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 temperature \(20.00{ }^{\circ} \mathrm{C}\) 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

Answer: To find the temperatures at each interface, follow these steps: 1. Calculate the temperature drop across the glass panes and air gap using Fourier's Law. 2. Sum the temperature drops to find the temperatures at each interface. The interface temperatures are as follows: 1. First glass pane - air interface: T1 = 20.0°C - ΔTG 2. Air gap - glass pane interface: T2 = T1 - ΔTA 3. Second glass pane - air interface: T3 = T2 - ΔTG 4. The outside temperature: T4 = 0.00°C.

Step-by-step solution

Calculate the temperature drop across the glass panes

We will use Fourier's law of thermal conduction to calculate the temperature drop across the glass panes: \(Q = -kA\frac{dT}{dx}\), where \(Q\) is the heat flux, \(k\) is the thermal conductivity of the material, \(A\) is the area of the material, and \(\frac{dT}{dx}\) is the temperature gradient. Rearrange the formula to calculate the temperature gradient: \(\frac{dT}{dx} = -\frac{Q}{kA}\) The heat flux, area, and thermal conductivity are constant for the given dimensions and materials, so we can write the temperature drop for each pane of glass as follows: \(\Delta T_G = -\frac{Q}{k_G A} d_{G}\), where \(k_G=1.00 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) and \(d_G = 3.00 \mathrm{~mm}\) are the thermal conductivity and thickness of the glass, respectively.

Calculate the temperature drop across the air gap:

Using the same approach, we will calculate the temperature drop across the air gap: \(\Delta T_A = -\frac{Q}{k_A A} d_{A}\), where \(k_A=0.0260 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) and \(d_A = 1.00 \mathrm{~cm}\) are the thermal conductivity and thickness of the air gap, respectively.

Sum all temperature drops to find the interface temperatures:

Now that we have the temperature drop across the glass panes and the air gap, we can sum them up to find the temperatures at each interface. From the inside to the outside, we have: 1. The first glass pane - air interface: \(T1 = 20.0^{\circ}C - \Delta T_G\) 2. The air gap - glass pane interface: \(T2 = T1 - \Delta T_A\) 3. The second glass pane - air interface: \(T3 = T2 - \Delta T_G\) 4. The outside temperature is \(T4 = 0.00^{\circ}C\). #b) At what rate is heat lost from the room, per square meter of window?#

Apply Fourier's law to find the heat flux for the entire window assembly:

Assuming the two glass panes and the air gap are in series, we can rewrite Fourier's law as follows. \(Q = -A\frac{k_G \Delta T_G + k_A \Delta T_A}{d_G + d_A}\) We can find the heat flux per square meter of window by dividing the total heat flux by the area \(A\): \(Q_{loss} = -\frac{k_G \Delta T_G + k_A \Delta T_A}{d_G + d_A}\) #c) What would the rate of heat loss per square meter be then, under the same temperature conditions?#

Apply the same approach for a single layer of glass:\(

We will assume a single layer of glass with a thickness of \)6.00\mathrm{~mm}\(. Let's denote the temperature drop across this layer as \)\Delta T_S$. We can calculate the heat loss per square meter: \(Q_{loss}^{'} = -\frac{k_G \Delta T_S}{2d_G}\) #d) Why is evacuating the space between the glass panes not done?# To answer this question, we need to consider practical implications, such as the cost and difficulty of maintaining a vacuum, as well as the potential structural issues that could arise due to the pressure difference between the evacuated space and the surrounding atmosphere. This makes the evacuation of air between the panes not a feasible solution in real-world scenarios.

Key concepts

Fourier's Law of Thermal Conduction

Fourier's Law of Thermal Conduction is a fundamental principle that describes how heat passes through materials. It is critical to solving problems involving heat transfer and is given by the equation \(Q = -kA\frac{dT}{dx}\), where \(Q\) is the heat flux or the amount of heat per unit time passing through an area \(A\), \(k\) is the thermal conductivity of the material which is a measure of its ability to conduct heat, and \(\frac{dT}{dx}\) is the temperature gradient, the rate at which the temperature changes with distance.

When applying Fourier's law to an instance such as a thermal window, the negative sign in the equation indicates that heat flows in the direction of decreasing temperature, from the warm room to the colder outside. It's essential to remember that the efficiency of heat transfer through different materials in the window (like glass and air in our exercise) is governed by their respective conductivity values and thickness.

Heat Flux

Heat flux is a measure of the rate of thermal energy flow per unit area, with units of Watts per square meter (\(W/m^2\)). In the context of our exercise, we seek to understand how much heat is being lost through the thermal window. Heat flux is directly proportional to the temperature difference across a material and inversely proportional to the material's thickness.

Using Fourier's law, the magnitude of the heat flux through each layer of the window can be determined. It provides a clear understanding of how efficiently heat is escaping. The calculation of heat flux is essential for optimizing thermal insulation, as in the design of energy-efficient windows.

Temperature Gradient

The temperature gradient refers to the rate of temperature change over a distance within a conducting material. It is a vector quantity, which, in the exercise, is represented by the symbol \(\frac{dT}{dx}\), and it tells us how rapidly the temperature varies from one spatial point to another. The temperature gradient is crucial in determining the heat flow direction and magnitude across different media such as the glass panes and air gap of a thermal window.

In real-world applications, insulating materials are often used to minimize the temperature gradient, effectively reducing the amount of heat lost through the material over time, which is instrumental in thermal regulating processes.

Heat Loss Per Square Meter

When calculating the rate of heat loss from a space, such as a room through a window, we refer to this quantity as heat loss per square meter. This measurement is quite practical, as it helps to quantify the energy performance of building materials. In our exercise, this concept was explored by solving how much heat energy was lost through different configurations of the window, under the same temperature conditions.

To put it into perspective, this rate of heat loss affects both the energy efficiency of a building and the comfort of its occupants. Understanding how thermal properties like conductivity and the structure of materials impact this rate can contribute significantly to designing more efficient buildings and saving on energy costs.