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Chapter 19: Problem 5

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 (K). Estimate \(\theta_{\mathrm{D}}\) for aluminum, given that the specific heat is \(4.60 \mathrm{J} / \mathrm{kg}-\mathrm{K}\) at \(15 \mathrm{K}\).

Short Answer

Expert verified
Question: Estimate the Debye temperature (θD) for aluminum, given the following information: the constant A in the equation is given by \(A = 12 \pi^{4} R / 5 \theta_{\mathrm{D}}^{3}\), specific heat of aluminum at 15K is \(4.60\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}\), and the gas constant R is \(8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\). Answer: The Debye temperature for aluminum is approximately \(432\,\mathrm{K}\).

Step by step solution

01

Given information

The given information is: Constant A: \(A = 12 \pi^{4} R / 5 \theta_{\mathrm{D}}^{3}\) Specific heat at 15 K: \(C = 4.60\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1}\) Gas constant R: \(R = 8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\)
02

Identify the specific heat formula

We must identify the specific heat formula relating A, C, and θD. In this exercise, we are given that: \(C = A T^{3}\)
03

Convert and substitute given values

Convert the specific heat from J/kg-K to J/mol-K by using the molar mass of aluminum (27 g/mol): \(4.60\,\mathrm{J}\,\mathrm{kg}^{-1}\,\mathrm{K}^{-1} \cdot \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} \cdot \frac{27\,\mathrm{g}}{1\,\mathrm{mol}} = 0.1242\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\) Substitute the given values into the equation for C and A at T=15 K: \(0.1242\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1} = (12\pi^{4}(8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}) / 5\theta_{\mathrm{D}}^{3})(15\,\mathrm{K})^{3}\)
04

Solve for θD

Now, we need to solve the equation above for θD: \(\theta_{D}^3 = \frac{12 \pi^{4} R \cdot (15\,\mathrm{K})^3}{5 \times 0.1242\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}}\) Calculate the expression for θD³: \(\theta_{D}^3 = 781965.7\,\mathrm{K}^{3}\) Now, take the cube root to find θD: \(\theta_{D} = \sqrt[3]{781965.7\,\mathrm{K}^{3}}\) \(\theta_{D} \approx 431.96\,\mathrm{K}\) Therefore, the Debye temperature for aluminum is approximately \(432\,\mathrm{K}\).

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Most popular questions from this chapter

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

Nonsteady-state heat flow may be described by the following partial differential equation: $$\frac{\partial T}{\partial t}=D_{T} \frac{\partial^{2} T}{\partial x^{2}}$$ where \(D_{T}\) is the thermal diffusivity; this expression is the thermal equivalent of Fick's second law of diffusion (Equation 5.4b). The thermal diffusivity is defined according to $$D_{T}=\frac{k}{\rho c_{p}}$$ In this expression, \(k, \rho,\) and \(c_{p}\) represent the thermal conductivity, the mass density and the specific heat at constant pressure, respectively. (a) What are the SI units for \(D_{T} ?\) (b) Determine values of \(D_{T}\) for copper brass, magnesia,fused silica, polystyrene, and polypropylene using the data in Table 19.1. Density values are included in Table B.1, Appendix B.

(a) Calculate the heat flux through a sheet of brass \(7.5 \mathrm{mm}(0.30 \text { in. })\) thick if the temperatures at the two faces are 150 and \(50^{\circ} \mathrm{C}\) \(\left(302 \text { and } 122^{\circ} \mathrm{F}\right) ;\) assume steady-state heat flow. (b) What is the heat loss per hour if the area of the sheet is \(0.5 \mathrm{m}^{2}\left(5.4 \mathrm{ft}^{2}\right) ?(\mathrm{c}) \mathrm{What}\) will be the heat loss per hour if soda-lime glass instead of brass is used? (d) Calculate the heat loss per hour if brass is used and the thickness is increased to \(15 \mathrm{mm}(0.59 \text { in. })\).

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) 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.
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