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10/10/2011
Chem 105
Mon 10 Oct 2011
Announcements
1. Blackboard gradebook is combined for both 105 sections. For Keller
(JWK) students, the entries labeled FH (Fenton Heirtzler) should be
blank .
2. OWL reminder. All OWL sections that are due or past due may be
worked on at any time. If a catastrophe happened and you could not
complete an OWL section, GO AHEAD and work on it.
1.
2.
3.
4.
Chapt 5: Thermochemistry: Kinds of energy
Heat flow and heat measurement
Specific heat and heat capacity
Heating/cooling curves
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Chapt 5: Energy and Chemical Reactions
Kinetic energy = motion
Thermal energy - atoms vibrating or moving in gas,
liquid, or even solid (slightly)
Mechanical energy - macroscopic motion (baseball, car
moving etc)
Electrical energy – electrons moving in a wire (current)
Sound energy - (really just #1)
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Chapt 6: Energy and Chemical Reactions
Potential Energy = stored energy
Chemical - Energy stored in compounds that can
be released by a chemical reaction
Electrostatic – separation of + and – charges
released as charges come closer to each other.
Gravitational – separation of macroscopic bodies
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Conservation of energy
Physicists give us the First Law of Thermodynamics:
“Total energy of the universe is constant”.
For the rest of us, we know that energy is not “lost” - it
flows from one form to another.
sound
Thermal energy
Mechanical energy of flying parts
Mechanical energy of
speeding car
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Conservation of energy
sound
Thermal energy
Mechanical energy of
speaker parts
electrical
energy
chemical energy
of battery
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Conservation of energy
electrical
energy
A hand-cranked
version of this
device was used in
1820 by James
Joule to define the
“Joule” energy unit.
Mechanical
energy of stirrer
thermal energy of water
(increased water
temperature)
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Quantitative definitions
1 calorie (cal) = amount of energy required to raise the
temperature of 1.00 g water by 1.00 °C.
1 kilocalorie (kcal) = amount of energy required to raise the
temperature of 1.00 kg water by 1.00 °C.
1 calorie (cal) = 4.184 joule (J)
The exact conversion of electrical to
thermal energy was determined in
the stirred-water experiment:
1 kilocalorie (kcal) = 4.184 kilojoule (kJ)
1 kJ = 0.2778 watt-hour
(No need to memorize these numbers; they are in text and on exam
reference sheets)
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Heat flows from higher temperature body to a lower
temperature body by conduction.
Warm object
At high temps,
atoms vibrate
faster with
larger
amplitudes
At thermal
equilibrium,
atomic motions
the same &
temps are equal.
Cool object
Motion is
transferred to
atoms in
adjacent body.
Heat flow
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Heat transfer by conduction animation
(This works similarly for gases, liquids, or solids.)
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DEFINE
q = heat transferred to an object
- positive for heat gained
- negative for heat lost
Tinitial
Tinitial
Initial state
Heat flow q
Final state
Tfinal
Tf < Ti
q<0
Tfinal
Tf > Ti
q>0
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q =Heat transferred to
object (joules)
“Joules per gram per Kelvin”
Specific heat capacity (joule/gram ·K)
q = m x C x (Tf – Ti)
You DO need to
remember this equation.
Mass of the body (solid,
liquid, or gas) (g)
Final temperature (K or °C)
Initial temperature (K or °C)
q = m x C x (Tf – Ti)
J  g
J
J
 K  g  o  oC
gK
g C
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10/10/2011
Say we have a beaker of warm water
144 g water
at 55.5 °C
… and a 29.5-g sample
of metal was taken from
a boiling water bath
and was placed in the
first beaker of water.
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What will be the SIGN of qwater?
142
1. +
13
2. 1
4
2
3
3. Don’t know
1
1
14
27
40
53
66
79
92
105
118
131
144
157
170
183
196
2
3
4
5
6
7
8
9
10
11
12
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Answer = 1. +
q is defined as “the heat that flows INTO an object”.
If a hot object is placed in the water, then heat flows
from the one with higher temperature to the one
with lower temp. I.e. q > 0 for water.
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We can use temperature changes and the known
specific heat capacity of water to identify the
metal.
144 g water
at 55.5 °C
A 29.5-g metal
block was taken
from boiling water
bath.
Final temp = 57.8 °C
For water: Tf – Ti = 57.8-55.5 °C = +2.3°C
For metal: Tf – Ti = 57.8-100.0 °C = -42.2°C
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Perhaps this metal is magnesium. Note
that the data allow calculation of
specific heat to only 2 sig figs.
q w  m w C w ΔTw  144 g 
4.184 J
o
o


x
57.8
C
55.5
C
o
g C
q w  1385.7 J
Heat flow
into water
Heat flow
into metal
J/g-°C
Mg 1.03
Al
0.897
Co
0.695
Fe
0.449
Cu
0.385
Au
0.129
We can solve for Cm
We know these..
q m  q w  m m C m ΔTw
- qw
 1385.7J
Cm 

m m  ΔTw 29.5g  57.8o C  100.0o C 
Cm 
1.113 J 1.1 J
 o
o
g C
g C
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Heating/Cooling Curves
Includes heat flow during phase changes:
Evaporation (l  g)
Condensation (g  l)
Melting
(s  l)
Freezing
(l  s)
[Sublimation
(s  g)
(chem 106)]
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Melting requires added heat.
Changes of state occur at constant temperature.
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In boiling water, the vapor phase in bubbles of vapor is in equilibrium with
the surrounding liquid phase.
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H2O (s) -----> H2O (l)
ΔH = +5.99 kJ
ΔHfus = +5.99 kJ/mol (“molar heat of fusion)
= +333 J/g (“mass heat of fusion”)
q = (mass)(heat of fusion) = mΔHfus
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Heating Curve for Water
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What quantity of heat is required to melt 500. g of ice
at 0 oC, and heat the water to vapor at 100 oC?
Given data:
Heat of fusion of ice = 333 J/g
Specific heat of water =
4.2 J/g•K
Heat of vaporization
= 2260 J/g
+333 J/g
+2260 J/g
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What quantity of heat is required
to melt 500. g of ice (at 0 oC) and
convert the water to vapor at
100 oC?
q = (500. g)(333 J/g) = 1.67x 105 J
2.To warm water from 0oC to100 oC
q = (500. g)(4.184 J/g•K)(100- 0)K
= 2.092 x 105 J
Temperature
1. To melt ice
100°
0°
3. To evaporate water at 100 oC
q = (500. g)(2260 J/g)
= 1.130 x 106 J
4. Total heat = sum = 1.506 x 106 J
= 1.51 x 103 kJ
Heat added
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One clicker question on heating/cooling curve
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Identify the line segment on the following diagram that would require the heat of
vaporization and specific heat of liquid to calculate an energy change.
Within this segment, you
would need to use the
specific heat capacity of
WATER VAPOR.
42%
29%
Within this segment, you would need
to use ONLY the heat of
vaporization of water, because there
is NO temperature change.
...
tF
tA
...
se
gm
en
tD
...
ne
Li
Li
ne
se
gm
en
tG
...
Li
ne
se
gm
en
...
tE
se
gm
en
ne
5.
Li
4.
13%
2%
se
gm
en
3.
13%
ne
2.
Line segment E-F-G
Line segment G-H-J
Line segment D-E-F
Line segment A-B-C
Line segment F-G-H
Li
1.
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OWL:
Ch 5- 3b Homework: Phase Change Energetics
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Typical heating/cooling curve
Evaporate
Temp 
2270 °C -->
 Condense
560 °C -->
232 °C -->
211 °C -->
<Freeze
Melt->
Heat flow in 
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q = mCΔT=25.9 g x 0.243 J/g°C x -328°C
q = -2064.3 J
q = m ΔHfus = 25.9 g x 59.60 J/g
Temp 
560 °C -->
232 °C -->
Freeze
q = -1543.6 J (negative „cause its
freezing)
q = mCΔT=25.9 g x 0.2260 J/g°C x -21°C
211 °C -->
q = -122.9 J
Heat flow in 
qtot = - 3730 J
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Arrgghhh!
- 3730
Don‟t forget to calculate “kJ” if it asks for it.
“Removed” takes care of the sign of q.
“q” IS negative, but “heat removed” cannot be negative.
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The End