two thin sheets of glass, we create a lightweight, transparent, and highly effective thermal barrier.

Visualized Solution

$T_{h o t} = 2 7^{\circ} C, T_{co l d} = 0^{\circ} C$

$R = \frac{l}{K A}$

$R_{1} = R_{3} = \frac{0.01}{0.8 \times 1}$

$R_{2} = \frac{0.05}{0.08 \times 1}$

$R_{e q} = R_{1} + R_{2} + R_{3}$

$H = \frac{Δ T _{t o t a l}}{R _{e q}}$

$Δ T_{1} = H \times R_{1}$

$Δ T_{3} = H \times R_{3}$

$θ_{1} - θ_{2} = 25.9 6^{\circ} C$

The Sigma Insight: Heat Transfer

The Electrical Analogy of Heat

Calculating the Resistances

The Master Equation

Unveiling the Interface Temperatures

The Real Hero

Similar Questions

A heat source at $T = 1 0^{3}$ K is connected to another heat reservoir at $T = 1 0^{2}$ K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is $0.1 WK^{- 1} m^{- 1}$ , the energy flux through it in the steady state is

The temperature $θ$ at the junction of two insulating sheets, having thermal resistances $R_{1}$ and $R_{2}$ as well as top and bottom temperatures $θ_{1}$ and $θ_{2}$ (as shown in figure) is given by

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thot​=27∘C,Tcold​=0∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=KAl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​=R3​=0.8×10.01​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​=0.08×10.05​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=R1​+R2​+R3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H=Req​ΔTtotal​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔT1​=H×R1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔT3​=H×R3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ1​−θ2​=25.96∘C