When the iron and the water reach thermal equilibrium, they have same temperature, [tex]T=58.5^{\circ}C[/tex].
We can consider this as an isolated system, so the heat released by the water is equal to the heat absorbed by the iron.
The hear released by the water is:
[tex]Q_w =m_w C_{sw} \Delta T_w[/tex]
where [tex]m_w[/tex] is the water mass, [tex]C_{sw}=4.186 J/(g^{\circ}C)[/tex] is the specific heat of the water, and [tex]\Delta T_w = 63.6^{\circ}-58.5^{\circ}=5.1^{\circ}C[/tex] is the variation of temperature of the water.
Similarly, the heat absorbed by the iron is:
[tex]Q_i = m_i C_{si} \Delta T_i[/tex]
where [tex]m_i = 33.0 g[/tex] is the iron mass, [tex]C_{si}=0.444 J/(g^{\circ}C)[/tex] is the iron specific heat, and [tex]\Delta T_i = 58.5^{\circ}-22.7^{\circ}=35.8^{\circ}C[/tex] is the variation of temperature of the iron.
Writing [tex]Q_w=Q_i[/tex] and replacing the numbers, we can solve to find mw, the mass of the water:
[tex]m_w= \frac{m_i C_{si} \Delta T_i}{C_{sw} \Delta T_w} =24.6 g[/tex]
