A) Set up an energy balance around the steam 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{gh}_{\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{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}
From the steam tables,
Steam at 1 bar and 100°C \rightarrow 2506.18 \frac{\mathrm{kJ}}{\mathrm{kg}}=\widehat{\mathrm{U}}_{\text {initial }}
Steam at 1 bar and 200°C \rightarrow 2658.21 \frac{\mathrm{kJ}}{\mathrm{kg}}=\widehat{\mathrm{U}}_{\text {final }}
\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=152.03 \frac{\mathrm{kJ}}{\mathrm{kg}}
Now that we have the change in internal energy, we may 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)\end{aligned}
Steam at 1 bar and 100^{\circ} \mathrm{C} \rightarrow 1.6959 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}=\widehat{\mathrm{V}}_{\text {initial }}
Steam at 1 bar and 200^{\circ} \mathrm{C} \rightarrow 2.1724 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}=\widehat{\mathrm{V}}_{\text {final }}
\begin{aligned}&\mathrm{V}_{\text {final }}=\widehat{\mathrm{V}}_{\text {final }} \times\rm Mass =\left(2.1724 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\right)(1 \mathrm{~kg})=2.1724 \mathrm{~m}^{3}\\&V_{\text {initial }}=\widehat{V}_{\text {initial }} \times\rm Mass =\left(1.6959 \frac{\mathrm{m}^{3}}{\mathrm{~kg}}\right)(1 \mathrm{~kg})=1.6959 \mathrm{~m}^{3}\end{aligned}
The gas is expanding, and P is the pressure opposing the motion, which is not given explicitly. But if the pressure within the container started at 1 bar and remained constant throughout the process, then it is reasonable to assume that the pressure opposing the expansion is also equal to 1 bar.
\begin{gathered}\mathrm{W}_{\mathrm{EC}}=-1 \operatorname{bar}\left(2.1724 \mathrm{~m}^{3}-1.6959 \mathrm{~m}^{3}\right)\left\{\left(\frac{10^{5} \mathrm{~Pa}}{1 \mathrm{~bar}}\right)\left(\frac{1 \mathrm{~J}}{1 \mathrm{~Pa\,m}^{3}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)\right\} \\=\bf -47.65 ~kJ\end{gathered}
Returning to the energy balance
\begin{gathered}M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q\\1 \mathrm{~kg}\left(152.03 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=-47.65 \mathrm{~kJ}+\mathrm{Q}\\Q=\bf 199.7 \,kJ\end{gathered}
B) Set up an energy balance around the steam in the cylinder
\begin{aligned}\Delta\left\{\mathrm { M } \left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2} +\mathrm{gh}\right)\right\} \\& =\sum^{\mathrm{j}=\mathrm{J}}_{j=1}\left\{\mathrm{m}_{\mathrm{j}, \mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{j}}+\frac{\mathrm{v}_{\mathrm{j}}^{2}}{2}+\mathrm{gh}_{\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} \\& \mathrm{M}\left(\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}\right)=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q}\end{aligned}
Find \mathrm{W}_{\mathrm{EC}}
\mathrm{W}_{\mathrm{EC}}=-\mathrm{PdV}
Solve for initial and final volume using the ideal gas law
\begin{aligned}& V=\frac{n R T}{P}=\frac{\mathrm{mRT}}{(\mathrm{MW}) \mathrm{P}}\\& \mathrm{V}_{\text {initial }}=\frac{(1 \mathrm{~kg})\left(8.314 \times 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)\left(\frac{1000 \mathrm{~mol}}{\mathrm{kmol}}\right)(373 \mathrm{~K})}{\left(18 \frac{\mathrm{kg}}{\mathrm{kmol}}\right) 1 \mathrm{bar}}=1.72 \mathrm{~m}^{3}\\& \mathrm{V}_{\text {final }}=\frac{(1 \mathrm{~kg})\left(8.314 \times 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)\left(\frac{1000 \mathrm{~mol}}{\mathrm{kmol}}\right)(473 \mathrm{~K})}{\left(18 \frac{\mathrm{kg}}{\mathrm{kmol}}\right) 1 \mathrm{bar}}=2.18 \mathrm{~m}^{3}\end{aligned}
Return to our \mathrm{W}_{\mathrm{EC}} equation
\mathrm{W}_{\mathrm{EC}}=-\int_{\mathrm{V}_{\text {initial }}}^{\mathrm{V}_{\text {final }}} \mathrm{PdV}
Since this is constant pressure, we may remove P from the integral
\begin{aligned}& W_{\mathrm{EC}}=-P \int_{\mathrm{V}_{\text {initial }}}^{\mathrm{V}_{\text {final }}} \mathrm{dV}=-P\left(\mathrm{~V}_{\text {final }}-\mathrm{V}_{\text {initial }}\right) \\& \mathrm{W}_{\mathrm{EC}}=-\mathbf{4 6} \mathbf{k J}\end{aligned}
Now find \widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}
\begin{aligned}&\mathrm{d} \widehat{\mathrm{U}}=\mathrm{C}_{\mathrm{v}} \mathrm{dt}\\&\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}=\int_{\text {initial }}^{\text {final }} \mathrm{C}_{\mathrm{v}} d t\\&\mathrm{C}_{\mathrm{v}}=\mathrm{C}_{\mathrm{p}}-\mathrm{R}\\& \widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}=\int_{\text {initial }}^{\text {final }}\left(\mathrm{C}_{\mathrm{p}}-\mathrm{R}\right) \mathrm{dt}\end{aligned}
From Appendix D
\frac{C_{\mathrm{P}}^{*}}{\mathrm{R}}=\mathrm{A}+\mathrm{BT}+\mathrm{CT}^{2}+\mathrm{DT}^{3}+\mathrm{ET}^{4}
\begin{array}{|c|c|c|c|c|c|c|c|} \hline \bf{Name} & \bf{Formula} & \bf A & \bf B\times10^3 & \bf C\times10^5 & \bf D\times10^8 & \bf E\times10^{11} & \bf {T \, range (K)}\\ \hline \text {Water} & \rm H_2O & 4.395 & -4.186 & 1.405 & -1.564 & 0.632 & 50-100\\\hline\end{array}
\begin{aligned}& \widehat{U}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}=\int_{\text {initial }}^{\text {final }}\left(\mathrm{R}\left(\mathrm{A}+\mathrm{BT}+\mathrm{CT}^{2}+\mathrm{DT}^{3}+\mathrm{ET}^{4}\right)-\mathrm{R}\right) \mathrm{dT} \\& \widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}=\int_{\text {initial }}^{\text {final }}\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+\mathrm{RET}^{4}-\mathrm{R}\right) \mathrm{dT} \\& \widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }}=\mathrm{RAT}+\frac{1}{2} \mathrm{RBT}^{2}+\frac{1}{3} \mathrm{RCT}^{3}+\frac{1}{4} \mathrm{RDT}^{4}+\frac{1}{5} \mathrm{RET}^{5}-\mathrm{RT} \\& \widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }} \\&\qquad\qquad =8.314 \frac{\mathrm{J}}{\mathrm{mol} \,\mathrm{K}}\left\{4.395(473-373)-\frac{1}{2}\left(4.186 \times 10^{-3}\right)\left(473^{2}-373^{2}\right)\right. \\&\qquad\qquad+\left(\frac{1}{3}\right)\left(1.405 \times 10^{-5}\right)\left(473^{3}-373^{3}\right) \\&\qquad\qquad-\left(\frac{1}{4}\right)\left(1.564 \times 10^{-8}\right)\left(473^{4}-373^{4}\right) \\&\qquad\qquad\left.+\left(\frac{1}{5}\right)\left(6.32 \times 10^{-12}\right)\left(473^{5}-373^{5}\right)-(473-373)\right\} \\& \widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }} \\&\qquad\qquad =\int_{\text {initial }}^{\text {final }}\left\{\mathrm { R } \left(4.395-\left(4.186 \times 10^{-3}\right) \mathrm{T}+\left(1.405 \times 10^{-5}\right) \mathrm{T}^{2}\right.\right. \\&\qquad\qquad \left.\left.-\left(1.564 \times 10^{-8}\right) \mathrm{T}^{3}+\left(6.32 \times 10^{-12}\right) \mathrm{T}^{4}\right)-\mathrm{R}\right\} \mathrm{dt}\end{aligned}
\begin{aligned}\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }} & \\& =\mathrm{R} \int_{\text {initial }}^{\text {final }}\left\{4.395-\left(4.186 \times 10^{-3}\right) \mathrm{T}+\left(1.405 \times 10^{-5}\right) \mathrm{T}^{2}\right. \\& \left.-\left(1.564 \times 10^{-8}\right) \mathrm{T}^{3}+\left(6.32 \times 10^{-12}\right) \mathrm{T}^{4}-\mathrm{R}\right\} \mathrm{dt} \\\widehat{\mathrm{U}}_{\text {final }}-\widehat{\mathrm{U}}_{\text {initial }} & \\& =\mathrm{R}\left\{(4.395)(473-373)-\left(\frac{1}{2}\right)\left(4.186 \times 10^{-3}\right)\left(473^{2}-373^{2}\right)\right. \\& +\left(\frac{1}{3}\right)\left(1.405 \times 10^{-5}\right)\left(473^{3}-373^{3}\right) \\& -\left(\frac{1}{4}\right)\left(1.564 \times 10^{-8}\right)\left(473^{4}-373^{4}\right) \\& \left.+\left(\frac{1}{5}\right)\left(6.32 \times 10^{-12}\right)\left(473^{5}-373^{5}\right)-\mathrm{R}\right\}=146.04 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}
Returning to the energy balance
\begin{aligned}& M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q} \\& 1 \mathrm{~kg}\left(146.04 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=-46 \mathrm{~kJ}+\mathrm{Q} \\& \bf Q=192.04 ~kJ\end{aligned}
Our answers for part B) compared to part A) were correct within <5 %.
C) If we had not accounted for expansion work, we would have said Q = ∆U and our answer for Q would have been about 25% off. That would not have been a good approximation.