Respuesta :
Answer:
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
Explanation:
Consider 100 grams 3.5% C cast iron.
Mass of carbon in 100 grams of cast iron = [tex]m_1=100 g\times \frac{3.5 }{100}=3.5 g[/tex]
Mass of iron in cast iron =[tex]m_2[/tex] = 100 g - 3.5 g = 96.5 g
Density of graphite = [tex]d_1=2.3g/cm^3[/tex]
Volume of the graphite = [tex]v_1[/tex]
[tex]v_1=\frac{m_1}{d_1}=\frac{3.5 g}{2.3g/cm^3}=1.52 cm^3[/tex]
Density of iron= [tex]d_2=7.9 g/cm^3[/tex]
Volume of the iron = [tex]v_2[/tex]
[tex]v_2=\frac{m_2}{d_2}=\frac{96.5 g}{7.9 g/cm^3}=12.22 cm^3[/tex]
Total volume of the cast iron ,V=[tex]v_1+v_2=1.52 cm^3+12.22 cm^3=13.74 cm^3[/tex]
Volume percent of the graphite :
[tex]=\frac{v_1}{V}\times 100[/tex]
[tex]=\frac{1.52 cm^3}{13.74 cm^3}\times 100=11.06\%\approx 11\%[/tex]
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
The volume percent of graphite VGr in a 3.5 wt% C cast iron is;
V_gr = 11.08%
We are given;
Density of Ferrite; ρₐ = 7.9 g/cm³
Density of Graphite; [tex]\rho _{g}[/tex] = 2.3 g/cm³
Weight percent of C cast iron; C₀ = 3.5%
Now the formula for the mass fraction of ferrite is;
Wₐ = [tex]\frac{C_{g} - C_{0}}{C_{g} - C_{a}}[/tex]
If we consider 100 g of 3.5 wt% C cast iron, then [tex]C_{g}[/tex] = 100 and Cₐ = 0
Thus;
Wₐ = [tex]\frac{100 - 3.5}{100 - 0}[/tex]
Wₐ = 0.965
Thus, mass fraction of graphite is;
[tex]W _{g}[/tex] = 1 - Wₐ
[tex]W _{g}[/tex] = 1 - 0.965
[tex]W _{g}[/tex] = 0.035
Formula for the volume percent of graphite is;
[tex]V_{gr} = \frac{\frac{W_{g}}{\rho_{g}}}{\frac{W_{a}}{\rho_{a}} + \frac{W_{g}}{\rho_{g}}} * 100%[/tex]
Plugging in the relevant values gives;
V_gr = [(0.035/2.3)/((0.965/7.9) + (0.035/2.3))]
V_gr = 0.01521739/(0.1221518987 + 0.01521739)
V_gr = 11.08%
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