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m1= 2kg
t=0C
Q=117 kJ
m2=?
ct=333,7 kJ/kg
ct=Q/m2
m2= Q/ct
m2=117kJ/333,7kJ/kg≈o,35 kJ×kg/kJ=0,35kg
%masy stopionej=m2/m×100%
%masy stopionej=0,35kg/2kg×100%=17,5%
Odp.: Stopieniu uległo ok. 17,5% masy lodu.