The piece of iron is brought from a temperature of 100C to a temperature of 22C. In this process, the heat released by this piece of iron is
[tex]Q=m C_s \Delta T[/tex]
where [tex]m=200 g[/tex] is the mass of the piece of iron, [tex]C_s = 0.444 J/(g C)[/tex] is the iron's specific heat, and the temperature variation is [tex]\Delta T = 22^{\circ}-100^{\circ}=-78^{\circ}C[/tex]. Using these values, the amount of heat released by the iron is
[tex]Q=(200 g)(0.441 J/(gC))(-78C)=-6926J=-6.9 kJ[/tex]
With negative sign because it is amount of heat released. This heat is then trasferred to the water, so the amount of heat absorbed by the water is Q=6.9 kJ.
