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( T) 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 E(glass) = (840 J/kg-K)(2 kg)(80 K) = 1.34 x 10 5 E(HDPE) = (1850 J/kg-K)(2 kg)(80 K) = 2.96 x 10 5 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 = Q 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 Btu/lb m - ° F 1J/kg-K = 0.090 Btu/lb m - ° F Thus 1
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T = 65 Btu 10 lb m () 0.090 Btu/lb m - ° F = 72.2 ° F and T f = T o + T = 77 ° F + 72.2 ° F = 149.2 ° F (65.1 ° C) 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 2
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Therefore, at 50 K C v = AT 3 = 3 x 10 -5 J/mol-K 4 () (50 K) 3 = 3.75 J/mol-K 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 12 π 4 R 5 θ D 3 Or, solving for θ D 3
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θ D = 12 π 4 R 5A 1/3 (12)( π ) 4 (8.31 J/mol -K) (5) 4.96 x10 5 J/ mol- K 4 () 1/ 3 = 340 K 19.6 (a) The reason that C v rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited.
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Chap 19 Solns-6E - CHAPTER 19 THERMAL PROPERTIES PROBLEM...

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