A *change in entropy* for a reversible process that takes place at **constant temperature** can be calculated using this equation

#DeltaS = Q/T#, where

#Q# - heat absorbed/released during the process;

#T# - the temperature at which this takes place.

In your case, temperature is **not constant** for neither the piece of aluminium, nor the water. However, if the change in temperature is small enough, you can use the **average temperature** to express the change in entropy for a process.

#DeltaS = Q/bar(T)#, where

#bar(T)# - the average temperature.

So, the idea behind this problem is that the heat **lost** by the metal will be **absorbed** by the water.

#-q_"metal" = q_"water"#

The equation that establishes a relationship between heat gained/lost and change in temperature looks like this

#q = m * c * DeltaT#, where

#m# - the mass of the substance;

#c# - its specific heat;

#DeltaT# - the change in temperature, defined as #T_"final" - T_"initial"#;

So, plug in your values into the above equation and solve for #T_"final"#.

#-m_"metal" * c_"metal" * DeltaT_"metal" = m_"water" * c_"water" * DeltaT_"water"#

#3.8cancel("kg") * 900"J"/(cancel("kg") cancel(""^@cancel("C"))) * (30 - T_"final")cancel(""^@"C") = 1cancel("kg") * 4184"J"/(cancel("kg") cancel(""^@"C")) * (T_"final" - 20)cancel(""^@"C")#

#3420cancel("J") * (30-T_"final") = 4184cancel("J") * (T_"final" - 20)#

#7604 * T_"final" = 186280 => T_"final" = 186280/7604 = 24.5^@"C"#

The heat *lost by the metal* will be

#q_"metal" = 3.8 * 900 * (24.5-30) = -"18,810 J"#

The heat *gained by the water* will be

#q_"water" = -q_"metal" = +"18,810 J"#

Since the change in temperature is not that significant, you can determine the average temperature for the metal and for water by

#{ (T_"initial metal" = (273.15 + 30)"K" = "303.15 K"), (T_"final" = (273.15 + 24.5)"K" = "297.65 K") :}#

#bar(T)_"metal" = ((303.15 + 297.65)"K")/2 = "300.4 K"#

and

#{ (T_"initial water" = (273.15 + 20)"K" = "293.15 K"), (T_"final" = "297.65 K") :}#

#bar(T)_"water" = ((293.15 + 297.65)"K")/2 = "295.4 K"#

So, the net change in entropy for the metal will be

#DeltaS_"metal" = Q_"metal"/bar(T)_"metal" = (-"18,810 J")/("300.4 K") = -"62.6 J/K"#

The net change in entropy for the water will be

#DeltaS_"water" = Q_"water"/bar(T)_"water" = (+"18,810 J")/("295.4 K") = +"63.7 J/K"#

The **net change in entropy** for the system will thus be

#DeltaS_"sys" = DeltaS_"metal" + DeltaS_"water"#

#DeltaS_"sys" = -62.6 + 63.7 = color(green)(+"1.1 J/K")#

**SIDE NOTE** *I'll leave the answer rounded to two sig figs*.