Specific heat

How much heat must be added to convert 25 kg of ice at −15 °C into water at 23 °C? The specific heat of fusion of ice is 333 kJ/kg.

Final Answer:

Q =  12296 kJ

Step-by-step explanation:

m=25 kg t1=15 °C t2=23 °C l=333 1000=333000 J/kg c=4180 J/kg/°C  Q1=m l=25 333000=8325000 J Q2=m c (t2t1)=25 4180 (23(15))=3971000 J  Q3=Q1+Q2=8325000+3971000=12296000 J  Q=Q3 kJ=Q3:1000  kJ=12296000:1000  kJ=12296 kJ=12296 kJ



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