Answer
$T=109°C$
$\Delta S=0.251\ kJ/K$
Work Step by Step
For the iron-aluminum system:
$\Delta U_{Fe}+\Delta U_{Al}=0$
$[mc(T-T_1)]_{Fe}+[mc(T-T_1)]_{Al}=0$
Given $m_{Fe}=40\ kg,\ m_{Al}=30\ kg,\ c_{Al}=0.949\ kJ/kg.°C,\ c_{Fe}= 0.45\ kJ/kg.°C$,
$T_{1_{Fe}}=60°C=333\ K,\ T_{1_{Al}}=140°C=413\ K$, we can solve for the final temperature:
$T=109°C=382\ K$
For the entropy change:
$\Delta S_{Fe}=\left[mc\ln\left(\frac{T}{T_1}\right)\right]_{Fe}=2.472\ kJ/K$
$\Delta S_{Al}=\left[mc\ln\left(\frac{T}{T_1}\right)\right]_{Al}=-2.221\ kJ/K$
$\Delta S=\Delta S_{Fe}+\Delta S_{Al}=0.251\ kJ/K$
