Question

The mass fraction of eutectoid ferrite in a hypoeutectoid iron-carbon alloy (just below its eutectoid temperature) is 0.871. On the basis of this information, determine the composition of the alloy. If it is not possible to determine the composition from the information provided, enter 0.

Answer 1

0.7447%

Step-by-step explanation:

From the diagram below:

We consider the
`Fe-Fe_3C` phase transformation

From the lever rule, mass fraction of eutectoid ferrite in a hypo eutectoid iron-carbon alloy

`0.871 =(\frac{C_{Fe_(3C)}-C_0}{C_{Fe_(3C)}-C_(Fe_3)} )-((0.76-C_0)/(0.76-0.022) )`

`0.871 =((6.7-C_0)/(6.7-0.022))-((0.76-C_0)/(0.76-0.022) )`

`0.871 =((6.7-C_0)/(6.678))-((0.76-C_0)/(0.738) )`

`(0.871*6.678*0.738)=(6.7-C_o)0.738-(0.76-C_o)6.678`

`4.29261 = (0.738*6.7-6.678*0.76)+C_o(6.678-0.738)`

`4.29261 =(-0.13068)+C_o(5.94)`

`4.29261+0.13068 =C_o(5.94)`

`4.4239=C_o(5.94)`

`C_o=(4.4239)/(5.94)`

`C_o=0.7447%` % of C

Hence, the weight% of Carbon in the alloy is
`C_o=0.7447%` % of C.