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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)?

Short Answer

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

Step by step solution

01

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.
02

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.
03

Calculate the Area of the Window

Given the window dimensions 1.40 m by 2.50 m, calculate the area:\[A = 1.40 \times 2.50 = 3.50 \text{ m}^2\]
04

Calculate the Heat Transfer Rate Through the Glass

Using the values:- \(k_{glass} = 1.0 \text{ W/m} \cdot \text{K}\),- \(d_{glass} = 5.20 \times 10^{-3} \text{ m}\),- \(T_{inside} = 19.5 \text{ }^\circ\text{C}\), \(T_{outside} = -20.0 \text{ }^\circ\text{C}\),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:\[Q_{glass} \approx 26548.08 \text{ W}\]
05

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:\[Q_{paper} \approx 9216.67 \text{ W}\]
06

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%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

At a chemical plant where you are an engineer, a tank contains an unknown liquid. You must determine the liquid's specific heat capacity. You put 0.500 kg of the liquid into an insulated metal cup of mass 0.200 kg. Initially the liquid and cup are at 20.0\(^\circ\)C. You add 0.500 kg of water that has a temperature of 80.0\(^\circ\)C. After thermal equilibrium has been reached, the final temperature of the two liquids and the cup is 58.1\(^\circ\)C. You then empty the cup and repeat the experiment with the same initial temperatures, but this time with 1.00 kg of the unknown liquid. The final temperature is 49.3\(^\circ\)C. Assume that the specific heat capacities are constant over the temperature range of the experiment and that no heat is lost to the surroundings. Calculate the specific heat capacity of the liquid and of the metal from which the cup is made.

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0\(^\circ\)C. How many kilograms of ice at \(-\)20.0\(^\circ\)C must be dropped into the water to make the final temperature of the system 40.0\(^\circ\)C?

To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its room-temperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.

A machinist bores a hole of diameter 1.35 cm in a steel plate that is at 25.0\(^\circ\)C. What is the cross-sectional area of the hole (a) at 25.0\(^\circ\)C and (b) when the temperature of the plate is increased to 175\(^\circ\)C? Assume that the coefficient of linear expansion remains constant over this temperature range.

In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0\(^\circ\)C is mixed with a mass m of water that has an initial temperature of 80.0\(^\circ\)C. No heat is lost to the surroundings. If the final temperature of the system is 28.0\(\circ\)C, what is the mass m of the water that was initially at 80.0\(^\circ\)C?

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