Answer:
0.7447%
Explanation:
From the diagram below:
We consider the [tex]Fe-Fe_3C[/tex] phase transformation
From the lever rule, mass fraction of eutectoid ferrite in a hypo eutectoid iron-carbon alloy
[tex]0.871 =(\frac{C_{Fe_{3C}}-C_0}{C_{Fe_{3C}}-C_{Fe_3}} )-(\frac{0.76-C_0}{0.76-0.022} )[/tex]
[tex]0.871 =(\frac{6.7-C_0}{6.7-0.022})-(\frac{0.76-C_0}{0.76-0.022} )[/tex]
[tex]0.871 =(\frac{6.7-C_0}{6.678})-(\frac{0.76-C_0}{0.738} )[/tex]
[tex](0.871*6.678*0.738)=(6.7-C_o)0.738-(0.76-C_o)6.678[/tex]
[tex]4.29261 = (0.738*6.7-6.678*0.76)+C_o(6.678-0.738)[/tex]
[tex]4.29261 =(-0.13068)+C_o(5.94)[/tex]
[tex]4.29261+0.13068 =C_o(5.94)[/tex]
[tex]4.4239=C_o(5.94)[/tex]
[tex]C_o=\frac{4.4239}{5.94}[/tex]
[tex]C_o=0.7447%[/tex] % of C
Hence, the weight% of Carbon in the alloy is [tex]C_o=0.7447%[/tex] % of C.
