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Chapter 19: Problem 20

For some ceramic materials, why does the thermal conductivity first decrease and then increase with rising temperature?

Short Answer

Expert verified
Answer: In some ceramic materials, the thermal conductivity first decreases due to increased phonon scattering caused by lattice defects at lower temperatures. However, as the temperature continues to rise, phonon-phonon scattering (Umklapp scattering) becomes dominant, allowing more heat-carrying phonons to propagate without being scattered by lattice defects, resulting in an increase in thermal conductivity.

Step by step solution

01

Understanding Thermal Conductivity

Thermal conductivity is the ability of a material to conduct heat. It depends on the material's composition, structure, and temperature. In general, metals have high thermal conductivity, while insulators like ceramics have lower thermal conductivity.
02

Factors Affecting Thermal Conductivity

There are three main factors that influence the thermal conductivity of a material: 1. Phonons (lattice vibrations): The primary mode of heat conduction in insulators and non-metallic materials. 2. Free electrons: The main heat carriers in metals, which contribute to higher thermal conductivity. 3. Impurities and defects: These can scatter phonons and electrons, reducing the overall thermal conductivity of a material.
03

Temperature Dependence of Phonon Scattering

In ceramic materials, heat is mainly conducted through phonon vibrations. As temperature increases, the phonon population and vibrational energy also increase, causing more frequent collisions between phonons and lattice defects. These increasing collisions lead to an increased scattering of phonons, resulting in a decrease in thermal conductivity.
04

Why Thermal Conductivity Increases After a Certain Temperature

The decrease in thermal conductivity with increasing temperature continues until the phonon scattering due to lattice defects becomes less significant compared to the phonon-phonon scattering (Umklapp scattering). Umklapp scattering is a temperature-dependent process that occurs when multiple phonons collide, which leads to a new phonon being created with a different direction. At higher temperatures, this process becomes more significant, and more heat-carrying phonons can propagate through the material without being scattered by lattice defects, causing an increase in thermal conductivity.
05

Conclusion

For some ceramic materials, the thermal conductivity first decreases and then increases with rising temperature because of the temperature-dependent behavior of phonon vibrations. With increasing temperature, phonon scattering due to lattice defects becomes more prominent, causing a decrease in thermal conductivity. However, after reaching a certain temperature, phonon-phonon scattering (Umklapp scattering) becomes dominant, allowing more heat-carrying phonons to propagate through the material without being scattered by lattice defects, resulting in an increase in thermal conductivity.

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Most popular questions from this chapter

(a) Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative. (b) Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.

(a) Briefly explain why \(C_{v}\) rises with increasing temperature at temperatures near \(0 \mathrm{K}\) (b) Briefly explain why \(C_{v}\) becomes virtually independent of temperature at temperatures far removed from \(0 \mathrm{K}\).

A bimetallic strip is constructed from strips of two different metals that are bonded along their lengths. Explain how such a device may be used in a thermostat to regulate temperature.

For copper, the heat capacity at constant volume \(C_{v}\) at \(20 \mathrm{K}\) is \(0.38 \mathrm{J} / \mathrm{mol}\) - \(\mathrm{K}\), and the Debye temperature is \(340 \mathrm{K}\). Estimate the specific heat (a) at \(40 \mathrm{K}\) and (b) at \(400 \mathrm{K}\).

A steel wire is stretched with a stress of \(70 \mathrm{MPa}(10,000 \mathrm{psi})\) at \(20^{\circ} \mathrm{C}\left(68^{\circ} \mathrm{F}\right) .\) If the length is held constant, to what temperature must the wire be heated to reduce the stress to \(17 \mathrm{MPa}(2500 \mathrm{psi}) ?\) v
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