Answer
$T= 663.7$ F
$P= 105.5$ psia
Work Step by Step
The static temperature and pressure of steam are determined from
$$
\begin{aligned}
& T=T_0-\frac{V^2}{2 c_p}=700^{\circ} \mathrm{F}-\frac{(900 \mathrm{ft} / \mathrm{s})^2}{2 \times 0.4455 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}}\left(\frac{1 \mathrm{Btu} / \mathrm{lbm}}{25,037 \mathrm{ft}^2 / \mathrm{s}^2}\right)=\mathbf{6 6 3 . 7} \mathrm{F} \\
& P=P_0\left(\frac{T}{T_0}\right)^{k /(k-1)}=(120 \mathrm{psia})\left(\frac{1123.7 \mathrm{R}}{1160 \mathrm{R}}\right)^{1.329 /(1.329-1)}=\mathbf{1 0 5 . 5 p s i a}
\end{aligned}
$$
