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