Question

How much heat is evolved in converting 1.00 mol of steam at 135.0 ∘C to ice at -55.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Express your answer in units of kilojoules Assume the system is at atmospheric pressure.

Answer 1

1.00 mol of H2O is 18 g

Ti = 135.0 oC

Tf = -55.0 oC

Cg = 2.01 J/goC

Heat released to convert vapour from 135.0 oC to 100.0 oC

Q1 = m*Cg*(Ti-Tf)

= 18 g * 2.01 J/goC *(135-100) oC

= 1266.3 J

Lv = 2260.0 J/g

Heat released to convert gas to liquid at 100.0 oC

Q2 = m*Lv

= 18.0g *2260.0 J/g

= 40680 J

Cl = 4.184 J/goC

Heat released to convert liquid from 100.0 oC to 0.0 oC

Q3 = m*Cl*(Ti-Tf)

= 18 g * 4.184 J/goC *(100-0) oC

= 7531.2 J

Lf = 333.0 J/g

Heat released to convert liquid to solid at 0.0 oC

Q4 = m*Lf

= 18.0g *333.0 J/g

= 5994 J

Cs = 2.09 J/goC

Heat released to convert solid from 0.0 oC to -55.0 oC

Q5 = m*Cs*(Ti-Tf)

= 18 g * 2.09 J/goC *(0--55) oC

= 2069.1 J

Total heat released = Q1 + Q2 + Q3 + Q4 + Q5

= 1266.3 J + 40680 J + 7531.2 J + 5994 J + 2069.1 J

= 57540.6 J

= 57.5 KJ

Answer: 57.5 KJ