# 9.2 At 500°C (930°F), what is the maximum solubility (a) of Cu in Ag

```9.2
At 500C (930F), what is the maximum solubility (a) of Cu in Ag?
(b) Of Ag in Cu?
Solution
(a)
From Figure 9.7, the maximum solubility of Cu in Ag at 500C corresponds to the position
of the –( + ) phase boundary at this temperature, or to about 2 wt% Cu.
(b)
From this same figure, the maximum solubility of Ag in Cu corresponds to the position of
the –( + ) phase boundary at this temperature, or about 1.5 wt% Ag.
9.5
Consider a specimen of ice that is at 210C and 1 atm pressure.
Using Figure 9.2, the
pressure–temperature phase diagram for H2O, determine the pressure to which the specimen must be
raised or lowered to cause it (a) to melt, and (b) to sublime.
Solution
The figure below shows the pressure-temperature phase diagram for H2O, Figure 10.2;
a
vertical line has been constructed at -10C, and the location on this line at 1 atm pressure (point B) is
also noted.
(a)
Melting occurs, (by changing pressure) as, moving vertically (upward) at this temperature,
we cross the Ice-Liquid phase boundary.
This occurs at approximately 570 atm;
thus, the pressure
of the specimen must be raised from 1 to 570 atm.
(b)
In order to determine the pressure at which sublimation occurs at this temperature, we move
vertically downward from 1 atm until we cross the Ice-Vapor phase boundary.
occurs at approximately 0.0023 atm.
This intersection
9.7
Given here are the solidus and liquidus temperatures for the germanium-silicon system.
Construct the phase diagram for this system and label each region.
Composition
Solidus
Liquidus
(wt% Si)
Temperature
Temperature
(&deg;C)
(&deg;C)
0
938
938
10
1005
1147
20
1065
1226
30
1123
1278
40
1178
1315
50
1232
1346
60
1282
1367
70
1326
1385
80
1359
1397
90
1390
1408
100
1414
1414
Solution
The germanium-silicon phase diagram is constructed below.
9.15
A 1.5-kg specimen of a 90 wt% Pb–10 wt% Sn alloy is heated to 250C (480F);
temperature it is entirely an -phase solid solution (Figure 9.8).
at this
The alloy is to be melted to the
extent that 50% of the specimen is liquid, the remainder being the  phase.
This may be
accomplished either 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 1.5-kg specimen at 250C to achieve this state?
Solution
(a)
Probably the easiest way to solve this part of the problem is by trial and error--that is, on the
Pb-Sn phase diagram (Figure 9.8), moving vertically at the given composition, through the  + L
region until the tie-line lengths on both sides of the given composition are the same.
This occurs at
approximately 295C (560F).
(b)
We can also produce a 50% liquid solution at 250C, by adding Sn to the alloy.
At 250C
and within the  + L phase region
C = 14 wt% Sn-86 wt% Pb
CL = 34 wt% Sn-66 wt% Pb
Let C0 be the new alloy composition to give W = WL = 0.5. Then,
W = 0.5 =
CL  C0
34  C0
=
CL  C
34  14
And solving for C0 gives 24 wt% Sn. Now, let mSn be the mass of Sn added to the alloy to achieve

this new composition. The amount of Sn in the original alloy is
(0.10)(1.5 kg) = 0.15 kg
Then, using a modified form of Equation 4.3
0.15 kg  m 
Sn 100 = 24


 1.5 kg  mSn 
And, solving for mSn (the mass of tin to be added), yields mSn = 0.276 kg.

9.27
A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature from an elevated
temperature in such a way that the high-temperature microstructure is preserved.
This
microstructure is found to consist of the α phase and Mg 2Pb, having respective mass fractions of 0.65
and 0.35.
Determine the approximate temperature from which the alloy was quenched.
Solution
We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% Mg
alloy was quenched, given the mass fractions of  and Mg2Pb phases. We can write a lever-rule
expression for the mass fraction of the  phase as
W = 0.65 =
CMg 2Pb  C0
CMg 2Pb  C
The value of C0 is stated as 45 wt% Pb-55 wt% Mg, and CMg Pb is 81 wt% Pb-19 wt% Mg, which is

2
independent of temperature (Figure 9.20);
thus,
0.65 =
81  45
81  C
which yields

C = 25.6 wt% Pb
The temperature at which the –( + Mg2Pb) phase boundary (Figure 9.20) has a value of 25.6 wt%