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Chap 19 Solns-6E

# Chap 19 Solns-6E - CHAPTER 19 THERMAL PROPERTIES PROBLEM...

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CHAPTER 19 THERMAL PROPERTIES PROBLEM SOLUTIONS 19.1 The energy, E , required to raise the temperature of a given mass of material, m , is the product of the specific heat, the mass of material, and the temperature change, T , as E = c p m( ∆Τ 29 The T is equal to 100 ° C - 20 ° C = 80 ° C (= 80 K), while the mass is 2 kg, and the specific heats are presented in Table 19.1. Thus, E(aluminum) = (900 J/kg-K)(2 kg)(80 K) = 1.44 x 10 5 J E(steel) = (486 J/kg-K)(2 kg)(80 K) = 7.78 x 10 4 J E(glass) = (840 J/kg-K)(2 kg)(80 K) = 1.34 x 10 5 J E(HDPE) = (1850 J/kg -K)(2 kg)(80 K) = 2.96 x 10 5 J 19.2 We are asked to determine the temperature to which 10 lb m of brass initially at 25 ° C would be raised if 65 Btu of heat is supplied. This is accomplished by utilization of a modified form of Equation (19.1) as T = DQ mc p in which Q is the amount of heat supplied, m is the mass of the specimen, and c p is the specific heat. From Table 19.1, c p = 375 J/kg-K, which in Customary U.S. units is just c p = (375 J/kg- K) 2.39 x 10 -4 Βτυ/λβ μ - Φ 1 ϑ/κγ -Κ = 0.090Βτυ/λβ μ - Φ Thus 244

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T = 65 Btu 10 lb m ( ) 0.090 Btu/lb m - F ( ) = 72.2 F and T f = T o + ∆Τ = 77 Φ + 72.2 Φ = 149.2 Φ (65.1 Χ 29 19.3 (a) This problem asks that we determine the heat capacities at constant pressure, C p , for copper, iron, gold, and nickel. All we need do is multiply the c p values in Table 19.1 by the atomic weight, taking into account the conversion from grams to kilograms. Thus, for Cu C p = (386 J/kg- K)(1 kg/1000 g)(63.55 g/mol) = 24.5 J/mol - K For Fe C p = (448 J/kg- K)(1 kg/1000 g)(55.85 g/mol) = 25.0 J/mol - K For Au C p = (128 J/kg- K)(1 kg/1000 g)(196.97 g/mol) = 25.2 J/mol - K For Ni C p = (443 J/kg- K)(1 kg/1000 g)(58.69 g/mol) = 26.0 J/mol - K (b) These values of C p are very close to one another because room temperature is considerably above the Debye temperature for these metals, and, therefore, the values of C p should be about equal to 3R [(3)(8.31 J/mol-K) = 24.9 J/mol-K], which is indeed the case for all four of these metals. 19.4 (a) For aluminum, C v at 50 K may be approximated by Equation (19.2), since this temperature is significantly below the Debye temperature. The value of C v at 30 K is given, and thus, we may compute the constant A as A = C v T 3 = 0.81 J/ mol- K (30 K) 3 = 3 x 10 -5 J/mol- K 4 245
Therefore, at 50 K C v = AT 3 = 3 x 10 -5 J/mol-K 4 ( 29 (50Κ 29 3 = 3.75ϑ/μολ and c v = (3.75 J/mol- K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg - K (b) Since 425 K is above the Debye temperature, a good approximation for C v is C v = 3R = (3)(8.31 J/mol -K) = 24.9 J/mol -K And, converting this to specific heat c v = (24.9 J/mol- K)(1 mol/26.98 g)(1000 g/kg) = 925 J/kg - K 19.5 For copper, we want to compute the Debye temperature, θ D , given the expression for A in Equation (19.2) and the heat capacity at 10 K. First of all, let us determine the magnitude of A , as A = C v T 3 = (0.78 J /mol- K)(1 kg/1000 g)(63.55 g/mol) (10 K) 3 = 4.96 x 10 -5 J/mol- K 4 As stipulated in the problem A = 12 π 4 Ρ 3 Or, solving for θ D 246

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θ D = 12 p 4 R 5A 1/3 = (12)( π29 4 (8.31 ϑ/μολ-Κ 29 (5 29 4.96 ξ 10 -5 ϑ/ μολ-Κ 4 ( 29 1/ 3 = 340 Κ 19.6 (a) The reason that C v rises with increasing temperature at temperatures near 0 K is because, in
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