Answer
$T_{2}=104.7^{\circ}F$
$h_{2}=124.7\frac{Btu}{lbm}$
Work Step by Step
From table A-11E
$P_{1}=9.869psia$
$\upsilon_{1}=\upsilon_{f}+x\upsilon_{fg}=0.01143\frac{ft^3}{lbm}+0.8*(4.4286\frac{ft^3}{lbm}-0.01143\frac{ft^3}{lbm})=3.5452\frac{ft^3}{lbm}$
Then:
$V_{1}=m\upsilon_{1}=0.13lbm*3.5452\frac{ft^3}{lbm}=0.4609ft^3$
$V_{2}=1.5V_{1}=1.5*0.4609ft^3=0.6914ft^3$
To calculate the distance traveled by the piston:
$A=\frac{\pi D^2}{4}=\frac{\pi*(1ft)^2}{4}=0.7854ft^2$
$\Delta x=\frac{\Delta V}{A}=\frac{0.6914ft^3-0.4609ft^3}{0.7854ft^2}=0.2935ft*\frac{12in}{1ft}=3.522in$
Then the change of pressure is:
$A=\frac{\pi D^2}{4}=\frac{\pi*(12in)^2}{4}=113.1in^2$
$P=\frac{F}{A}=\frac{k\Delta x}{A}=\frac{37\frac{lbf}{in}*3.522in}{113.1in^2}=1.1522psia$
Now to calculate the final temperature and enthalpy:
$P_{2}=P_{1}+\Delta P=9.869psia+1.1522psia=11.0212psia$
$\upsilon_{2}=\frac{V_{2}}{m}=\frac{0.6914ft^3}{0.13lbm}=5.3185\frac{ft^3}{lbm}$
As $\upsilon\gt \upsilon_{g}$ is a superheated vapor
From EES:
$T_{2}=104.7^{\circ}F$
$h_{2}=124.7\frac{Btu}{lbm}$
