Question

How much heat is evolved in converting 1.00 mol of steam at 160.0 ∘C to ice at -45.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

First of all, we convert the amount of item given in moles into grams. Each mol of steam is 18 grams of steam. Such that the heat released during the evolution of per gram of steam is,
                H = (2.01 J/goC)(160 - 100) + 2260 J/g + (4.179 J/gC)(100 - 0) + (333.5 J/g) + (2.09 J/gC)(45oC)
                 H  = 3226.05 J/g
Then, we multiply this value by 18 g to get the final answer which is equal to 58068.9 Joules.