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How much heat is necessary to change 30 g of ice at -20°C into water at 20°C?

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    c(ice)= 2 060 J/kg•degr
    c(water)= 4180 J/kg•degr
    λ = 335000 J/kg

    Q=Q1+Q2+Q3
    Q1=m•c(ice) •ΔT= 0.03•2060 • 20 =1236 J.
    Q2=λm=335000•0.03=10050 J
    Q3 = m•c(water) •ΔT=0.03•4180•20=2508 J
    Q=1236+10050+2508 =13794 J

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