Question

A thermal window consists of two panes of glass separated by an air gap. Each pane of glass is \(3.00 \mathrm{~mm}\) thick, and the air gap is \(1.00 \mathrm{~cm}\) thick. Window glass has a thermal conductivity of \(1.00 \mathrm{~W} /(\mathrm{m} \mathrm{K}),\) and air has a thermal conductivity of \(0.0260 \mathrm{~W} /(\mathrm{m} \mathrm{K})\). Suppose a thermal window separates a room at \(20.00^{\circ} \mathrm{C}\) from the outside at \(0.00^{\circ} \mathrm{C} .\) a) What is the temperature at each of the four air-glass interfaces? b) At what rate is heat lost from the room, per square meter of window? c) Suppose the window had no air gap but consisted of a single layer of glass \(6.00 \mathrm{~mm}\) thick. What would the rate of heat loss per square meter be then, under the same temperature conditions? d) Heat conduction through the thermal window could be reduced essentially to zero by evacuating the space between the glass panes. Why is this not done?

Short answer

Question: Calculate the temperature at each air-glass interface and find the rate of heat loss per square meter of the window for a double-pane thermal window. Explain why the space between the glass panes is not evacuated. Answer: The temperatures at the air-glass interfaces are 17.00°C, 16.74°C, 0.260°C, and 3.260°C. The rate of heat loss for the double-pane thermal window is 15.43 W/m². The space between the glass panes is not evacuated due to reasons like difficulty in maintaining the vacuum, risk of glass breaking due to pressure differences, the need for whole window replacement if vacuum breaks, and high production and maintenance costs.

Step-by-step solution

a) Temperature at each air-glass interfaces

For this part, we need to find the temperatures at the air-glass interfaces. We will apply the concept of temperature gradient. The temperature gradient can be defined by the formula: \(\Delta T = k \times \dfrac{\Delta x}{A}\) Where \(\Delta T\) is the change in temperature, \(k\) is the thermal conductivity, \(\Delta x\) is the thickness, and \(A\) is the area. We will first find the temperature drop across each pane of glass: \(\Delta T_{glass} = 1.00\dfrac{3.00 \times 10^{-3}}{1} = 3.00 K\) The temperature drop across the air gap is: \(\Delta T_{air} = 0.0260\dfrac{10^{-2}}{1} = 0.260 K\) The temperature at the first interface (inside) will be: \(T_1 = 20.00 - 3.00 = 17.00 ^{\circ}C\) The temperature at the second interface (air-glass) will be: \(T_2 = 17.00 - 0.260 = 16.74 ^{\circ}C\) The temperature at the third interface (air-glass) will be: \(T_3 = 0.00 + 0.260 = 0.260 ^{\circ}C\) The temperature at the fourth interface (outside) will be: \(T_4 = 0.260 + 3.00 = 3.260 ^{\circ}C\) So, the temperatures at the air-glass interfaces are \(17.00^{\circ}C\), \(16.74^{\circ}C\), \(0.260^{\circ}C\) and \(3.260^{\circ}C\).

b) Rate of heat loss from the room, per square meter of window

To find the rate of heat loss, we need to find the temperature difference and apply the concept of thermal resistance. The total temperature difference, \(\Delta T_{total}\), is the sum of the changes in temperature across the glass and air: \(\Delta T_{total} = 2\Delta T_{glass} + \Delta T_{air} = 2 \times 3.00 + 0.260 = 6.260 K\) Thermal resistance, \(R\), can be calculated as follows: \(R = \dfrac{\Delta x}{kA}\) Now, let's add the thermal resistance of the two glasses and the air, assuming the area, \(A = 1\,\text{m}^2\), \(R_{total} = \dfrac{2 \times 10^{-3}}{1.00} + \dfrac{10^{-2}}{0.0260} = 0.020 + 0.385 = 0.405\) Using Ohm's law for heat conduction, we get: \(\text{Rate of heat loss} = \dfrac{\Delta T_{total}}{R_{total}} = \dfrac{6.260}{0.405} = 15.43 \mathrm{~W}\) Therefore, the rate of heat loss per square meter of window is \(15.43 \mathrm{~W}\).

c) Rate of heat loss with a single layer of glass

Now, we will calculate the rate of heat loss for a single layer of glass with thickness \(6.00\,\text{mm}\). The temperature gradient for a single \(6\,\text{mm}\) layer of glass: \(\Delta T_{glass} = 1.00 \times \dfrac{6.00 \times 10^{-3}}{1} = 6.00 K\) The temperature difference will be: \(\Delta T_{total} = 20.00 - 0.00 = 20.00 K\) Now, we can calculate the thermal resistance of this single layer: \(R_{single} = \dfrac{6.00 \times 10^{-3}}{1.00} = 0.006\) Finally, let's find the rate of heat loss: \(\text{Rate of heat loss} = \dfrac{\Delta T_{total}}{R_{single}} = \dfrac{20.00}{0.006} = 3333.33 \mathrm{~W}\) The rate of heat loss per square meter with a single layer of glass is \(3333.33 \mathrm{~W}\).

d) Reason for not evacuating the space between the panes

Evacuating the space between the glass panes would indeed significantly reduce heat conduction through the thermal window. However, there are a few reasons why this is not done: 1. Evacuation is difficult to maintain due to air leakage. 2. Vacuum between panes would create a pressure difference, which might cause the glass to flex inward and break. 3. If a vacuum were to break, the entire window would need replacement. 4. The cost of production and maintenance for such a window would be high. Therefore, practical considerations such as cost, safety, and maintenance prevent the complete evacuation of the space between glass panes.