Asked by susan
How many joules of heat are removed to completely covert 80.0 grams of water to ice at zero Celcius?
For water: c = 4.179 J/g °C Hfusion = 333.528J/g Hvap = 2258.601 J/g
For water: c = 4.179 J/g °C Hfusion = 333.528J/g Hvap = 2258.601 J/g
Answers
Answered by
DrBob222
q = mass H2O x heat fusion
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