For a copper-silver alloy of composition 25 \(\mathrm{wt} \% \mathrm{Ag}-75 \mathrm{wt} \% \mathrm{Cu}\) and at \(775^{\circ} \mathrm{C}\left(1425^{\circ} \mathrm{F}\right)\), do the following: (a) Determine the mass fractions of \(\alpha\) and \(\beta\) phases. (b) Determine the mass fractions of primary \(\alpha\) and eutectic microconstituents. (c) Determine the mass fraction of eutectic \(\alpha\).

Find the Cu-Ag phase diagram and locate the given composition and temperature on the diagram. The diagram will help us to determine the phases present at the given conditions and their corresponding compositions.

From the Cu-Ag phase diagram at 775°C, we can see that the given composition (25 wt% Ag-75 wt% Cu) lies in the two-phase region, where both alpha (\(\alpha\)) and beta (\(\beta\)) phases coexist.

From the phase diagram, identify the compositions of the \(\alpha\) and \(\beta\) phases at 775°C: \(C_{\alpha} = C_{1} \approx 8\, \% \, Ag\) \(C_{\beta} = C_{2} \approx 72\, \% \, Ag\)

The lever rule is given by: \(W_{\alpha} = \frac{C_{2} - C_{0}}{C_{2} - C_{1}} \) \(W_{\beta} = \frac{C_{0} - C_{1}}{C_{2} - C_{1}}\) Plugging in the given composition (\(C_{0} = 25\, \% \, Ag\)) and the values from step 3: \(W_{\alpha} = \frac{72 - 25}{72 - 8} \approx 0.78\) \(W_{\beta} = \frac{25 - 8}{72 - 8} \approx 0.22\) Thus, the mass fractions of the \(\alpha\) and \(\beta\) phases are approximately \(78\%\) and \(22\%\), respectively.

According to the phase diagram, primary \(\alpha\) phase is the only phase to crystallize first. The remaining alloy then consists of eutectic microconstituents (eutectic mixture of \(\alpha\) and \(\beta\) phases). The mass fractions are determined as: \(W_{\text{primary }\alpha} = W_{\alpha} - W_{\text{eutectic }\alpha}\) \(W_{\text{eutectic mixture}} = W_{\text{eutectic }\alpha} + W_{\text{eutectic }\beta}\) Since the eutectic composition at 775°C consists of approximately 45% of \(\alpha\) and 55% of \(\beta\), the mass fractions of eutectic \(\alpha\) and eutectic \(\beta\) can be expressed as: \(W_{\text{eutectic }\alpha} = 0.45W_{\text{eutectic mixture}} \) \(W_{\text{eutectic }\beta} = 0.55W_{\text{eutectic mixture}} \) As we already found that the mass fractions of \(\alpha\) phase \(W_{\alpha}\) = 0.78, therefore: \(W_{\text{primary }\alpha} = W_{\alpha} - 0.45W_{\text{eutectic mixture}} = 0.78 - 0.45W_{\text{eutectic mixture}}\) Given \(W_{\beta} = 0.22\), we have: \(0.55W_{\text{eutectic mixture}} = 0.22\) Now we can solve for \(W_{\text{eutectic mixture}}\) and find the mass fractions of primary \(\alpha\) and the eutectic mixture: \(W_{\text{eutectic mixture}} = \frac{0.22}{0.55} \approx 0.40\) \(W_{\text{primary }\alpha} = 0.78 - 0.45 \times 0.40 \approx 0.60\) Thus, the mass fractions of primary \(\alpha\) and eutectic microconstituents are approximately \(60\%\) and \(40\%\), respectively.

Since we already know that the eutectic mixture is composed of 45% \(\alpha\) and 55% \(\beta\): \(W_{\text{eutectic }\alpha} = 0.45 \times W_{\text{eutectic mixture}} = 0.45 \times 0.40 \approx 0.18\) Therefore, the mass fraction of eutectic \(\alpha\) is approximately \(18\%\).