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Chapter 19: Problem 17

Briefly explain why metals are typically better thermal conductors than ceramic materials.

Short Answer

Expert verified
Answer: Metals are generally better thermal conductors than ceramic materials due to their highly ordered lattice structure with closely packed atoms and the presence of free electrons in their metallic bonds, which allows for efficient heat transfer. In contrast, ceramics have less ordered structures with predominantly ionic bonding, which results in localized atomic vibrations and hinders heat flow through the material.

Step by step solution

01

Understand Thermal Conductivity

Thermal conductivity is the ability of a material to transfer heat. Heat is transferred mainly through three mechanisms: conduction, convection, and radiation. In solids, conduction is the dominant method of heat transfer. The thermal conductivity of a material depends on several factors, such as its atomic/molecular structure, the type of bonding between atoms, and the presence of impurities or defects in the material.
02

Analyze the Atomic/Molecular Structure of Metals and Ceramics

Metals have a highly ordered lattice structure made up of closely packed atoms. These atoms can easily transmit heat through their lattice network. Ceramic materials, on the other hand, are made up of a combination of metal and non-metal atoms that form complex structures. These structures are typically less ordered and have a lower packing density than metals.
03

Examine the Bonding in Metals and Ceramics

In metals, the bonding between atoms is typically metallic, with a 'sea' of free electrons distributed among positively charged metal ions. These free electrons can move easily through the lattice and are excellent carriers of heat. In ceramics, the bonding is predominantly ionic (and/or covalent in some cases), which leads to the formation of localized bonds, and there are no free electrons to carry heat.
04

Consider the Effect of Impurities and Defects

Both metals and ceramics can contain impurities or defects. However, due to the close packing and ordered arrangement of atoms in metals, defects or impurities may have a lesser effect on the thermal conductivity compared to ceramics. In ceramics, defects and impurities can further localize atomic vibrations and hinder the flow of heat through the material.
05

Conclusion

Metals are typically better thermal conductors than ceramic materials mainly because of their highly ordered lattice structure with closely packed atoms and the presence of free electrons in their metallic bonds, which allows for efficient heat transfer. Ceramics, on the other hand, have less ordered structures with predominantly ionic bonding, which results in localized atomic vibrations and hinders heat flow through the material.

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Most popular questions from this chapter

For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your choices. (a) Pure silver; sterling silver \((92.5\) wt\% \(\mathrm{Ag}-7.5 \mathrm{wt} \% \mathrm{Cu})\) (b) Fused silica; polycrystalline silica (c) Linear and syndiotactic poly(vinyl chloride) \((D P=1000) ;\) linear and syndiotactic polystyrene \((D P=1000)\) (d) Atactic polypropylene \(\left(\bar{M}_{w_{5}}=10^{6} \mathrm{~g} / \mathrm{mol}\right)\); isotactic polypropylene \(\left(\bar{M}_{w}=10^{5} \mathrm{~g} / \mathrm{mol}\right)\)

Briefly explain why the thermal conductivities are higher for crystalline than for noncrystalline ceramics.

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

What measures may be taken to reduce the likelihood of thermal shock of a ceramic piece?

The constant \(A\) in Equation \(19.2\) is \(12 \pi^{4} R / 5 \theta_{\mathrm{D}}^{3}\), where \(R\) is the gas constant and \(\theta_{\mathrm{D}}\) is the Debye temperature \((\mathrm{K})\). Estimate \(\theta_{\mathrm{D}}\) for aluminum, given that the specific heat is \(4.60 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) at \(15 \mathrm{~K}\).
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