Question

• A mixture of 92 kg of ZrO2 with 8 kg of CaO is melted into a liquid at high temperature.
• a) On slow cooling, at what temperature will solid first appear?
• b) What is the composition of this solid?
• c) At what temperature will the sample become completely solid?
• c) As the sample continues to cool, at what temperature will a second solid phase begin to precipitate?
• d) What is the composition of this solid phase?
• e) What is the expected composition and relative amounts of the phases expected to be present at room temperature?

Answer 1

ZrO2-CaO Phase diagram:

Molar mass of ZrO2=123 g/mol

=>92 Kg ZrO2=92000 g= 92000/123=748 moles

Molar Mass of CaO=56 g/mol

=> 8 Kg CaO=8000g=8000/56=143 moles

Mole % of CaO = Moles of CaO/(Mole of CaO+Mole of ZrO2)*100=143/(143+748)*100=16%

a)

=> By drawing one vertical line in the given plot at CaO Mole%=16, This can be seen that at higher temperature, Mass will be liquid. At around 2700 K, Liquid will turn into Liquid + CSS Phase.

b) To find the composition of this first particle of CSS, draw Horizontal line from 1st intersection point of above drawn vertical line from top. This horizontal line intersects dash line at left hand side where CaO mole%=~10

=> 1st particle of CSS formed will have composition of ~10% CaO

c) As we move downwrards on the same vertical line, the line intersects with dash curve, below which only CSS phase exist which is whole solid. Temperature corresponding to this intersection = ~2400 K

d) As we move down vertically on the same line, at T=~1200 K, TSS+CSS phase will start to develop.

e) at the intersection obtained in d), Drawing horizontal line, later line intersects at left hand side at ~5 mole% CaO. Therefore, TSS phase will have ~5 mole% CaO. CSS as seen in c) part will have 16 mole% CaO