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Chapter 19: Problem 4

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 \(400 \mathrm{K}\).

Short Answer

Expert verified
Question: Estimate the specific heat capacity of copper at (a) 40K and (b) 400K, given the Debye temperature (Θ_D) is 340K and the specific heat capacity at 20K is 0.38 J/mol-K. Answer: The specific heat capacity of copper at (a) 40K is approximately 1.52 J/mol-K and (b) at 400K is approximately 208.07 J/mol-K.

Step by step solution

01

Recall Debye’s Law formula

Debye's Law for estimating the specific heat capacity at constant volume is given by the following equation: Cv(T) = (9 * N * k * (T/Θ_D)^3) * ∫(0 to x) (u^4 * e^u) / (e^u - 1)^2 du where, Cv(T) is the specific heat at constant volume at temperature T, N is the number of atoms in one mole, k is the Boltzmann constant, Θ_D is the Debye temperature, T is the temperature in Kelvin, and u = x * (Θ_D / T). The limits of integration are from 0 to x, where x → ∞.
02

Calculate the ratio T/Θ_D for both temperatures

We are to estimate the specific heat at 40K and 400K. For T = 40K, Ratio_40K = T/Θ_D = 40/340 = 0.1176 For T = 400K, Ratio_400K = T/Θ_D = 400/340 = 1.1765
03

Use the given heat capacity at 20K to find a constant

We are given that at 20K, the heat capacity for copper is 0.38 J/mol-K. We can use this value to find a constant that will be used to estimate the specific heat capacity at the other temperatures. From Debye's Law equation, we have: Cv(20K) = A * (20/340)^3 0.38 = A * (1/289) A = 0.38 * 289 A = 109.82 J/mol Now, we can use this constant to find the specific heat capacity at the other temperatures.
04

Estimate the specific heat capacity at 40K and 400K

Now that we have the constant A, we can use the formula to estimate the specific heat capacity at the desired temperatures: Cv(40K) = A * (40/340)^3 Cv(40K) = 109.82 * (1/72.25) Cv(40K) ≈ 1.52 J/mol-K Cv(400K) = A * (400/340)^3 Cv(400K) = 109.82 * (1.1765)^3 Cv(400K) ≈ 208.07 J/mol-K
05

Answer

The specific heat capacity of copper at (a) 40K is approximately 1.52 J/mol-K and (b) at 400K is approximately 208.07 J/mol-K.

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Most popular questions from this chapter

We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower limits for the room- temperature thermal conductivity of an aluminum oxide material having a volume fraction of 0.25 of pores that are filled with still air.

(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 \(0 \mathrm{K}\).

The difference between the specific heats at constant pressure and volume is described by the expression $$c_{p}-c_{v}=\frac{\alpha_{v}^{2} v_{0} T}{\beta}\quad\quad\quad\quad\quad(19.10)$$ where \(\alpha_{v}\) is the volume coefficient of thermal expansion, \(v_{0}\) is the specific volume (i.e., volume per unit mass, or the reciprocal of density \(), \beta\) is the compressibility, and \(T\) is the absolute temperature. Compute the values of \(c_{v}\) at room temperature \((293 \mathrm{K})\) for aluminum and iron using the data in Table 19.1, assuming that \(\alpha_{v}=3 \alpha_{l}\) and given that the values of \(\beta\) for \(A l\) and \(F e\) are \(1.77 \times 10^{-11}\) and \(2.65 \times 10^{-12}(\mathrm{Pa})^{-1},\) respectively.

Compute the density for iron at \(700^{\circ} \mathrm{C}\), given that its room- temperature density is 7.870 \(\mathrm{g} / \mathrm{cm}^{3} .\) Assume that the volume coefficient of thermal expansion, \(\alpha_{v},\) is equal to \(3 \alpha_{l}\).
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