Problem 65 - University Physics with Modern Physics

Question

A picture window has dimensions of 1.40 m \(\times\) 2.50 mand is made of glass 5.20 mm thick. On a winter day, the temperature of the outside surface of the glass is -20.0\(^\circ\)C, while the temperature of the inside surface is a comfortable 19.5\(^\circ\)C. (a) At what rate is heat being lost through the window by conduction? (b) At what rate would heat be lost through the window if you covered it with a 0.750-mm-thick layer of paper (thermal conductivity 0.0500 W/m \(\cdot\) K)?

Step-by-Step Solution

Answer

Without paper: ~26548 W; with paper: ~9217 W.

Step 1: Understand the Problem and Given Data

We need to find the rate of heat transfer through the window glass and compare it with the heat transfer rate if the window is covered with paper. We are provided with the dimensions of the window and the thicknesses of the glass and paper. Additionally, we have the thermal conductivity of the glass and paper, along with the temperatures of the inside and outside surfaces.

Step 2: Identify the Formula for Heat Transfer

The rate of heat transfer by conduction is calculated using Fourier's law, given by:\[Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d}\]where:- \(Q\) is the heat transfer rate (in watts, W),- \(k\) is the thermal conductivity,- \(A\) is the area of the window,- \(T_{inside}\) and \(T_{outside}\) are the temperatures of the inside and outside surfaces, respectively,- \(d\) is the thickness of the material.

Step 3: Calculate the Area of the Window

Given the window dimensions 1.40 m by 2.50 m, calculate the area:\[A = 1.40 \times 2.50 = 3.50 \text{ m}^2\]

Step 4: Calculate the Heat Transfer Rate Through the Glass

Using the values:- \(k_{glass} = 1.0 \text{ W/m} \cdot \text{K}\),- \(d_{glass} = 5.20 \times 10^{-3} \text{ m}\),- \(T_{inside} = 19.5 \underline{\phantom{xxx}}^\circ\text{C}\), \(T_{outside} = -20.0 \underline{\phantom{xxx}}^\circ\text{C}\),calculate:\[Q_{glass} = \frac{1.0 \times 3.50 \times (19.5 - (-20.0))}{5.20 \times 10^{-3}} = \frac{1.0 \times 3.50 \times 39.5}{5.20 \times 10^{-3}}\]Calculating the above expression gives:\[Q_{glass} \approx 26548.08 \text{ W}\]

Step 5: Calculate the Heat Transfer Rate Through the Paper Layer

Now consider the paper with:- \(k_{paper} = 0.0500 \text{ W/m} \cdot \text{K}\),- \(d_{paper} = 0.750 \times 10^{-3} \text{ m}\).The heat transfer rate is:\[Q_{paper} = \frac{0.0500 \times 3.50 \times (19.5 - (-20.0))}{0.750 \times 10^{-3}} = \frac{0.0500 \times 3.50 \times 39.5}{0.750 \times 10^{-3}}\]Calculating the above expression gives:\[Q_{paper} \approx 9216.67 \text{ W}\]

Step 6: Compare Heat Transfer Rates

Compare the rates:- Without paper: \(Q_{glass} \approx 26548.08 \text{ W}\)- With paper: \(Q_{paper} \approx 9216.67 \text{ W}\)The paper significantly reduces the rate of heat loss through the window by approximately 65%.

Key Concepts

Fourier's Law of Heat Conduction

Fourier's Law of Heat Conduction is an essential principle when exploring how heat moves through materials. It essentially describes how thermal energy, also known as heat, is transferred from the hotter side of a material to its cooler side. In the exercise regarding the window, this principle allows us to understand how heat transfers from the warm interior of a building to the cold exterior through the window glass.

According to this law, the heat transfer rate (Q) can be calculated using the formula:\[Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d}\]Here, \(k\) represents the thermal conductivity of the material, \(A\) is the area through which the heat is passing, \(T_{inside} - T_{outside}\) is the temperature difference between the inside and outside, and \(d\) is the thickness of the material. This formula is quite effective for understanding how changing any variable affects heat loss.

For example, a higher thermal conductivity results in more heat being transferred, emphasizing the importance of material choice in managing heat retention or loss.

Thermal Conductivity

Thermal conductivity is a property that indicates how well a material can conduct heat. In simple terms, it measures the ability of a material to pass heat through it. Materials with high thermal conductivity transfer heat quickly, while those with low thermal conductivity do not.

In our window exercise, the glass used has a thermal conductivity of 1.0 W/m·K, meaning it readily allows heat to pass through. On the other hand, the paper has a much lower thermal conductivity of 0.0500 W/m·K, making it a good insulator. This is why adding a paper layer significantly reduces heat loss compared to when only glass is used.

Choosing materials with suitable thermal conductivities is crucial in construction and other applications, as it directly impacts energy efficiency. Think about how different materials are used in insulation based on their ability to conduct heat.

Thermal Resistance

Thermal resistance is the ability of a material to resist the flow of heat. The interplay between thermal conductivity and resistance is essential in managing heat transfer. Simply put, the higher the thermal resistance, the less heat that will pass through the material.

You can think of thermal resistance as the opposite of thermal conductivity: while conductivity measures how quickly heat flows, resistance measures how much gear is needed to stop it. In the exercise, the paper layer provides additional thermal resistance compared to the glass alone.

Calculating total thermal resistance involves understanding the thickness of the material and its thermal conductivity. The effectiveness of insulation is largely due to its capacity to provide greater resistance to heat flow, which is why insulating materials are fundamental in energy conservation efforts.

Heat Transfer Rate

The heat transfer rate is a measure of how quickly heat moves through a material. It is determined using Fourier's Law, considering factors like thermal conductivity, area, temperature difference, and material thickness. In essence, it tells us the quantity of heat that flows from a warmer area to a cooler one over a specified time.

In our window example, two different heat transfer rates are calculated — one for the glass and one for when the paper layer is added. This allows us to see how different materials and thicknesses can change the rate at which heat is lost from inside the building to the external environment.

The calculations showed that the rate of heat loss through the glass was around 26548.08 W, while it dropped to about 9216.67 W when covered with paper. This example demonstrates how a simple change, like adding a layer with lower thermal conductivity, can lead to substantial energy savings. Understanding heat transfer rates is crucial for designing efficient heating and cooling systems in homes and buildings.