Answer
$T_2=118.94°C$
Work Step by Step
The energy balance for this system reduces to:
$-W_e=\Delta U=m(u_2-u_1)$
The electrical work is given by (with $V=50 V,\ I=10\ A,\ \Delta t=30min=1800s$):
$-W_e=V.I.\Delta t$
$-W_e=900\ kJ$
At the initial state (Table A-4):
$u_1=167.53\ kJ/kg,\ v_1=0.001008\ m³/kg$
From the energy balance ($m=3\ kg$):
$u_2=\frac{-W_e}{m}+u_1$
$u_2=467.53\ kJ/kg$
Since the volume is constant: $v_2=v_1$
From a NIST table (https://www.nist.gov/sites/default/files/documents/srd/NISTIR5078-Tab3.pdf):
we can interpolate to find:
$T_2=118.94°C$, $P_2=112.74\ MPa$
