Answer
$T_2=12.4°C$
Work Step by Step
The energy balance for this system reduces to:
$\Delta U_{ice}+\Delta U_{water}=0$
$\Delta U_{ice}=m_{ice}(c_{solid}(0-T_{1,solid})+\Delta h_{fusion}+c(T_2-0))$
$\Delta U_{water}=m_{water}c(T_2-T_{1,water})$
Given $m_{ice}=80\ kg,\ m_{water}=1000\ kg, c=4.18\ kJ/kg.K, c_{solid}=2.11\ kJ/kg.K$,
$\Delta h_{fusion}=333.7\ kJ/kg,\ T_{1,solid}=-5°C,\ T_{1,water}=20°C$
Solving for the final temperature: $T_2=12.4°C$
