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1.12 Heat capacity
a. Calculate the heat capacity per mole and per gram of N 2 gas, neglecting the vibrations of the molecule. How does this compare with the experimental value of 0.743 J g-1^ K-1? b. Calculate the heat capacity per mole and per gram of CO 2 gas, neglecting the vibrations of the molecule. How does this compare with the experimental value of 0.648 J K-1^ g-1? Assume that CO 2 molecule is linear (O-C-O), so that it has two rotational degrees of freedom. c. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of solid silver. How does this compare with the experimental value of 0.235 J K-1^ g-1? d. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of the silicon crystal. How does this compare with the experimental value of 0.71 J K- g-1?
Solution
a. N 2 has 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 2 × 14.01 g/mol = 28.02 g/mol. Let Cm = heat capacity per mole, cs = specific heat capacity (heat capacity per gram), and R = gas constant, then: ∴ cs = Cm / Mat = (20.8 J K-1^ mol-1)/(28.02 g/mol) = 0.742 J K-1^ g- This is close to the experimental value. b. CO 2 has the linear structure O=C=O. Rotations about the molecular axis have negligible rotational energy as the moment of inertia about this axis is negligible. There are therefore 2 rotational degrees of freedom. In total there are 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 12.01 + 2 × 16 = 44.01 g/mol. ∴ cs = Cm / Mat = (20.8 J K-1^ mol-1)/(44.01 g/mol) = 0.47 J K-1^ g- This is smaller than the experimental value 0.648 J K-1^ g-1. The 5 degrees of freedom assigned to the CO 2 molecule did not include molecular vibrations and “vibrations” of the atoms that include “flexing” or bond bending.
c. For solid silver, there are 6 degrees of freedom: 3 vibrational KE and 3 elastic PE terms. Its molar mass is, Mat = 107.87 g/mol. ∴ cs = Cm / Mat = (24.9 J K-1^ mol-1)/(107.87 g/mol) = 0.231 J K-1^ g- This is very close to the experimental value. d. For a solid, heat capacity per mole is 3 R. The molar mass of Si is Mat = 28.09 g/mol. ∴ cs = Cm / Mat = (24.9 J K-1^ mol-1)/(28.09 g/mol) = 0.886 J K-1^ g- The experimental value is substantially less and is due to the failure of classical physics. One has to consider the quantum nature of the atomic vibrations and also the distribution of vibrational energy among the atoms. The student is referred to modern physics texts (under heat capacity in the Einstein model and the Debye model of lattice vibrations).
∴ EA = 3.47 eV/atom b. To find Do , use one of the equations for the diffusion coefficients: ∴ ∴ D 0 = 8.12 × 10 -5^ m^2 /s c. Given: time ( t ) = (1 hr) × (3600 s/hr) = 3600 s At 1000 °C, rms diffusion distance ( L 1000 °C) in time t is given by: L 1000 °C ∴ L 1000 ° C = 1.04 × 10 -7^ m or 0.104 μ m At 1200 °C: L 1200 °C ∴ L 1200 ° C = 8.90 × 10 -7^ m or 0.89 μ m (almost 10 times longer than at 1000 °C) d. Diffusion in polycrystalline Si would involve diffusion along grain boundaries, which is easier than diffusion in the bulk. The activation energy is smaller because it is easier for an atom to break bonds and jump to a neighboring site; there are vacancies or voids, broken bonds, and strained bonds in a grain boundary.
1.23 BCC and FCC Crystals
a. Tungsten (W) has the BCC crystal structure. The radius of the W atom is 0.1371 nm. The atomic mass of W is 183.8 amu (g mol-1). Calculate the number of W atoms per unit volume and density of W. b. Platinum (Pt) has the FCC crystal structure. The radius of the Pt atom is 0.1386 nm. The atomic mass of Pt is 195.09 amu (g mol-1). Calculate the number of Pt atoms per unit volume and density of Pt.
Solution
a. Consider a cube diagonal with two corner atoms and the central atom in contact. The length of the diagonal is 4 R. Therefore a^2 + a^2 + a^2 = (4 R ) 2 or Thus for W, 0.3166 nm In the BCC structure, the total number of atoms = = 2 atoms. The density of W is therefore = 19.23 × 103 kg m-3^ or 19.23 g cm- The atomic concentration of W with 2 atoms per unit cell is = 6.303 × 1028 m-3^ or 6.303 × 1022 cm- b. Consider a face diagonal with corner atoms and the central atom in contact. The length of the diagonal is 4 R. Therefore a^2 + a^2 = (4 R ) 2 or
*2.19 Thermal resistance Consider a coaxial cable operating under steady state
conditions when the current flow through the inner conductor generates Joule heat at a rate P = I^2 R. The heat generated per second by the core conductor flows through the dielectric;. The inner conductor reaches a temperature Ti whereas the outer conductor is at To. Show that the thermal resistance θ of the hollow cylindrical insulation for heat flow in the radial direction is Thermal resistance of hollow cylinder where a is the inside (core conductor) radius, b is the outside radius (outer conductor), κ is the thermal conductivity of the insulation, and L is the cable length. Consider a coaxial cable that has a copper core conductor and polyethylene (PE) dielectric with the following properties: Core conductor resistivity ρ = 19 nΩ m, core radius, a = 4 mm, dielectric thickness, b - a = 3.5 mm, dielectric thermal conductivity κ = 0.3 W m-1^ K-1. The outside temperature To is 25 °C. The cable is carrying a current of 500 A. What is the temperature of the inner conductor?
Solution
Consider a thin cylindrical shell of thickness dr as shown in Figure 2Q12-1. The temperature difference across dr is dT. The surface area of this shell is 2π rL. Thus, from Fourier’s law, which we can integrate with respect to r from r = a where T = Ti to r = b where T = To , i.e. Thus the thermal resistance of the hollow cylindrical insulation is
Figure 2Q12-1: Thermal resistance of a hollow cylindrical shell. Consider an infinitesimally thin cylindrical shell of radius r and thickness dr in the dielectric and concentrically around the inner conductor. The surface area is 2π rL. The actual length of the conductor does not affect the calculations as long as the length is sufficiently long such that there is no heat transfer along the length; heat flows radially from the inner to the outer conductor. We consider a portion of length L of a very long cable and we set L = 1 m so that the calculations are per unit length. The joule heating per unit second (power) generated by the current I through the core conductor is = 94.5 W The thermal resistance of the insulation is, = 0.33 ° C/W Thus, the temperature difference Δ T due to Q ′ flowing through θ is, Δ T = Q ′θ = (94.5 W)(0.33 °C/W) = 31.2 ° C. The inner temperature is therefore, Ti = To + Δ T = 25 + 31.2 = 56.2 ° C. Note that for simplicity we assumed that the inner conductor resistivity ρ and thermal conductivity κ are constant (do not change with temperature).
Now the bulb’s operating temperature ( T 1 ) can be found using our obtained values in the equation for resistivity of W (assuming room temperature To = 293 K and given n = 1.2): isolate: b. First we need the surface area A of the Tungsten filament. Since it is cylindrical in shape: A = L (π D ) = (0.579 m)(π)(63.5 × 10 -6^ m) = 0.0001155 m^2 Now the temperature of the filament T 1 can be found by isolating it in Stefan’s equation and substituting in the given values for emissivity (∈ = 0.35), Stefan’s constant (σ s = 5. × 10 -8^ W m-2^ K-4) and room temperature ( To = 293 K). ∴ ∴ ∴ T 1 = 2570 K Note: We can even ignore To to get the same temperature since To << T 1 : ∴ ∴ T 1 = 2570 K These values are fairly close to the answer obtained in part (a). c. Let V be the voltage and R be the resistance when the filament is at temperature Tm. We are given the temperature Tm = 3407 °C + 273 = 3680 K. Since we know the following:
and We can make a substitution for ρ and use the values given for the light bulb filament to find the resistance of the filament at temperature Tm. ∴ R = 213.3 Ω Assuming that all electrical power is radiated from the surface of the bulb, we can use Stefan’s law and substitute in V^2 / R for power P : ∴ ∴ ∴ V = 299 V
