Answer
$T_2=16.4°C=289.4\ K$
$\Delta S_{total}=1.52\ kJ/K$
Work Step by Step
From the energy balance:
$[mc(T_2-T_1)]_{Cu}+[mc(T_2-T_1)]_{H_2O}=0$
$m_{H_2O}=\rho V,\ \rho=997\ kg/m^3,\ V=0.090\ m^3$
$m_{H_2O}=89.73\ kg$
Given $c_{H_2O}=4.18\ kJ/kg.K,\ c_{Cu}=0.386\ kJ/kg.K$(Table A-3),
$m_{Cu}=50\ kg, T_{1,Cu}=140°C,\ T_{1,H_2O=10°C}$
solving for the final temperature
$T_2=16.4°C=289.4\ K$
Entropy generated
$\Delta S=mc\ln\left(\frac{T_2}{T_1}\right)$
Cu: $\Delta S=-6.864\ kJ/K$
$H_2O$: $\Delta S=8.388\ kJ/K$
$\Delta S_{total}=\Delta S_{Cu}+\Delta S_{H_2O}$
$\Delta S_{total}=1.52\ kJ/K$
