Compute the volume percent of graphite \(V_{\mathrm{Gr}}\) in a \(3.5 \mathrm{wt} \% \mathrm{C}\) cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of \(7.9\) and \(2.3 \mathrm{~g} / \mathrm{cm}^{3}\) for ferrite and graphite, respectively.

First, let's define the weight percentages as follows: - The cast iron sample has a weight percentage of carbon (as graphite) of 3.5 wt%; let's denote this as \(W_{\mathrm{C}}\). - Since there are only two components in this cast iron sample, the weight percentage of the ferrite component will be the remaining percentage, which is \((100 - W_{\mathrm{C}})\) wt%; let's denote this as \(W_{\mathrm{F}}\).

Now, let's calculate the weight ratio of graphite to ferrite (W). We will define this ratio as the weight percent of graphite divided by the weight percent of ferrite: \(W = \frac{W_{\mathrm{C}}}{W_{\mathrm{F}}} = \frac{3.5}{100 - 3.5} = \frac{3.5}{96.5}\)

To establish a volume relationship between ferrite and graphite, we will use their densities to express this ratio in terms of the volume percents of ferrite and graphite. Let's denote the volume percent of ferrite as \(V_{\mathrm{F}}\), and the volume percent of graphite as \(V_{\mathrm{Gr}}\). Using the definition of density and the weight ratio, we can form the following expression: \(W = \frac{\rho_{\mathrm{Gr}} \cdot V_{\mathrm{Gr}}}{\rho_{\mathrm{F}} \cdot V_{\mathrm{F}}}\) In this expression, \(\rho_{\mathrm{F}}\) is the density of ferrite and \(\rho_{\mathrm{Gr}}\) is the density of graphite. Since we are given the densities as \(\rho_{\mathrm{F}} = 7.9 \mathrm{~g/cm}^3\) and \(\rho_{\mathrm{Gr}} = 2.3 \mathrm{~g/cm}^3\), we can plug in these values and solve the equation for \(V_{\mathrm{Gr}}\): \(V_{\mathrm{Gr}} = W \cdot \frac{\rho_{\mathrm{F}} \cdot V_{\mathrm{F}}}{\rho_{\mathrm{Gr}}}\) However, we do not know \(V_{\mathrm{F}}\) directly, but we can express it in terms of \(V_{\mathrm{Gr}}\) as follows: \(V_{\mathrm{F}} = 1 - V_{\mathrm{Gr}}\) This is because the sum of volume percents of both ferrite and graphite must equal 100%.

Now, we can substitute the expression for \(V_{\mathrm{F}}\) into the equation for \(V_{\mathrm{Gr}}\) and solve for the latter: \(V_{\mathrm{Gr}} = W \cdot \frac{\rho_{\mathrm{F}} (1 - V_{\mathrm{Gr}})}{\rho_{\mathrm{Gr}}}\) We plug in W and the values for the densities to get: \(V_{\mathrm{Gr}} = ( \frac{3.5}{96.5} ) \cdot \frac{7.9 (1 - V_{\mathrm{Gr}})}{2.3}\) Let's solve for \(V_{\mathrm{Gr}}\): \(V_{\mathrm{Gr}} = 1 - \frac{2.3 \cdot 96.5}{3.5 \cdot 7.9}\) \(V_{\mathrm{Gr}} = 1 - \frac{222.95}{27.65}\) \(V_{\mathrm{Gr}} = 1 - 8.064\) \(V_{\mathrm{Gr}} = -7.064\) Since it is impossible to have a negative volume percent, we made an error in our calculation. We must have made the error when expressing the volume percents of graphite and ferrite relative to each other. Let's go back and correct that. We should express \(V_{\mathrm{F}}\) in terms of \(V_{\mathrm{Gr}}\) as follows: \(V_{\mathrm{F}} = \frac{1 - V_{\mathrm{Gr}}}{V_{\mathrm{Gr}}}\) Now, let's plug this in the equation for \(V_{\mathrm{Gr}}\) and solve for the latter: \(V_{\mathrm{Gr}} = W \cdot \frac{\rho_{\mathrm{F}} \cdot V_{\mathrm{F}}}{\rho_{\mathrm{Gr}}}\) \(V_{\mathrm{Gr}} = ( \frac{3.5}{96.5} ) \cdot \frac{7.9 \cdot (\frac{1 - V_{\mathrm{Gr}}}{V_{\mathrm{Gr}}})}{2.3}\) Now, solve for \(V_{\mathrm{Gr}}\): \(V_{\mathrm{Gr}}^2 = ( \frac{3.5}{96.5} ) \cdot \frac{7.9 (1 - V_{\mathrm{Gr}})}{2.3}\) \(V_{\mathrm{Gr}}^2 = \frac{221.46 (1 - V_{\mathrm{Gr}})}{96.5}\) \(96.5 \cdot V_{\mathrm{Gr}}^2 = 221.46 - 221.46 \cdot V_{\mathrm{Gr}}\) Now, solving the quadratic equation for \(V_{\mathrm{Gr}}\), we get two possible values: \(V_{\mathrm{Gr}} = 0.0528\) or \(V_{\mathrm{Gr}} = 0.9472\) Since the volume percent of graphite must be less than the total volume percent (100%), we choose the smaller value: \(V_{\mathrm{Gr}} = 0.0528\) To express this as a percentage, we multiply by 100: \(V_{\mathrm{Gr}} = 5.28 \%\) Therefore, the volume percent of graphite in the cast iron is approximately \(5.28\%\).