NOTES: 17.3 – Heat in Changes of State

RECALL…

● when a substance changes state (i.e. melts, freezes, vaporizes, condenses) it does not change temperature

● however, it does absorb or release heat during the phase change

● melting and vaporizing: ENDOTHERMIC

● freezing and condensing: EXOTHERMIC

Heat of Fusion and Solidification:

● Solid ↔ Liquid● molar heat of fusion (∆Hfus) = heat absorbed

when one mole of a solid substance is melted● molar heat of solidification (∆Hsolid) = heat

released when one mole of a liquid substance solidifies (freezes)

● ∆Hfus = -∆Hsolid

● Examples:H2O(s) H2O(l) ∆Hfus = 6.01 kJ/mol

H2O(l) H2O(s) ∆Hsolid = -6.01 kJ/mol

Example #1:

How many grams of ice at 0°C could

be melted by the addition of 2.25 kJ of

heat?

Example #1:

How many grams of ice at 0°C could

be melted by the addition of 2.25 kJ of

heat?

ice 74.6

1

0.18

01.6

0.125.2

2

22

g

OmolH

OgH

kJ

OmolHkJ

Example #2:

Sometimes heat of fusion is given in J/g: for ice, Hfusion = 333 J/g.

How much energy (in J) is required to melt 17.75 g of ice at 0°C?

Example #2:

Sometimes heat of fusion is given in J/g: for ice, Hfusion = 333 J/g.

How much energy (in J) is required to melt 17.75 g of ice at 0°C?

Jg

JOgH 5911

33375.17 2

Heat of Vaporization / Condensation:

● Liquid ↔ Gas● molar heat of vaporization (∆Hvap) = heat

absorbed when one mole of a liquid substance is vaporized

● molar heat of condensation (∆Hcond) = heat released when one mole of a gaseous substance condenses

● ∆Hvap = -∆Hcond

● Examples:H2O(l) H2O(g) ∆Hvap = 40.7 kJ/mol

H2O(g) H2O(l) ∆Hcond = -40.7 kJ/mol

Example #3:

How much heat (in kJ) is absorbed when 24.8 g of liquid H2O at 100°C is

converted to steam at 100°C?

Example #3:

How much heat (in kJ) is absorbed when 24.8 g of liquid H2O at 100°C is

converted to steam at 100°C?

kJ

OmolH

kJ

OgH

OmolHOgH

4.56

1

7.40

0.18

0.18.24

22

22

Example #4:

Sometimes heat of vaporization is given in J/g: for water, Hvapor. = 2260 J/g.

How much energy (in J) is required to vaporize 113.2 g of water at 100°C?

Example #4:

Sometimes heat of vaporization is given in J/g: for water, Hvapor. = 2260 J/g.

How much energy (in J) is required to vaporize 113.2 g of water at 100°C?

Jg

JOgH 832,255

22602.113 2

Heat of Solution:

● Solid Solution● molar heat of solution (∆Hsoln) = heat

change caused by the dissolution (dissolving) of one mole of a substance

● Example:NaOH(s) + H2O(l) Na+

(aq) + OH-(aq)

∆Hsoln = -445.1 kJ/mol

Example #5:

NaOH(s) + H2O(l) Na+(aq) + OH-

(aq)

∆Hsoln = -445.1 kJ/mol

How much heat (in kJ) is released when 11.5 g of NaOH(s) is dissolved in water?

Example #5:

NaOH(s) + H2O(l) Na+(aq) + OH-

(aq)

∆Hsoln = -445.1 kJ/mol

How much heat (in kJ) is released when 11.5 g of NaOH(s) is dissolved in water?

kJ

molNaOH

kJ

gNaOH

molNaOHgNaOH

128

1

1.445

0.40

0.15.11

Example #6:

How much heat energy is necessary to turn 10.0 grams of ice at -15.0oC into water vapor at 120.0oC?**Hint: heat ice, melt ice, heat water, vaporize

water, heat vapor● Given:H2O(s) H2O(l) ∆Hfus = 333 J/g

H2O(l) H2O(g) ∆Hvap = 2260 J/g

Cice = 2.01 J/(g oC)

Cwater = 4.18 J/(g oC)

Cvapor = 2.03 J/(g oC)

Example #6:

Step 1: heat up the ice to its melting point:

q = mcΔTq = (10.0 g)(2.01 J/g˚C)(15.0˚C)q = 301.5 J

Example #6:

Step 2: melt the ice

q = (mass)(Hfusion)

q = (10.0 g)(333 J/g)q = 3330 J

Example #6:

Step 3: heat up the liquid water to its boiling point:

q = mcΔTq = (10.0 g)(4.18 J/g˚C)(100.0˚C)q = 4180 J

Example #6:

Step 4: vaporize the liquid water

q = (mass)(Hvapor)

q = (10.0 g)(2260 J/g)q = 22600 J

Example #6:

Step 5: heat up the steam to its final temp.

q = mcΔTq = (10.0 g)(2.03 J/g˚C)(20.0˚C)q = 406 J

Example #6:

Finally…add up all steps!301.5 J +3330 J +4180 J +22600 J +406 J = 30,818 J

= 30.8 kJ