Answer
$S_{\text {gen total }}=0.414\text{Btu/R}$
Work Step by Step
$
T=T_{\sin @ 40 \text { psia }}=267.2^{\circ} \mathrm{F}=727.2 \mathrm{R}
$
Taking the contents of the cylinder as the system and noting that the temperature of water remains constant, the entropy change of the system during this isothermal, internally reversible process becomes
$
\Delta S_{\text {system }}=\frac{Q_{\text {sysin }}}{T_{\text {sys }}}=\frac{600 \mathrm{Btu}}{727.2 \mathrm{R}}=0.8251 \mathrm{Btu} / \mathrm{R}
$
Similarly, the entropy change of the heat source is determined from
$
\Delta S_{\text {source }}=-\frac{Q_{\text {souroeput }}}{T_{\text {source }}}=-\frac{600 \mathrm{Btu}}{1000+460 \mathrm{R}}=-0.4110 \mathrm{~Btu} / \mathrm{R}
$
Now consider a combined system that includes the cylinder and the source. Noting that no heat or mass crosses the boundaries of this combined system, the entropy balance for it can be expressed as
$
\begin{aligned}
& \underbrace{S_{\text {in }}-S_{\text {out }}}_{\begin{array}{c}
\text { Net entropy tinsfer } \\
\text { byheat and mass }
\end{array}}+\underbrace{S_{\text {gen }}}_{\begin{array}{c}
\text { Entropy } \\
\text { geneation }
\end{array}}=\underbrace{\Delta S_{\text {system }}}_{\begin{array}{c}
\text { Change } \\
\text { in entropy }
\end{array}} \\
& 0+S_{\text {gentotal }}=\Delta S_{\text {water }}+\Delta S_{\text {source }}
\end{aligned}
$
Therefore, the total entropy generated during this process is
$
S_{\text {gen total }}=\Delta S_{\text {witer }}+\Delta S_{\text {source }}=0.8251-0.4110=\mathbf{0 . 4 1 4 B t u} / \mathbf{R}
$
