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```Thermochemical Calculations
CA Standards
Students know energy is released when a material
condenses or freezes and is absorbed when a material
evaporates or melts.
Students know how to solve problems involving heat flow
and temperature changes, using known values of specific
heat and latent heat of phase change.
Units for Measuring Heat
The Joule is the SI system unit for
measuring heat:
1 kg  m
1 Joule1 newton m eter
2
s
2
The calorie is the heat required to raise
the temperature of 1 gram of water by
1 Celsius degree
1calorie 4.18 Joules
Specific Heat
The amount of heat
required to raise
the temperature of
one gram of
substance by one
degree Celsius.
5
4
6
5
7
4
3
8
3
2
9
2
1
11
6
7
8
9
1
10
Calculations Involving Specific Heat
Q  m  T  c p
OR
Q
cp 
m  T
cp = Specific Heat
Q = Heat lost or gained
T = Temperature change
m = Mass
Specific Heat
The amount of heat required to raise the temperature
of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
0.129
Gold (solid)
0.129
Latent Heat of Phase Change
Molar Heat of Fusion
The energy that must be absorbed in
order to convert one mole of solid to
liquid at its melting point.
Molar Heat of Solidification
The energy that must be removed in
order to convert one mole of liquid to
solid at its freezing point.
Latent Heat of Phase Change #2
Molar Heat of Vaporization
The energy that must be absorbed in
order to convert one mole of liquid to
gas at its boiling point.
Molar Heat of Condensation
The energy that must be removed in
order to convert one mole of gas to
liquid at its condensation point.
Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert
60 grams of ice at 0C to liquid water at 0C?
60 g H 2O 1 m olH 2O 6.009kJ
 20 kiloJoules
18.02 g H 2O 1 m olH 2O
Mass
of ice
Molar
Mass of
water
Heat
of
fusion
A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water
at 25oC. If the system's final temperature is 46oC, what was the initial
Pb
T = ? oC
mass = 322 g
Ti = 25oC
mass = 264 g
Tf = 46oC
Pb
- LOSE heat = GAIN heat
- [(Cp,Pb) (mass) (T)]
Drop Units:
=
(Cp,H O) (mass) (T)
2
- [(0.138 J/goC) (322 g) (46oC - Ti)]
=
(4.184 J/goC) (264 g) (46oC- 25oC)]
- [(44.44) (46oC - Ti)]
=
(1104.6) (21oC)]
- 2044 + 44.44 Ti
=
23197
44.44 Ti =
Ti
=
25241
568oC
Calorimetry Problems
2
question #12
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at
500oC). When thermal equilibrium is reached, the system has a temperature of 42oC.
Find the mass of the iron.
Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g
- LOSE heat = GAIN heat
- [(Cp,Fe) (mass) (T)]
=
(Cp,H O) (mass) (T)
2
- [(0.4495 J/goC) (X g) (42oC - 500oC)]
Drop Units:
- [(0.4495) (X) (-458)]
205.9 X
X
=
=
=
(4.184 J/goC) (240 g) (42oC - 20oC)]
(4.184) (240 g) (22)
22091
= 107.3 g Fe
Calorimetry Problems 2
question #5
A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial
temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the
final temperature of the mixture? Assume that the gold experiences no change in
state of matter.
Au
T = 785oC
mass = 97 g
T = 15oC
mass = 323 g
- LOSE heat = GAIN heat
- [(Cp,Au) (mass) (T)]
Drop Units:
=
(Cp,H O) (mass) (T)
2
- [(0.129 J/goC) (97 g) (Tf - 785oC)]
=
- [(12.5) (Tf - 785oC)] =
-12.5 Tf + 9.82 x 103
3 x 104 =
Tf
=
=
(4.184 J/goC) (323 g) (Tf - 15oC)]
(1.35 x 103) (Tf - 15oC)]
1.35 x 103 Tf - 2.02 x 104
1.36 x 103 Tf
22.1oC
Calorimetry Problems
2
question #8
If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final
temperature of the system.
T = 72oC
mass = 87 g
T = 13oC
mass = 59 g
- LOSE heat = GAIN heat
- [(Cp,H O) (mass) (T)]
=
2
Drop Units:
(Cp,H O) (mass) (T)
2
- [(4.184 J/goC) (87 g) (Tf - 72oC)] =
(4.184 J/goC) (59 g) (Tf - 13oC)
- [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC)
-364 Tf + 26208 = 246.8 Tf - 3208
29416 =
Tf
=
610.8 Tf
48.2oC
Calorimetry Problems
2
question #9
ice
T = -11oC
mass = 38 g
A
T = 56oC
mass = 214 g
D
D
water cools
B
warm
water
C
melt ice
warm
Temperature (oC)
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC.
Find the system's final temperature.
140
120
100
80
60
40
20
0
-20
-40
-60
-80
-100
H = mol x Hvap
H = mol x
Hfus
Heat = mass x t x Cp,
gas
Heat = mass x t x Cp,
liquid
Heat = mass x t x Cp,
solid
Time
- LOSE heat = GAIN heat
A
B
C
- [(Cp,H2O(l)) (mass) (T)] = (Cp,H2O(s)) (mass) (T) + (Cf) (mass) + (Cp,H2O(l)) (mass) (T)
-[(4.184 J/goC)(214g)(Tf-56oC)] = (2.077J/goC)(38g)(11oC) + (333J/g)(38g) + (4.184J/goC)(38g)(Tf-0oC)
- [(895) (Tf - 56oC)] =
868 + 12654
+
(159) (Tf)]
- 895 Tf + 50141
=
868 + 12654 + 159 Tf
- 895 Tf + 50141
=
13522 + 159 Tf
36619
=
1054 Tf
Tf
=
34.7oC
Calorimetry Problems
2
question #10
Heat of Solution
The Heat of Solution is the amount of heat
energy absorbed (endothermic) or released
(exothermic) when a specific amount of
solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH4NO3
+25.69
KNO3
+34.89
HCl
-74.84
```