4/20/14
Specific Heat
p  Specific
heat: The amount of heat that
must be added to a stated mass of a
substance to raise it’s temperature, with
no change in state.
Thermochemistry
Part 4: Phase Changes &
Enthalpies of Formation
(q = mcΔT )
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Example: How much heat is released by 250.0 g of H2O
as it cools from 85.0oC to 40.0oC? (Remember, specific
heat of water = 4.18 J/goC)
But what if there is a phase change?
q = mcΔT
q = (250.0 g)(4.18 J/goC)(40.0-85.0)
q = -47,000 J = -47.0 kJ
LATENT HEAT OF FUSION, ΔHfus
the enthalpy change (energy
absorbed) when a compound is converted
from a solid to a liquid without a change in
temperature.
LATENT HEAT OF FUSION, ΔHfus
p  Definition:
p  “Latent”
means hidden; the heat absorbed/
released during a phase change does not
cause the temperature to change.
p  Note:
A = solid
B = melting (solid + liquid)
C = liquid
D = boiling (liquid + gas)
E = gas
ΔHfus for water is 334 J/g
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LATENT HEAT OF VAPORIZATION, ΔHvap
the enthalpy change (energy
absorbed) when one mole of the compound
is converted from a liquid to a gas without a
change in temperature.
LATENT HEAT OF VAPORIZATION,
ΔHvap
p  Definition:
p  Note:
for water ΔHvap is 2260 J/g
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0°C to -40.0°C?
Five steps…
1.  Cool the steam
2.  Condense
3.  Cool the liquid water
4.  Freeze
5.  Cool the solid ice
m∙csteam∙ΔT
m(-ΔHvap)
m∙cwater∙ΔT
m(-ΔHfus)
m∙cice∙ΔT
A = solid
B = melting (solid + liquid)
C = liquid
D = boiling (liquid + gas)
E = gas
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0°C to -40.0°C?
When substances change state, they often
have different specific heats:
—  cice
= 2.09 J/goC
—  cwater = 4.18 J/goC
—  csteam = 2.03 J/goC
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0°C to -40.0°C?
cooling = exothermic → negative heat values
qsteam = mcΔT = (250.0g)(2.03J/goC)(100.0–125.0) = -12,700 J
Now YOU try a few…
qvap = mHvap = (250.0g)(-2260J/g) = -565,000 J
qwater = mcΔT = (250.0g)(4.18J/goC)(0-100) = -105,000 J
qfus = mHfus = (250.0g)(-334J/g) = -83,500 J
qice = mcΔT = (250.0g)(2.09J/goC)(-40.0-0) = -20,900 J
qtotal = -787,000J
-787 kJ
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Example 2: How much heat energy is required to bring
135.5 g of water at 55.0oC to it’s boiling point (100.oC)
and then vaporize it?
Water must be heated to it’s boiling point.
q = mcΔT = (135.5 g)(4.18J/goC)(100-55)
q = 25.5 kJ
2. Water must be vaporized:
q = mΔHvap = (135.5 g)(2260 J/g)
q = 306 kJ
3. 
Add them together:
q = 25.5 kJ + 306 kJ = 332 kJ
1. 
Example 3: How much energy is required to convert 15.0
g of ice at -12.5oC to steam at 123.0oC?
1. 
2. 
3. 
4. 
5. 
Heat the ice (from -12.5 °C → 0 °C)
Melt the ice (at 0 °C)
Heat the water (from 0 °C → 100 °C)
Vaporize the water (at 100 °C)
Heat the steam. (from 100 °C → 123 °C)
Example 3: How much energy is required to convert 15.0
g of ice at -12.5oC to steam at 123.0oC?
qice = mcΔT = (15.0g)(2.09J/goC)(0.0 - -12.5) = 392 J
qfus = mHfus = (15.0g)(334J/g) = 5,010 J
Enthalpies of Formation
qwater = mcΔT = (15.0g)(4.18J/goC)(100-0) = 6,270 J
enthalpy
qvap = mHvap = (15.0g)(2260J/g) = 33,900 J
qtotal = 392 J + 5,010 J + 6,270 J + 33,900 J + 700. J = 46,300J
46.3 kJ
Enthalpies of Formation
p  usually
exothermic
p  see table for ΔHf° value
p  enthalpy of formation of an element
in its stable state = 0
p  these can be used to calculate ΔH°
for a reaction
standard
conditions
ΔH°f
qsteam = mcΔT = (15.0g)(2.03J/goC)(123.0-100.0) = 700. J
change
(delta)
formation
Standard Enthalpy Change
Standard enthalpy change, ΔH°, for a
given thermochemical equation is = to the
sum of the standard enthalpies of
formation of the product – the standard
enthalpies of formation of the reactants.
ΔH °rxn = Σ(n ⋅ ΔH °f
) - Σ(n ⋅ ΔH °f
products
)
reactants
“sum of”
(sigma)
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Standard Enthalpy Change
p 
elements in their standard states can be omitted:
Standard Enthalpy Change
p 
2 Al(s) + Fe2O3(s) → 2 Fe(s) + Al2O3(s)
ΔHrxn =
Σ(ΔHf°products) - Σ(ΔHf°reactants)
the coefficient of the products and reactants in
the thermochemical equation must be taken into
account:
2 Al(s) + 3 Cu2+(aq)
→ 2 Al3+(aq) + 3 Cu(s)
ΔHrxn = ΔHf°Al2O3 - ΔHf°Fe2O3
ΔHrxn =
Σ(ΔHf°products) - Σ(ΔHf°reactants)
ΔHrxn = (-1676.0 kJ) – (-822.1 kJ)
ΔHrxn = 2ΔHf°Al3+ - 3ΔHf°Cu2+
ΔHrxn = -853.9 kJ
ΔHrxn = 2(-531.0 kJ) – 3(64.8 kJ)
ΔHrxn = -1256.4 kJ
Standard Enthalpy Change
p 
Example: Calculate ΔH for the combustion of
one mole of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔHrxn =
Σ(ΔHf°products) - Σ(ΔHf°reactants)
ΔHrxn = [3ΔHf°CO2+4ΔHf°H2O] - ΔHf°C3H8
ΔHrxn = [3(-393.5kJ)+4(-285.8kJ)]–(-103.8 kJ)
ΔHrxn = -2219.9 kJ
Standard Enthalpy Change
p 
Example: The thermochemical equation for the
combustion of benzene, C6H6, is:
C6H6(l) + 15/2 O2(g) → 6CO2(g) + 3H2O(l)
ΔH° = -3267.4 kJ
Calculate the standard heat of formation of benzene.
-3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–ΔHf
°C H
6 6
-3267.4kJ
= -3218.4–ΔHf°C6H6
-49.0kJ = –ΔHf°C6H6
ΔHf°C6H6 = +49.0 kJ
Standard Enthalpy Change
p 
Example: When hydrochloric acid is added to a
solution of sodium carbonate, carbon dioxide gas
is formed. The equation for the reaction is:
2H+(aq) + CO32-(aq) → CO2(g) + H2O(l)
Calculate ΔH° for this thermochemical equation.
ΔH° = [(-393.5kJ)+(-285.8kJ)]–[2(0 kJ)+(-677.1kJ)
ΔH° = (-679.3kJ)–(-677.1kJ)
ΔH° = -2.2 kJ
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