Answer
$η_{II}=21.6\%$
Work Step by Step
First we determine the mole fraction of pure water in seawater using Eqs. 13-4 and 13-5. Noting that $\mathrm{mf}_{\mathrm{s}}=0.032$ and $\mathrm{mf}_{\mathrm{w}}=1-\mathrm{mf}_{\mathrm{s}}=0.968$,
$$
\begin{aligned}
& M_{\mathrm{m}}=\frac{1}{\sum \frac{\mathrm{mf}_i}{M_i}}=\frac{1}{\frac{\mathrm{mf}_s}{M_s}+\frac{\mathrm{mf}_w}{M_w}}=\frac{1}{\frac{0.032}{58.44}+\frac{0.968}{18.0}}=18.41 \mathrm{~kg} / \mathrm{kmol} \\
& y_i=\mathrm{mf}_i \frac{M_m}{M_i} \rightarrow y_w=\mathrm{mf}_w \frac{M_m}{M_w}=(0.968) \frac{18.41 \mathrm{~kg} / \mathrm{kmol}}{18.0 \mathrm{~kg} / \mathrm{kmol}}=0.9900
\end{aligned}
$$ The maximum work output associated with mixing $1 \mathrm{~kg}$ of seawater (or the minimum work input required to produce $1 \mathrm{~kg}$ of freshwater from seawater) is $$
w_{\text {max }, \text { out }}=R_w T_0 \ln \left(1 / y_w\right)=(0.4615 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(283.15 \mathrm{~K}) \ln (1 / 0.990)=1.313 \mathrm{~kJ} / \mathrm{kg} \text { fresh wate r }
$$ The power that can be generated as $1.4 \mathrm{~m}^3 / \mathrm{s}$ fresh water mixes reversibly with seawater is $$
\dot{W}_{\max \text { out }}=\rho \dot{W} \dot{W}_{\max \text { out }}=\left(1000 \mathrm{~kg} / \mathrm{m}^3\right)\left(1.4 \mathrm{~m}^3 / \mathrm{s}\right)(1.313 \mathrm{~kJ} / \mathrm{kg})\left(\frac{1 \mathrm{~kW}}{1 \mathrm{~kJ} / \mathrm{s}}\right)=1.84 \mathrm{~kW}
$$ Then the second law efficiency of the plant becomes $$
\eta_{\mathrm{II}}=\frac{\dot{W}_{\min , \text { in }}}{\dot{W}_{\text {in }}}=\frac{1.83 \mathrm{MW}}{8.5 \mathrm{MW}}=0.216=\mathbf{2 1 . 6} \%
$$
