Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of the ordinary thermal conductivity?

Short Answer

Expert verified
Answer: Apparent thermal conductivity is more suitable for insulators because it accounts for all modes of heat transfer (conduction, convection, and radiation) and provides a more accurate representation of the insulator's performance in real-world applications, considering the complex structure and multiple factors affecting heat transfer in insulating materials.

Step by step solution

01

Define ordinary thermal conductivity

Thermal conductivity is a property of a material that measures its ability to conduct heat. It is typically denoted by the symbol 'k' and is measured in watts per meter per degree Kelvin (W/m·K). The higher the thermal conductivity, the better the material is at conducting heat.
02

Define apparent thermal conductivity

The apparent thermal conductivity is a measure of heat conduction that takes into account the complex structure of a material and its ability to transfer heat through conduction, convection, and radiation. In insulating materials, the effective heat transfer often depends on several factors, such as the density, structure, and orientation of the material.
03

The importance of taking into account all forms of heat transfer

In many practical cases, especially with insulating materials, the heat transfer occurs through a combination of conduction, convection, and radiation. In such cases, it is necessary to account for all these modes of heat transfer to obtain an accurate value of the material's ability to conduct heat. The apparent thermal conductivity helps to achieve this by considering all mechanisms that contribute to the overall heat transfer process.
04

Explain why apparent thermal conductivity is more suitable for insulators

Apparent thermal conductivity is more suitable for insulators for the following reasons: 1. Insulating materials often have complex and heterogeneous structures, which means that the heat transfer in such materials is not solely dependent on conduction. Therefore, using the ordinary thermal conductivity may give inaccurate results for such materials. 2. The apparent thermal conductivity takes into account all modes of heat transfer (i.e., conduction, convection, and radiation), giving a more accurate measurement of an insulator's ability to conduct heat and perform under different environmental conditions. 3. Insulating materials may have different effective heat transfer capabilities based on their density, structure, and orientation, which are better described by the apparent thermal conductivity. In conclusion, the apparent thermal conductivity is a more suitable parameter for characterizing the heat conduction ability of insulators because it accounts for all modes of heat transfer and provides a more accurate representation of the insulator's performance in real-world applications.

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Key Concepts

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

Ordinary Thermal Conductivity
Understanding thermal conductivity is critical for grasping how materials interact with heat. Ordinary thermal conductivity, represented by the symbol \( k \), quantifies a material's inherent ability to conduct heat. It is expressed in units of watts per meter per degree Kelvin (W/m·K). This property is pivotal in engineering and construction, as it influences how quickly or slowly heat moves through materials.

For example, metals typically have a high thermal conductivity and efficiently transfer heat. This property can be demonstrated by considering a metal spoon that becomes hot quickly when placed in a pot of boiling water. The high conductivity means that the heat from the water is rapidly transferred through the material of the spoon.

However, it's important to bear in mind that ordinary thermal conductivity is an idealized measurement. It mainly accounts for heat transfer by conduction, one of the modes of heat transfer, which occurs at the microscopic level as energy is transferred between adjacent atoms or molecules.
Heat Conduction in Insulators
When we delve into the heat conduction in insulators, we deal with a different scenario compared to conductive materials like metals. Insulators are designed to impede heat flow, which is why they often play a crucial role in energy conservation and thermal management systems.

The effectiveness of these materials doesn't rely on the ordinary thermal conductivity alone because they generally have a complex structure that affects heat flow. For instance, insulation used in building walls often consists of materials that trap air within a fibrous matrix. The air pockets significantly reduce heat transfer by conduction, making the material an excellent insulator.

Since heat transfer in insulators can occur via multiple mechanisms and can be influenced by factors such as density and structure, the concept of apparent thermal conductivity becomes essential. It encapsulates all the different ways heat can traverse insulative materials, offering a more comprehensive picture than simple conduction models.
Modes of Heat Transfer
In the natural world, heat can travel in three primary ways: conduction, convection, and radiation. These modes of heat transfer are fundamental to understanding how thermal energy moves through different environments and materials.

Conduction occurs when heat moves through a solid material, as seen with a heated iron rod; the heat travels from the hotter end to the cooler end through the rod's material. Convection, on the other hand, involves the movement of heat by the physical movement of fluid (which includes liquids and gases). An example of convection is the warm air rising from a radiator. Lastly, radiation is the transfer of heat in the form of electromagnetic waves, which is how we receive heat from the sun despite the cold vacuum of space.

Each mode of heat transfer operates based on different principles and can be influenced by various factors. For instance, conduction is affected by the nature of the material and its temperature, while convection currents can be influenced by the shape and orientation of the heating surface. This complexity is why the use of apparent thermal conductivity becomes imperative, as it integrates all these modes, allowing for a more accurate understanding and prediction of how heat will behave in real-world applications.

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Most popular questions from this chapter

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for \(10 \mathrm{~min}\). During the process, \(500 \mathrm{~kJ}\) of heat is lost from the water. The temperature rise of water is (a) \(5.6^{\circ} \mathrm{C}\) (b) \(9.6^{\circ} \mathrm{C}\) (c) \(13.6^{\circ} \mathrm{C}\) (d) \(23.3^{\circ} \mathrm{C}\) (e) \(42.5^{\circ} \mathrm{C}\)

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Consider a house in Atlanta, Georgia, that is maintained at \(22^{\circ} \mathrm{C}\) and has a total of \(20 \mathrm{~m}^{2}\) of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-125 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.

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