Answer
$m_{ice}=0.0294\ kg$
Work Step by Step
The energy balance for the constant-pressure system is:
$\Delta H_{water}+\Delta H_{ice}=0$
For the water (120°C, saturated):
$v_L=0.001060\ m³/kg,\ v_G=0.89133\ m³/kg,\ h_L=503.81\ kJ/kg,\ h_G=2202.1\ kJ/kg$:
With a quality of 20% initially and 0% in the end:
$v_1=0.17911\ m³/kg,\ h_1=944.24\ kJ/kg,\ v_2=v_L,\ h_2=h_L$
The mass is given by (with $V_1=0.01\ m³$):
$V_1=v_1.m$
$m=0.05583\ kg$
Hence $\Delta H_{water}=-24.59\ kJ$
For the ice(starting at 0°C, with $c=4.18\ kJ/kg.K,\ \Delta h_{fusion}=333.7\ kJ/kg$):
$\Delta H_{ice}=m_{ice}(\Delta h_{fusion}+c(120-0)K)$
Solving for the mass: $m_{ice}=0.0294\ kg$
