Question

How much heat is evolved in converting 1.00 mol of steam at 160.0 ∘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

Soluton: The process of converting steam to ice can be separated into five segments:
1)Cooling of steam from 160.0 ^∘C to 100 ^∘C (Q₁)
2)Conversion of steam to water. (Q2)
3)Cooling of water to from 100 ^∘C to 0 ^∘C. (Q3)
4)Conversion of water to ice. (Q4)
5)Cooling of ice from 0 ^∘C to -55.0^∘ C. (Q5)

Mass of H2O (m) =(1.00 mol H2O) x (18.01532 g H2O/mol) = 18.015 g H2O

Q1 =(2.01 J/(g°C)) x (18.015 g) x (160.0 - 100.0)°C = 2173 J from cooling the steam to its boiling point

Q2=(2260 J/g) x (18.015 g) = 40793 J from steam to water (Latent heat of vapourization of water = 2260 J/g)

Q3=(4.184 J/g·°C) x (18.015 g) x (100 - 0) °C) = 7537 J from cooling the condensed steam to its freezing point

Q4=(334 J/g) x (18.015 g) = 6017 J from freezing the water (Latent heat for fusion is 334 J/g)

Q5=(2.09 J/(g⋅°C) x (18.015 g) x (0 - (-45.0))°C = 1694 J from cooling the ice to the final temperature

Total heat evoved = Q1+Q2+Q3+Q4+Q5,

2173 J + 40793 J + 7537 J + 6017 J + 1694 J = 58214 J = 58.2 kJ