Solution: The thermal conductivity of the composite wire kc parallel to the wire length is given by kc = fCu kCu + fAg kAg, where fCu and fAg are the volume fractions of the copper and silver, respectively, and kCu and kAg are the thermal conductivities of the copper and silver, respectively. The thermal conductivity must be a minimum of 410 W/(m–K). Since the thermal conductivity of copper is less than 410 W/(m–K), there is a maximum volume fraction of copper that will permit this thermal conductivity criterion to be met. The sum of the volume fractions of copper and silver must equal one, i.e., fCu + fAg = 1, such that fAg = 1–fCu. Substituting and solving for fCu, 410 < fCu(401) + (1–fCu)(429) (429–401) fCu < 429–410 fCu < 0.68. The electrical conductivity of the composite wire σc parallel to the wire length is given by σc = fCu σCu + fAg σAg, The electrical conductivity must be a minimum of 60 × 106Ω–1m–1. Since the electrical conductivity of copper is less than 60 × 106Ω–1m–1, there is a maximum volume fraction of copper that will permit this electrical conductivity criterion to be met. Substituting and solving for fCu, 60 × 106 < fCu (59 × 106) + (1–fCu) (63 × 106) (63 × 106–59 × 106) fCu < (63 × 106–60 × 106) fCu < 0.75. A 1 m length of the composite wire with a diameter of 1 cm has a volume V of V = [π (1 cm)2/4] (100 cm) = 78.5 cm3. Thus the density of the composite wire must not exceed ρc = 750 g/(78.5 cm3) = 9.55 g/cm3. The density of the composite wire is given by ρc = fCu ρCu + fAg ρAg. Since the density of silver is greater than 9.55 g/cm3, there is a maximum volume fraction of silver that will permit this density criterion to be met. Substituting and solving for fCu, 9.55 > fCu (8.96) + (1–fCu) (10.49) (10.49–8.96)) fCu > (10.49–9.55) fCu > 0.61 Based on the three criteria listed in this problem, the allowed range for the volume fraction of copper is 0.61 < fCu < 0.68. Since the volume fraction of the copper is proportional to the square of the diameter of the copper core divided by the square of the wire diameter, 0.61 < [d2/(1 cm)2] < 0.68 or 0.78 cm < d < 0.82 cm. 365 © 2016 Cengage Learning. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.