A) Set up an energy balance around the water in the cylinder
\begin{aligned}\Delta\left\{M \left(\widehat{U}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\} \\& =\sum_{\mathrm{j}=1}^{\mathrm{j}=\mathrm{J}}\left\{\mathrm{m}_{\mathrm{j}, \mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{j}}+\frac{\mathrm{v}_{\mathrm{j}}^{2}}{2}+\mathrm{g} \mathrm{h}_{\mathrm{j}}\right)\right\}-\sum_{\mathrm{k}=1}^{\mathrm{k}=\mathrm{K}}\left\{\mathrm{m}_{\mathrm{k}, \text { out }}\left(\widehat{\mathrm{H}}_{\mathrm{k}}+\frac{\mathrm{v}_{\mathrm{k}}^{2}}{2}+\mathrm{gh}_{\mathrm{k}}\right)\right\}+\mathrm{W}_{\mathrm{EC}} \\& +\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling terms for this stationary, closed system
\begin{aligned}& \Delta\{\mathrm{M} \widehat{\mathrm{U}}\}=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q} \\& \mathrm{M}\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q}\end{aligned}
Solve for the change in enthalpy of the water to relate its enthalpy change to its Internal energy change
\begin{aligned}& \mathrm{T}_{\text {final }}=95^{\circ} \mathrm{C}=95+273=368 \mathrm{~K}\\& \mathrm{T}_{\text {initial }}=5^{\circ} \mathrm{C}=5+273=278 \mathrm{~K}\end{aligned}
\frac{C_{P}}{R}=A+B T+C T^{2}
\begin{array}{|c|c|c|c|c|}\hline\text{Compound }& \text{Structure }& \rm A & \rm B \times10^3 & \rm C\times10^6\\ \hline \text {Water}& \rm H_2O & 8.712 & 1.25 & -0.18 \\ \hline\end{array}
\begin{aligned}&\mathrm{d} \widehat{\mathrm{H}}=\mathrm{C}_{\mathrm{p}} \mathrm{dT}\\&\mathrm{d} \widehat{\mathrm{H}}=\left\{\mathrm{R}\left(\mathrm{A}+\mathrm{BT}+\mathrm{CT}^2\right)\right\} \mathrm{dT}\\&\int_{\text {initial }}^{\text {final }} \mathrm{d} \widehat{\mathrm{H}}=\mathrm{R} \int_{\text {initial }}^{\text {final }}\left(\mathrm{A}+\mathrm{BT}+\mathrm{CT}^2\right) \mathrm{dT}\\& \left(\widehat{\mathrm{H}}_{\text {final }}-\widehat{\mathrm{H}}_{\text {initial }}\right) \\& \qquad\qquad=\mathrm{R}\left\{\mathrm{A}\left(\mathrm{T}_{\text {final }}-\mathrm{T}_{\text {initial }}\right)+\frac{1}{2} \mathrm{~B}\left(\mathrm{~T}_{\text {final }}-\mathrm{T}_{\text {initial }}\right)^2+\frac{1}{3} \mathrm{C}\left(\mathrm{T}_{\text {final }}-\mathrm{T}_{\text {initial }}\right)^3\right\}\\&\left(\widehat{\mathrm{H}}_{\text {final }}-\widehat{\mathrm{H}}_{\text {initial }}\right) \\& \qquad\qquad =\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol}~ \mathrm{K}}\right)\left(\frac{\mathrm{kmol}}{18.01 \mathrm{~kg}}\right)\{(8.712)(368-278) \\& \qquad\qquad\left.+\frac{1}{2}\left(1.25 \times 10^{-3}\right)\left(368^{2}-278^{2}\right)+\frac{1}{3}\left(-1.8 \times 10^{-7}\right)\left(368^{3}-278^{3}\right)\right\} \\&\qquad\qquad =378.34 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}
Convert Enthalpy to Internal Energy
\begin{aligned}&\mathrm{H}=\mathrm{U}+\mathrm{PV}\\&\mathrm{d} \widehat{\mathrm{H}}=\mathrm{d} \widehat{\mathrm{U}}+\mathrm{Pd} \widehat{\mathrm{V}}\\&\left(\widehat{\mathrm{H}}_{\text {final }}-\widehat{\mathrm{H}}_{\text {initial }}\right)=\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)+\mathrm{P}\left(\widehat{\mathrm{V}}_{\text {final }}-\widehat{\mathrm{V}}_{\text {initial }}\right)\end{aligned}
Find the specific volume initially and finally (specific volume is the inverse of density)
\begin{aligned}& \widehat{V}_{\text {final }}=\frac{1}{\rho_{\text {final }}}=\frac{1}{962.3 \frac{\mathrm{g}}{\mathrm{L}}}=0.00103917 \frac{\mathrm{L}}{\mathrm{g}}\left(\frac{1000 \mathrm{~g}}{\mathrm{~kg}}\right)=1.03917 \frac{\mathrm{L}}{\mathrm{kg}} \\\\& \widehat{V}_{\text {initial }}=\frac{1}{\rho_{\text {initial }}}=\frac{1}{1000.4 \frac{\mathrm{g}}{\mathrm{L}}}=0.00099960 \frac{\mathrm{L}}{\mathrm{g}}\left(\frac{1000 \mathrm{~g}}{\mathrm{~kg}}\right)=0.99960 \frac{\mathrm{L}}{\mathrm{kg}}\end{aligned}
Using the volumes found to solve for the change in internal energy
\begin{aligned}&378.34 \frac{\mathrm{kJ}}{\mathrm{kg}}=\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)+(1 \mathrm{bar})\left(1.03917 \frac{\mathrm{L}}{\mathrm{kg}}-0.99960 \frac{\mathrm{L}}{\mathrm{kg}}\right) \\&\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right) \\& \qquad\qquad =378.34 \frac{\mathrm{kJ}}{\mathrm{kg}} \\& \qquad\qquad -\left(0.03957 \frac{\mathrm{L~bar}}{\mathrm{kg}}\right)\left\{\left(\frac{10^{5} \mathrm{~Pa}}{1~ \mathrm{bar}}\right)\left(\frac{1 \mathrm{~J}}{1 \mathrm{~Pa} \mathrm{~m}^{3}}\right)\left(\frac{1 \mathrm{~m}^{3}}{1000 \mathrm{~L}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)\right\} \\&\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)=378.34 \frac{\mathrm{kJ}}{\mathrm{kg}}-0.003957 \frac{\mathrm{kJ}}{\mathrm{kg}}=378.336 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}
Now that we have the change in internal energy, we may also find the \mathrm{W}_{\mathrm{EC}} term of the energy balance
\begin{aligned}& \mathrm{W}_{\mathrm{EC}}=-\mathrm{PdV}\\& \mathrm{W}_{\mathrm{EC}}=-\mathrm{P}\left(\mathrm{V}_{\text {final }}-\mathrm{V}_{\text {initial }}\right)\\& \mathrm{V}_{\text {final }}=\widehat{\mathrm{V}}_{\text {final }} \times\rm Mass =\left(1.03917 \frac{\mathrm{L}}{\mathrm{kg}}\right)(1 \mathrm{~kg})=1.03917 \mathrm{~L}\\& \mathrm{V}_{\text {initial }}=\widehat{\mathrm{V}}_{\text {initial }} \times\rm Mass =\left(0.99960 \frac{\mathrm{L}}{\mathrm{kg}}\right)(1 \mathrm{~kg})=0.99960 \mathrm{~L}\\& \mathrm{W}_{\mathrm{EC}}=-1 \operatorname{bar}(1.03917 \mathrm{~L}-0.99960 \mathrm{~L})\left\{\left(\frac{10^5 \mathrm{~Pa}}{1 \,\mathrm{bar}}\right)\left(\frac{1 \mathrm{~J}}{1 \mathrm{~Pa} \mathrm{~m}^3}\right)\left(\frac{1 \mathrm{~m}^3}{1000 \mathrm{~L}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)\right\} \\&\qquad\qquad=-\mathbf{0 . 0 0 3 9 5 7 \mathbf { k J }}\end{aligned}
Returning to the energy balance
\begin{aligned}& M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q\\& 1 \mathrm{~kg}\left(378.336 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=-0.003957 \mathrm{~kJ}+\mathrm{Q}\\& \mathrm{Q}=\bf 378.34 ~kJ\end{aligned}
B) Set up an energy balance around the water in the cylinder
\begin{aligned}\Delta\left\{M \left(\widehat{U}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\sum_{\mathrm{j}=\mathrm{J}}^{\mathrm{j}=\mathrm{J}}\left\{\mathrm{m}_{\mathrm{j}, \mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{j}}+\frac{\mathrm{v}_{\mathrm{j}}^{2}}{2}+g \mathrm{~h}_{\mathrm{j}}\right)\right\}-\sum_{\mathrm{k}=1}^{\mathrm{k}=\mathrm{K}}\left\{\mathrm{m}_{\mathrm{k}, \text { out }}\left(\widehat{\mathrm{H}}_{\mathrm{k}}+\frac{\mathrm{v}_{\mathrm{k}}^{2}}{2}+\mathrm{gh}_{\mathrm{k}}\right)\right\}+\mathrm{W}_{\mathrm{EC}} \\& +\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling terms
\begin{aligned}& \Delta\{\mathrm{M} \widehat{\mathrm{U}}\}=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q}\\& M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q\end{aligned}
Solve for the change in enthalpy of the water to relate its enthalpy change to its Internal energy change
\begin{aligned}& \mathrm{T}_{\text {final }}=95^{\circ} \mathrm{C}=95+273=368 \mathrm{~K}\\& \mathrm{T}_{\text {initial }}=5^{\circ} \mathrm{C}=5+273=278 \mathrm{~K}\\\\& d \widehat{H}=C_{p} d T\\& \left(\widehat{H}_{\text {final }}-\widehat{H}_{\text {initial }}\right)=\mathrm{C}_{\mathrm{p}}\left(\mathrm{T}_{\text {final }}-\mathrm{T}_{\text {initial }}\right)\\& \left(\widehat{\mathrm{H}}_{\text {final }}-\widehat{\mathrm{H}}_{\text {initial }}\right)=4.19 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}(368-278)\\& \left(\widehat{\mathrm{H}}_{\text {final }}-\widehat{\mathrm{H}}_{\text {initial }}\right)=377.1 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}
Since we are assuming \widehat{H} \approx \widehat{U},
\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }} \approx 377.1 \frac{\mathrm{kJ}}{\mathrm{kg}}
Now that we have the change in internal energy, we may apply find the \mathrm{W}_{\mathrm{EC}} term of the energy balance
\begin{aligned}& \mathrm{W}_{\mathrm{EC}}=-\mathrm{PdV}\\& \mathrm{W}_{\mathrm{EC}}=-\mathrm{P}\left(\mathrm{V}_{\text {final }}-\mathrm{V}_{\text {initial }}\right)\\& \mathrm{V}_{\text {final }}=\widehat{\mathrm{V}}_{\text {final }} \times \operatorname{Mass}=\left(1000 \frac{\mathrm{L}}{\mathrm{kg}}\right)(1 \mathrm{~kg})=0.001 \mathrm{~L}\\& \mathrm{V}_{\text {initial }}=\widehat{\mathrm{V}}_{\text {initial }} \times\rm Mass =\left(1000 \frac{\mathrm{L}}{\mathrm{kg}}\right)(1 \mathrm{~kg})=0.001 \mathrm{~L}\\& \mathrm{W}_{\mathrm{EC}}=-1 \operatorname{bar}(0.001 \mathrm{~L}-0.001 \mathrm{~L})=0\end{aligned}
Returning to the energy balance
\begin{aligned}& M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q \\& 1 \mathrm{~kg}\left(377.1 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=0+\mathrm{Q}\\& \mathrm{Q}=\bf377.1 \,\mathbf{k J}\end{aligned}
C) The answers between part A) and B) are very close. These approximations seem to be legitimate and practical.