Problem 101 - University Physics with Modern Physics

Question

Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m\(^2\) to that for a double-pane window with the same area. The glass of a single pane is 4.2 mm thick, and the air space between the two panes of the double-pane window is 7.0 mm thick. The glass has thermal conductivity 0.80 W /m \(\cdot\) K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m\(^2 \cdot\) K/W.

Step-by-Step Solution

Answer

The heat loss rate through a single-pane window is approximately 11.70 times higher than that through a double-pane window.

Step 1: Understand the Problem

We need to find the ratio of heat loss rates between two window types: single-pane and double-pane. Heat loss rate depends on thermal resistance, which is determined by thermal conductivities and structural thickness of the panes.

Step 2: Define Heat Transfer Equation

The rate of heat loss (Q) through a material is given by \[ Q = \frac{A \times \Delta T}{R_{total}} \]where \( A \) is the area, \( \Delta T \) is the temperature difference across the material, and \( R_{total} \) is the total thermal resistance.

Step 3: Calculate Thermal Resistance for Single-Pane Window

The single-pane thermal resistance is calculated by \[ R_{glass} = \frac{d_{glass}}{k_{glass} \times A} = \frac{0.0042}{0.8 \times 0.15} = 0.035 \text{ m}^2\cdot\text{K/W}. \]Adding the thermal resistance of the air films on both sides, we have\[ R_{total, single} = R_{glass} + 0.15 = 0.035 + 0.15 = 0.185 \text{ m}^2\cdot\text{K/W}. \]

Step 4: Calculate Thermal Resistance for Double-Pane Window

For the double-pane window, we consider two glass panes and one air gap:- Resistance per glass pane is the same as calculated for single-pane: \[ R_{glass} = 0.035 \text{ m}^2\cdot\text{K/W}. \]- Air gap resistance is: \[ R_{air} = \frac{0.007}{0.024 \times 0.15} = 1.943 \text{ m}^2\cdot\text{K/W}, \]given air conductivity \( k_{air} = 0.024 \text{ W/m} \cdot \text{K}. \)- Total resistance for the double-pane window is:\[ R_{total, double} = 2 \times R_{glass} + R_{air} + 0.15 = 0.035 \times 2 + 1.943 + 0.15 = 2.163 \text{ m}^2\cdot\text{K/W}. \]

Step 5: Compute the Ratio of Heat Loss Rates

The ratio of heat loss rates is given by:\[ \frac{Q_{single}}{Q_{double}} = \frac{R_{total, double}}{R_{total, single}} = \frac{2.163}{0.185}. \]Calculate the ratio:\[ \frac{Q_{single}}{Q_{double}} \approx 11.70. \]

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept that plays a crucial role in understanding how energy moves through different materials. It's the process through which heat energy is transferred from one place to another, and it can occur in three primary ways: conduction, convection, and radiation. In this exercise, we're focusing primarily on **conduction**, which occurs when heat moves through a solid material.
Conduction happens due to the vibration and movement of particles in the material. The faster-moving particles transfer their energy to slower ones, causing the heat to travel across the material. In the context of windows, heat flows from the warmer side of the glass to the cooler side.
To evaluate heat transfer, we use the formula \[ Q = \frac{A \times \Delta T}{R_{total}} \]where:

• \( Q \) is the rate of heat loss.
• \( A \) is the area through which the heat is being transferred.
• \( \Delta T \) represents the temperature difference across the material.
• \( R_{total} \) is the total thermal resistance.

With a higher thermal resistance, the rate of heat transfer decreases, helping us understand how well a structure, like a window, prevents heat loss.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It is denoted by the symbol \( k \) and is expressed in watts per meter per kelvin (\( W/m \cdot K \)). Materials with high thermal conductivity are good conductors of heat, meaning they transfer heat easily and quickly. Conversely, materials with low thermal conductivity are good insulators.
In the context of the given exercise, glass and air are two materials considered, each with its own thermal conductivity. Glass has a thermal conductivity of 0.80 \( W/m \cdot K \), indicating it is relatively good at conducting heat. In contrast, the air trapped between the two glass panes of the double-pane window has a much lower thermal conductivity of 0.024 \( W/m \cdot K \). This lower value indicates that air is a much poorer conductor of heat and serves as a better insulator.
The concept of thermal conductivity helps us understand why double-pane windows are more effective at reducing heat loss. The air gap adds an insulating layer, significantly increasing the overall thermal resistance of the window, thereby decreasing the heat transfer from inside to outside.

Single-Pane vs Double-Pane Windows

When comparing single-pane and double-pane windows, the key factor is how each affects heat transfer and energy efficiency.
**Single-pane windows** consist of just one layer of glass; they offer minimal insulation since there is no barrier (other than the thin glass) to slow down the heat transfer. As a result, single-pane windows tend to have a larger heat loss rate, leading to higher energy consumption for maintaining indoor temperatures.

• Structure: 1 layer of glass.
• Heat transfer: Direct and faster.
• Thermal resistance: Lower due to less insulation.

On the other hand, **double-pane windows** include two glass layers with an air space in between, which provides additional insulation. This setup results in much higher thermal resistance, as shown in the exercise's computations. The air space acts as a buffer, minimizing heat loss and making double-pane windows more energy-efficient.

• Structure: 2 layers of glass with an air gap.
• Heat transfer: Slower due to the air barrier.
• Thermal resistance: Higher, offering better insulation.

The exercise demonstrates that the rate of heat loss through double-pane windows is significantly lower than that through single-pane windows, providing a ratio that highlights the effectiveness of the additional layer in insulating the building.