Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°C.
m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
ΔT₂ = 293 - 1181 = -888°C.
C - specific heat capacity, C(Fe) = 0,44 J/g°C.
Q₂ = -224664 J = -224,66 kJ.
Q₃ =- heat of fusion, ΔH = 209 J/g.
Q₃ = 120175 J = 120,17 kJ.
Q = Q₁ + Q₂ + Q₃ = -963,8 kJ.
