Answer
$S_{\text {gen }}=4.08 kJ/K$
Work Step by Step
We take the entire contents of the tank, water + iron block, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as
$
\begin{aligned}
\underbrace{E_{\text {in }}-E_{\text {out }}}_{\begin{array}{c}
\text { Net energy trusfer } \\
\text { by heat, work, and mass }
\end{array}} & =\underbrace{\Delta E_{\text {system }}}_{\begin{array}{c}
\text { Changein intemal, kinetic, } \\
\text { potentialeftc. energies }
\end{array}} \\
0 & =\Delta U
\end{aligned}
$
or,
$
\begin{gathered}
\Delta U_{\text {iron }}+\Delta U_{\text {water }}=0 \\
{\left[m c\left(T_2-T_1\right)\right]_{\text {iron }}+\left[m c\left(T_2-T_1\right)\right]_{\text {water }}=0}
\end{gathered}
$
Substituting,
$
\begin{gathered}
(25 \mathrm{~kg})(0.45 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_2-350^{\circ} \mathrm{C}\right)+(100 \mathrm{~kg})(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_2-18^{\circ} \mathrm{C}\right)=0 \\
T_2=26.7^{\circ} \mathrm{C}
\end{gathered}
$
The entropy generated during this process is determined from
$
\begin{aligned}
& \Delta S_{\mathrm{inon}}=m c_{\mathrm{wg}} \ln \left(\frac{T_2}{T_1}\right)=(25 \mathrm{~kg})(0.45 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \left(\frac{299.7 \mathrm{~K}}{623 \mathrm{~K}}\right)=-8.232 \mathrm{~kJ} / \mathrm{K} \\
& \Delta S_{\text {water }}=m c_{\mathrm{wvg}} \ln \left(\frac{T_2}{T_1}\right)=(100 \mathrm{~kg})(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \left(\frac{299.7 \mathrm{~K}}{291 \mathrm{~K}}\right)=12.314 \mathrm{~kJ} / \mathrm{K}
\end{aligned}
$
Thus,
$
S_{\text {gen }}=\Delta S_{\text {total }}=\Delta S_{\text {ivon }}+\Delta S_{\text {water }}=-8.232+12.314=4.08 \mathrm{~kJ} / \mathbf{K}
$
