A \(2.0-\mathrm{kg}\) specimen of an \(85 \mathrm{wt} \% \mathrm{Pb}-15 \mathrm{wt} \%\) Sn alloy is heated to \(200^{\circ} \mathrm{C}\left(390^{\circ} \mathrm{F}\right)\); at this temperature it is entirely an \(\alpha\) -phase solid solution (Figure 9.8 ). The alloy is to be melted to the extent that \(50 \%\) of the specimen is liquid, the remainder being the \(\alpha\) phase. This may be accomplished by heating the alloy or changing its composition while holding the temperature constant. (a) To what temperature must the specimen be heated? (b) How much tin must be added to the \(2.0-\mathrm{kg}\) specimen at \(200^{\circ} \mathrm{C}\) to achieve this state?

Before we can apply the lever rule, we need to look at the phase diagram (Figure 9.8) for the given Pb-Sn alloy. The phase diagram provides information on the points where phase changes occur and the \(\alpha\) phase composition lines.

Since we want to have \(50 \%\) of the specimen in the liquid phase, we can refer to the same point on the phase diagram. The point where the liquid composition would be \(50 \%\) would be halfway between the initial composition at \(85 \%\) Pb and the intersection with the liquidus line (which is at \(61.9 \% wt. Pb\)). Therefore, we can estimate the required temperature using the intersection of the half point on the phase diagram which corresponds to a solid phase melting point at approximately \(183^{\circ}C\). So, we need to heat the specimen up to around \(183^{\circ} \mathrm{C}\).

To determine the amount of tin added to the \(2.0-\mathrm{kg}\) specimen at \(200^{\circ} \mathrm{C}\), we first need to find the total mass of Tin in the \(2.0-\mathrm{kg}\) specimen. We can calculate this as follows: Tin_mass_initial = \(2.0 \times 0.15 = 0.3 \mathrm{kg}\) We now want to find the point on the phase diagram at \(200^{\circ} \mathrm{C}\) where \(50 \%\) of the specimen is in the liquid phase and still in the \(\alpha\) phase. Denoting the mass fraction of lead as \(x_\mathrm{Pb}\), the lever rule states: \(\frac{x_\mathrm{Pb}^\mathrm{L} - x_\mathrm{Pb}^\mathrm{0}}{x_\mathrm{Pb}^\mathrm{0} - x_\mathrm{Pb}^\mathrm{\alpha}} = \frac{mass_{liquid}}{mass_{solid}}\) We are given that we want 50% of the sample to be liquid, so: \(\frac{mass_{liquid}}{mass_{solid}} = \frac{1}{1}\) The composition of the alloy at this point on the phase diagram can be found from the intersection point of the \(200^{\circ}C\) isotherm with the liquidus and solidus lines, which are approximately \(x_\mathrm{Pb}^\mathrm{L} = 67.5 \%\) and \(x_\mathrm{Pb}^\mathrm{\alpha} = 73.3 \%\). The weight fraction of the initial alloy is \(x_\mathrm{Pb}^0 = 85 \%\) Now, we can solve for mass of the liquid phase: \(\frac{0.675 - 0.85}{0.85 - 0.733} = \frac{mass_{liquid}}{mass_{solid}}\) mass_{liquid} = mass_{solid} Now, we have: mass_{liquid} = mass_{solid} = \(\frac{mass_\text{total}}{2}\) \(mass_\text{total} = mass_{liquid}+mass_{solid} = 2 \times mass_{liquid} = 2 \times 0.3 = 0.6 \mathrm{kg}\) Finally, in order to find out, how much tin we need to add, we can calculate the resulting mass fraction of tin in the alloy: \(x_\mathrm{Sn} = 1 - x_\mathrm{Pb}^L = 1 - 0.675 = 0.325\) Hence, to calculate the mass of tin to be added: New_mass_of_Tin = \(0.6 \times 0.325 = 0.195 \mathrm{kg}\) Tin_to_be_added = New_mass_of_Tin - Tin_mass_initial = \(0.195 - 0.3 = -0.105\) The negative value indicates we actually need to remove tin instead of adding it. Therefore, we need to remove approximately \(0.105 \mathrm{kg}\) of tin from the current specimen to achieve 50% liquid phase at \(200^{\circ} \mathrm{C}\).