Answer
$T_2=489.07°F$
Work Step by Step
Assuming the process is adiabatic $Q=0$, the energy balance equation would be:
$-W_{sh}=\Delta U = mc_v\Delta T$
From the ideal gas relation:
$PV=mRT$
Plugging in the values of $P_1=20psia, V_1=1ft³, T_1=100°F=559.67°F, R=0.3830 psia.ft³/lbm.°R$:
We have $m=0.0933\ lbm$
With the values of:
$W_{sh}= -5000lbf.ft \times \frac{1\ Btu}{778.2\ lbf.ft}=-6.425\ Btu, c_v=0.177\ Btu/lbm.°R$
We get to: $\Delta T=389.07\ °R$
Therefore: $T_2=489.07°F$
