Question

How much heat is evolved in converting 1.00 mol of steam at 160.0 ∘C to ice at -50.0 ∘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

specific heat capacity of water, C1 = 4.186 J/goC
specific heat capacity of ice, C2 = 2.09 J/goC
specific heat capacity of steam, C3 = 2.01 J/goC
Latent heat of fusion of ice, Lf = 334 J/g
latent heat of vaporization of water LV = 2264.76 J/g

1 mol = 18 g since molar mass of water is 18 g

heat required to take steam from 160 oC to 100 oC,
Q1 = m*C3*delta T
      =18*2.01*(160-100)
      = 2170.8 J

heat required to convert steam to liquid water,
Q2 = m* LV
      = 18*2264.76
      =40765.7 J

heat required to take water from 100 oC to 0 oC,
Q3 = m*C1*delta T
      =18*4.186*(100-0)
      = 7534.8 J

heat required to convert water to liquid ice,
Q4 = m* Lf
      = 18*334
      =6012 J

heat required to take ice from 0 oC to -50 oC,
Q5 = m*C2*delta T
      =18*2.09*(50)
      = 1881 J

Total heat required= Q1 + Q2 + Q3 + Q4 + Q5
      = 2170.8 + 40765.7 + 7534.8 + 6012 + 1881
     = 58364.3 J

Answer: 58364.3 J