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