Answer
$I=12.8\ A$
Work Step by Step
The energy balance for this constant-pressure process is:
$Q-W_e=\Delta H=m(h_2-h_1)$
From tables A-11 to A-13:
State 1 (240 kPa saturated vapor):$h_1=247.32\ kJ/kg$
State 2: (240 kPa, 70°C ): $h_2=314.53\ kJ/kg$
The electrical work is given by:
$-W_e=V.I.\Delta t$
Given $Q=300\ kJ,\ V=110\ V, \Delta t=6\ min=360s,\ m=12\ kg,$:
and solving for the current in the energy balance: $I=12.8\ A$
