Answer
a) $m_{ice}=0.0546\ kg$
b) $m_{ice}=0.0487\ kg$
c) $m_{cold}=0.9\ kg$
Work Step by Step
The mass of 0.3 L of water ($\rho=1kg/L$) is 0.3 kg.
The energy balance for this system is:
$\Delta U_{ice}+\Delta U_{water}=0$
With the ice cubes at 0°C:
$\Delta U_{ice}=m_{ice}(\Delta h_f+c(T_2-0))$
$\Delta U_{water}=m_{water}c(T_2-T_{1,water})$
Given $\Delta h_f=333.7\ kJ/kg,\ c=4.18\ kJ/kg.K,\ T_2=5°C,\ T_{1,water}=20°C$
Solving for the mass of ice: $m_{ice}=0.0546\ kg$
b) For ice cubes at -20°C, $c_{ice}=2.11\ kJ/kg.K$:
$\Delta U_{ice}=m_{ice}(c_{ice}(0- T_{1,ice})+\Delta h_f+c_{water}(T_2-0))$
Solving for the mass of ice: $m_{ice}=0.0487\ kg$
c) For cold water:
$\Delta U_{cold}=m_{cold}c_{water}(T_2-0)$
Hence $m_{cold}=0.9\ kg$
