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Last time…
 Course policies
 Thermodynamics
 Types of Energy
 First Law of Thermodynamics (Conservation of Energy)

E

0
u n iverse
Energy Units
J
1J=
kJ
cal
kcal
BTU
0.2390 2.390 x 10-4
kWh
1
0.001
1 kJ =
1000
1
239.0
0.2390
1.0557 2.778 x 10-4
1 cal =
4.184 4.184 x 10-3
1
0.001
252 1.162 x 10-6
1 kcal =
4184
4.184
1000
1
0.252 1.162 x 10-3
1 kWh =
3.6 x 106
3.6 x 103
8.604 x 105
860.4
 Read as 1J = 0.2390 cal
1055 2.778 x 10-7
2.93 x 10-4
1
Example-Energy
Unit Conversion
•The amount of energy required to
pump one sodium ion out of a cell
is 114 “milli electron Volts”
•114meV=114 x 10-3eV
•1eV = 1.602 x 10-19J
The Chocolate Chip Challenge:
How many grams of salt would you need to give you the number of sodium ions
that could be transported by the energy in one chocolate chip?
(1 chocolate chip has 3.5 Calories)
Temperature and Heat
 Heat (q) and Temperature (T) are not the same!
 The more thermal energy something has, the greater
the motion of its atoms
 The total thermal energy in an object is the sum of the
individual energies of all the atoms, molecules, or ions
Which statement below best describes the process of
placing a thermometer initially at 22ºC into ice water?
1.
Some of the thermal energy of the
ice water is transferred to the
thermometer.
2.
Some of the thermal energy of the
thermometer is transferred to the ice
water, melting some of the ice.
3.
The atoms of mercury begin to move
faster as a result of the thermal
energy transfer between the
thermometer and the ice water.
4.
The mercury in the thermometer
begins to expand as a result of the
thermal energy transferred.
43%
27%
16%
1
14%
2
3
4
What happens to thermal (heat) energy?
 Warms another object (transfer)
 Causes a change of state
 Is used in an endothermic reaction
Heat Transfer
 If heat (q) is transferred, in which direction does it go?
 From hotter to cooler (related to the 2nd Law, but we’ll
get to that later)
 Heat lost = Heat gained (1st Law)
q

q
warmer
co o l er
 Thermal equilibrium= when two objects in contact
reach the same temperature
System and Surroundings
 System = Thing or things being studied

Isolated: Neither energy or matter can be transferred to surroundings

Closed: Energy, but not matter can be transferred to surroundings
 Surroundings = Everything else in the universe
Energy transfer between system and
surroundings
 Internal energy=Energy of a closed system
 We can measure changes
 Change occurs if heat is transferred to/from system or if
work is done on/by system
Esystemqw
qheat
wwork
How to Determine the Sign of q and w
+
 Endothermic- system takes in heat
Ammonium thiocyanate and barium hydroxide hydrate
 Exothermic- system gives off heat
System
E=q+w
Heat transferred in
q>0
Heat transferred out
q<0
Work done on
w>0
Work done by
w<0
Endothermic
Exothermic
-
Internal Energy Change Example:
A gas is compressed and during this process the
surroundings do 143 J of work on the gas. At the same
time, the gas absorbs 212 J of heat from the surroundings.
What is the change in the internal energy of the gas?
Heat Transfer and Specific Heat Capacity
 When I heat an object, what happens and how much
energy does it require? It depends . . .
1. Quantity (How much stuff do I have?)
2. Amount of heat energy added
3. Identity of the material
 Heat capacity is the energy needed to raise 1g by 1ºC.
q

m

C


T
q

n

C


T
How much energy does it take to boil
water for tea?
The instructions for baking brownies say to heat the oven
to 350ºF if using an aluminum pan, but to heat the oven to
325ºF if using a glass pan. Why is this?
 Glass heats up faster because it has a lower heat
capacity.
 Glass heats more slowly because it has a lower heat
capacity.
 Aluminum heats more slowly because it has a lower
heat capacity.
 Aluminum heats up faster because it has a higher heat
capacity.
The instructions for baking brownies say to heat the oven
to 350ºF if using an aluminum pan, but to heat the oven to
325ºF if using a glass pan. Why is this?
1.
2.
3.
4.
Glass heats up faster because it
has a lower heat capacity.
Glass heats more slowly
because it has a lower heat
capacity.
Aluminum heats more slowly
because it has a lower heat
capacity.
Aluminum heats up faster
because it has a higher heat
capacity.
75%
16%
2%
1
2
3
7%
4
Enthalpy
Quantitative: Calculating Heat Exchange: Specific Heat Capacity
Temperature Changes from Heat Exchange
Example 1: 5 g wood at 0 oC
Example 2: 10 g wood at 0 oC
Example 3: 5 g copper at 0 oC
Example 4: 5 g wood at 0 oC
Clicker Choices:
1: 0 oC 2: 33 oC
3: 50 oC
+
+
+
+
4. 67 oC
5 g wood at 100 oC
5 g wood at 100 oC
5 g copper at 100 oC
5 g copper at 100 oC
5: 100 oC 6: other
What happens to thermal (heat) energy?
When objects of different temperature meet:
 Warmer object cools
 Cooler object warms
 Thermal energy is transferred
qwarmer = -qcooler
specific heat x mass x T = specific heat x mass x T
warmer object
cooler object
Heat transfer between substances:
J
o
q
=
1
.
8

5
g

(
1
8
C
)
w
o
o
d
o
g
C
=
1
6
0
J
J
o
q
=
0
.
3
8
5

5
g

(
+
8
2
C
)
C
u
o
g
C
=
+
1
6
0
J
Example
 If we mix 250 g H2O at 95 oC with 50 g H2O at 5 oC,
what will the final temperature be?
Thermal Energy and Phase
Changes
First: What happens?
Thermal Energy and Phase
Changes
First: What happens?
Thermal Energy and Phase
Changes
First: What happens?
But what’s really
happening?
Warming:
• Molecules move more rapidly
• Kinetic Energy increases
• Temperature increases
Melting/Boiling:
• Molecules do NOT move more rapidly
• Temperature remains constant
• Intermolecular bonds are broken
• Chemical potential energy (enthalpy) increases
Energy and Phase Changes:
Quantitative
Treatment
Melting:
Heat of Fusion (Hfus) for Water: 333 J/g
Boiling:
Heat of Vaporization (Hvap) for Water: 2256 J/g
Total Quantitative Analysis
Convert 40.0 g of ice at –30 oC to steam at 125 oC
Warm ice: (Specific heat = 2.06 J/g-oC)
Melt ice:
Warm water (s.h. = 4.18 J/g-oC)
Total Quantitative Analysis
Convert 40.0 g of ice at –30 oC to steam at 125 oC
Boil water:
Warm steam (s.h. = 1.92 J/g-oC)
Enthalpy Change and Chemical Reactions
H = energy needed to break bonds – energy released forming bonds
Example: formation of water:
H = ?
Enthalpy Change and Chemical Reactions
H is usually more complicated, due to solvent and
solid interactions.
So, we measure H experimentally.
Calorimetry
Run reaction in a way that the heat exchanged
can be measured. Use a “calorimeter.”
Calorimetry Experiment
N2H4 + 3 O2  2 NO2 + 2 H2O
Energy released = E absorbed by water +
E absorbed by calorimeter
Ewater =
Ecalorimeter =
Total E =
H = energy/moles =
0.500 g N2H4
600 g water
420 J/oC
Hess’s Law
If reactions can be “added”
so can their H values.
Fig. 5-4, p. 183
Table 5-2, p. 195