Question

Bronze is an alloy made of copper (Cu) and tin (Sn). Calculate the mass of a bronze cylinder of radius 6.44 cm and length 44.37 cm. The composition of the bronze is 79.42 percent Cu and 20.58 percent Sn and the densities of Cu and Sn are 8.94 g/cm^3 and 7.31 g/cm^3, respectively. What assumption should you make in this calculation?

Answer 1

Answer:

starting from binary alloy:

⇒ mass of alloy cylinder is 49430.896 g

Explanation:

be density of binary alloy (da) :

⇒ da = (MCu + MSn) / (VCu + VSn)

∴ VCu = MCu / dCu ∧ VSn = MSn/ dSn

⇒ da = (MCu + MSn) / ((MCu / dCu) + (MSn / dSn))

⇒ 1 / da = ((MCu / dCu) + (MSn/dSn)) / (MCu + MSn)

∴ MCu / (MCu + MSn) = XCu

∴ MSn / ( MSn + MCu ) = XSn

⇒ 1 / da = ( XCu /dCu ) + ( XSn / dSn )

∴ XCu = 0.7942;  XSn = 0.2058

⇒ XCu / dCu = 0.7942 / 8.94 = 0.0888

⇒ Xsn / dSn = 0.2058 / 7.31 = 0.02815

⇒ 1 / da = 0.0888 + 0,02815 = 0.116953

⇒ da = 8.55042 g/cm³

Cylinder Volume:

V =$\pi$ * r² * l

∴ r = 6.44cm;  ∧ l = 44.37cm

⇒⇒V = $\pi$ * (6.44cm)² * (44.37cm) = 5781.0738 cm³

mass cylinder:

• m = da * V

⇒ m = (8.55042 g/cm³) * (5781.0738 cm³)

⇒ m = 49430.896 g

Answer 2

Answer:

49,415.43 gram is the mass of a bronze cylinder.

Explanation:

Assuming that the given composition of bronze is mass percentage of copper and tin in bronze.

Let he mass of bronze be x.

Mass percentage of copper in bronze = 79.42%

Mass of copper = 79.42% of x=0.7942 x

Volume of the copper: $V_1=\frac{Mass}{density}=\frac{0.7942 x}{8.94 g/cm^3}$

Mass percentage of copper in bronze = 20.58%

Mass of copper = 20.58% of x=0.2058 x

Volume of the copper: $V_2=\frac{Mass}{density}=\frac{0.2058 x}{7.31 g/cm^3}$

Volume of the bronze cylinder = V

Radius of the cylinder = r = 6.44 cm

Height of the cylinder = h = 44.37 cm

$V=\pi r^2h=3.14\times (6.44 cm)^2\times 44.37 cm=5,781.1073 cm^3$

$V=V_1+V_2$

$5,781.1073 cm^3=\frac{0.7942 x}{8.94 g/cm^3}+\frac{0.2058 x}{7.31 g/cm^3}$

x = 49,415.43 g

49,415.43 gram is the mass of a bronze cylinder.