Energy and
Thermochemistry
1
Energy
The ability to do work
 2 types

Potential: stored energy
 Kinetic: energy in motion

2
Thermochemistry
Changes of heat content and heat transfer
 Follow Law of Conservation of Energy
 Or, 1st Law of Thermodynamics


Energy can neither be created nor destroyed
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Temperature & Heat

Heat not same as temperature


Heat = energy transferred to one system by another
due to temperature difference
Temperature = measure of heat energy content &
ability to transfer heat



Thermometer
Higher thermal energy, greater motion of
constituents
Sum of individual energies of constituents = total
thermal energy
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Systems and Surroundings




System = the object in question
Surrounding(s) = everything outside the system
When both system and surrounding at same
temperature  thermal equilibrium
When not




Heat transfer to surrounding = exothermic
(you feel the heat)  hot metal!
Heat transfer to system = endothermic
(you feel cold)  cold metal!
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Math!

Joules (J) used for
energy quantities


But usually kJ (1000 J)
used
Ye Royal Olde School
used calorie (cal)

cal = amt of heat required
to raise the temperature of
1.00 g of water by 1C


1 kg  m
Joule (J) =
2
s
1 cal = 4.184 J (SI-unit)
But…Calorie (Cal) = 1000
cal

Used in nutrition science
and on food labels
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2
Heat Capacity

Specific heat capacity

Quantity of heat
required to raise the
temp of 1 gram of any
substance by 1 K
J
C=
gK

Molar heat capacity

Quantity of heat
required to raise the
temp of 1 mole of any
substance by 1 K
J
c=
mol  K
4.184 J
specific heat capacity of water =
75.4 J
g  K molar heat capacity of water =
mol  K
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Calculating heat transfer
Q = C  m  T
Q = transferred heat, m = mass of substance, T = temperature change

FYI

Specific heat capacity
of metals is very low


 < 1.000 J/(gK)
What does this tell us
about heat transfer in
metals?
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Let’s do an example

In your backyard, you have a swimming
pool that contains 5.19 x 103 kg of water.
How many kJ are required to raise the
temperature of this water from 7.2 °C to
25.0 °C?
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Example solved
Q = C  m  T = (4.184

J
)  (5.19 x 106 g)  (298.2 K - 280.4 K) = 3.87  108 J = 3.87  105 kJ
gK
Trick:  T in K = T in °C
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Practice

How many kJ are required to raise the
temperature of 25.8 g of quicksilver from
22.5 °C to 28.0 °C? CHg = 0.1395 J/(gK)
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Solution
T  28.0 C-22.5 C  5.5 C
J
kJ
Q  C  m  T  (0.1395
)  25.8g  5.5 C  20.J 
 20. 103 kJ
gK
1000J
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