Initially, one lb-mol of the compound is placed in a piston-cylinder device at T=25°F and P=1 bar. The compound is heated at constant pressure. Find the heat (Q) added if the final state is:
A) 50% solid and 50% liquid at T=50°F.
B) Liquid at 125°F.
C) 25% vapor and 75% liquid at T=150°F.
D) Vapor at 200°F.
A) Set up an energy balance on the compound
\begin{aligned}\left\{M\left(\widehat{U}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\}=\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{S}}+ \mathrm{W}_{\mathrm{EC}}+\mathrm{Q}\end{aligned}
Cancelling terms
\begin{gathered}M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=\mathrm{Q}+\mathrm{W}_{\mathrm{EC}}\\\mathrm{M}(\Delta \widehat{\mathrm{U}})=\mathrm{Q}+\mathrm{W}_{\mathrm{EC}}\end{gathered}
By definition, \Delta \widehat{\mathrm{U}}=\Delta \widehat{\mathrm{H}}-\mathrm{P}\left(\widehat{\mathrm{V}}_{2}-\widehat{\mathrm{V}}_{1}\right) and, when \mathrm{P} is constant, \mathrm{W}_{\mathrm{EC}}=-\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right).
\begin{gathered}\mathrm{M}\left(\Delta \widehat{\mathrm{H}}-\mathrm{P}\left(\widehat{\mathrm{V}}_{2}-\widehat{\mathrm{V}}_{1}\right)\right)=\mathrm{Q}-\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)\\ \frac{\dot{\mathbf{Q}}}{\mathbf{M}}=\Delta \widehat{\mathbf{H}}+\mathbf{P}\left(\widehat{\mathbf{V}}_{2}-\widehat{\mathbf{V}}_{1}\right)-\mathbf{P}\left(\widehat{\mathbf{V}}_{2}-\widehat{\mathbf{V}}_{1}\right)=\Delta \widehat{\mathbf{H}}\end{gathered}
This expression will be used in all four parts of the problem.
\frac{\dot{\mathrm{Q}}}{\mathrm{M}}=\Delta \widehat{\mathrm{H}}=\frac{\Delta \widehat{\mathrm{H}}^{\mathrm{fus}}}{2}+\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \mathrm{C}_{\mathrm{p}}^{*} \mathrm{dT};
Note that Heat of fusion is halved because only half of the solid (M/2) melted
\begin{gathered}\dot{\mathrm{Q}}=\mathrm{M}(\Delta \widehat{\mathrm{H}})=\mathrm{M}\left(\frac{\Delta \widehat{\mathrm{H}}^{\mathrm{fus}}}{2}+\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}\right) \\\dot{\mathrm{Q}}=(1~ \mathrm{lbmol})\left(\frac{50 ~\mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}}\right)\left(\frac{75 \frac{\mathrm{BTU}}{\mathrm{lb} _{\mathrm{m}}}}{2}+\int_{484.67^{\circ} \mathrm{R}}^{509.67^{\circ} \mathrm{R}} 0.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}^{\circ} \mathrm{R}} \mathrm{dT}\right) \\\dot{\mathrm{Q}}=(1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}^{\mathrm{R}}}\right)\left(\frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+17.5 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)=\bf 2750 \text { BTU }\end{gathered}
B) The energy balance is the same as in part A, but here, the change in specific enthalpy is equal to the sum of the amount of heat needed to heat the ice to its melting point, plus the amount of heat required to change phase, and finally the amount of heat required to raise the temperature of the new liquid to 125°F.
\begin{aligned}& \dot{\mathrm{Q}}=\mathrm{M}(\Delta \widehat{\mathrm{H}})=\mathrm{M}\left(\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \mathrm{C}_{\mathrm{p}, \mathrm{solid}} \mathrm{dT}+\Delta \widehat{\mathrm{H}}^{\text {fus }}+\int_{\mathrm{T}_{2}}^{\mathrm{T}_{3}} \mathrm{C}_{\mathrm{p}, \text { liquid }} \mathrm{dT}\right) \\& \dot{\mathrm{Q}}=(1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}^{2}}\right)\left(\int_{484.67^{\circ} \mathrm{R}}^{509.67^{\circ} \mathrm{R}} 0.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}^{\circ} \mathrm{R}} \mathrm{dT}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right. \\&\qquad\qquad\left.+\int_{509.67^{\circ} \mathrm{R}}^{584.67^{\circ} \mathrm{R}} 0.85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}^{\circ} \mathrm{R}} \mathrm{dT}\right) \\& \dot{\mathrm{Q}}=(1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb} \mathrm{m}}{1~ \mathrm{lbmol}^{\mathrm{C}}}\right)\left(17.5 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+63.8 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)=\bf 7812 \text { BTU }\end{aligned}
C) Because all the solid needs to melt and be raised to the appropriate temperature, and then only a quarter of the total liquid is converted into vapor, the \dot{Q} equation is broken into two terms, the liquid and the vapor portions.
\begin{aligned}& \dot{\mathrm{Q}}=0.75~ \mathrm{M}\left(\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \mathrm{C}_{\mathrm{p}, \text { solid }} \mathrm{dT}+\Delta \widehat{\mathrm{H}}^{\text {fus }}+\int_{\mathrm{T}_{2}}^{\mathrm{T}_{3}} \mathrm{C}_{\mathrm{p}, \text { liquid }} \mathrm{dT}\right) \\&\qquad\qquad\quad+0.25~ \mathrm{M}\left(\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \mathrm{C}_{\mathrm{p}, \text { solid }} \mathrm{dT}+\Delta \widehat{\mathrm{H}}^{\text {fus }}+\int_{\mathrm{T}_{2}}^{\mathrm{T}_{3}} \mathrm{C}_{\mathrm{p}, \text { liquid }} \mathrm{dT}+\Delta \widehat{\mathrm{H}}^{\text {vap }}\right)\end{aligned}
For \mathrm{T}_{1}=484.67^{\circ} \mathrm{R}, \mathrm{T}_{2}=509.67^{\circ} \mathrm{R}, and \mathrm{T}_{3}=609.67^{\circ} \mathrm{R} :
\begin{aligned}& \dot{\mathrm{Q}}=0.75\left((1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb} \mathrm{m}_{\mathrm{m}}}{1~ \mathrm{lbmol}}\right)\right)\left(17.5 \frac{\mathrm{BTU}}{\mathrm{lb}}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right) \\&\qquad\qquad\quad +0.25\left((1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}^{\mathrm{m}}}\right)\right)\left(17.5 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right. \\&\qquad\qquad\quad\left.+\frac{125 \mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)\\\\& \bf\dot{\mathrm{Q}}=10,438 \,BTU\end{aligned}
D) Because all of the solid becomes a vapor by the end, the process is modeled in five steps: heating solid, melting, heating liquid, boiling and heating vapor:
\begin{aligned}& \dot{\mathrm{Q}}=(1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}}\right)\left(\int_{484.67^{\circ} \mathrm{R}}^{509.67^{\circ} \mathrm{R}} 0.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}{ }^{\circ} \mathrm{R}} \mathrm{dT}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+\int_{509.67^{\circ} \mathrm{R}}^{609.67^{\circ} \mathrm{R}} 0.85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}^{\circ} \mathrm{R}} \mathrm{dT}\right. \\& \qquad\qquad\quad\left.+125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+\int_{609.65^{\circ} \mathrm{R}}^{659.65^{\circ} \mathrm{R}} \mathrm{C}_{\mathrm{p}}^{*} \mathrm{dT}\right) \\& \dot{\mathrm{Q}}=\left((1~ \mathrm{lbmol})\left(\frac{50~ \mathrm{lb}_{\mathrm{m}}}{1~ \mathrm{lbmol}^{\mathrm{l}}}\right)\right)\left(17.5 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+43.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right) \\& \qquad\qquad\quad =\bf17,310 \,BTU\end{aligned}