Answer
$w_a=132.2\ Btu/lbm$
$T_2=274.6°F$
Work Step by Step
From tables A-5E and A-6E:
Initial ($P_1=100\ psia, T_1=650°F$): $u_1=1233.7\ Btu/lbm,\ s_1=1.7816\ Btu/lbm.°R$
Final ($P_2=10\ psia, s_{2,s}=s_1$): $x_2=0.9961,\ u_{2,s}=1068.4\ Btu/lbm$
From the energy balance, the isentropic specific work output of the system is:
$w_s=u_1-u_{2,s}=165.3\ Btu/lbm$
Therefore the actual work is:
$w_a=\eta w_s=132.2\ Btu/lbm$
Hence the actual final internal intergy is:
$w_a=u_1-u_{2,a}\rightarrow u_{2,a}=1101.4\ Btu/lbm$
From table A-6 at $P_2, u_{2,a}$: $T_2=274.6°F$
