HELP ASAP - How much energy is lost when 80.0 grams of steam at 100.C turns into liquid water at 100.C?

Q = 43,000 calories = 180,000 joules = 180 Kilojoules

Explanation:

Heat flow for phase change can be defined as Q = m·ΔHv  

ΔHv = Heat of Vaporization for water = 540 calories/gram = 2259 joules/gram = 2.259 Kilojoules/gram

m = mass of object of interest ( in this problem 80 grams steam 100°C converting to 80 grams of water at 100°C

Q = m·ΔHv = 80g x 540 cals/g =43.200 calories x 4.184 joules/cal = 180,748.8 joules (calculator answer) => 180,000 joules (2 sig. figs.) = 180 Kj

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