Answer
$\dot{S}_{\text {gen,wall }}=0.187 \mathrm{~W} / \mathrm{K}$
Work Step by Step
$$
\begin{aligned}
& \underbrace{\dot{S}_{\text {in }}-\dot{S}_{\text {out }}}_{\begin{array}{c}
\text { of net entropy tensfer } \\
\text { by heat and mass }
\end{array}}+\underbrace{\dot{S}_{\text {gen }}}_{\begin{array}{c}
\text { Rate of entropy } \\
\text { generation }
\end{array}}=\underbrace{\Delta \dot{S}_{\text {system }}}_{\begin{array}{c}
\text { Rateof change } \\
\text { of entropy }
\end{array}}=0 \\
& \frac{\dot{Q}_{\text {in }}}{T_{\text {b in }}}-\frac{\dot{Q}_{\text {out }}}{T_{\text {b }, \text { out }}}+\dot{S}_{\text {gen }, \text { wall }}=0 \\
& \frac{1250 \mathrm{~W}}{289 \mathrm{~K}}-\frac{1250 \mathrm{~W}}{277 \mathrm{~K}}+\dot{S}_{\text {gen }, \text { wall }}=0 \\
& \dot{S}_{\text {gen,wall }}=0.187 \mathrm{~W} / \mathrm{K} \\
&
\end{aligned}
$$Therefore, the rate of entropy generation in the wall is $0.187 \mathrm{~Wù} / \mathrm{K}$.
