Energy
Heat
Temperature Changes
 Thermochemistry is the study of
energy changes that occur during
chemical reactions and during
phase changes.
 The energy stored in the chemical
bonds of a substance is called
chemical potential energy.
 That energy can be used to do work
or produce heat.
2


Energy is the ability to do work or produce
heat.
Energy in comes in two forms: kinetic and
potential
◦ Kinetic energy is energy of motion; the more
motion, the more KE
◦ Potential energy is energy of position or
composition; balls on top of a hill have PE;
gasoline has PE.

Law of Conservation of Energy: Energy
cannot be created or destroyed; it changes
forms and can be transferred from one
object to another.
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Heat and Heat Transfer
4

An exothermic process is any process that
releases energy from the system to the
surroundings.
◦ The system is the specific part of the universe
under focus.
◦ The surroundings is everything
else outside the system.
5



An endothermic process is any
process that absorbs energy from
the surroundings into the system.
Ex: A chemical hot pack is an
exothermic process in which heat is
released from the pack (system) to the
person and air (surroundings)
Ex: An ice cube melting on the table is
an endothermic process in which the
cube (system) absorbs energy from the
table and air (surroundings)
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Measuring Heat
reaction
Exothermic reaction, energy (in the
form of heat) is given off &
temperature of surroundings rises.
Energy goes out of the system to the
surroundings.
reaction
Endothermic reaction, energy (in the
form of heat) is taken in &
temperature of surroundings drops.
Energy goes into the system from the
surroundings.
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Temperature vs. heat
• Temperature is
– average kinetic energy per particle
– high KE = high motion = high T
• Touching objects exchange thermal energy
– On average, energy flows one way from things
with high KE (high T) to things with low KE (low
T)
• Temperature predicts energy flow direction
– Energy flows from higher temperature (hotter)
to lower temperature (colder).
– No flow of E = thermal equilibrium = same T
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Heat
• Heat is: The total amount of energy contained in a
certain amount (mass) of a material; symbol “q”
• Heat is: The energy that flows between objects
because of their difference in temperatures. A
change in heat in an object = Δq (Δ = change)
• Heat is different from temperature because heat takes the
mass of the object into account;
• Temperature helps determine the direction of heat flow
(Higher T to Lower T) and is based on the average KE of
the particles of the substance, and does not account for
mass of substance.
• Ex: cup of coffee vs. pot of coffee.
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

The amount of heat required to raise 1.00 gram
of a material 1.00 °C is called specific heat
capacity (c).
It is a unit that indicates how much heat a
substance can store.
◦ Units =
◦ cwater =

𝐽𝑜𝑢𝑙𝑒𝑠
𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
or
𝑔°𝐶
𝑔°𝐶
𝑐𝑎𝑙𝑜𝑟𝑖𝑒
𝐽𝑜𝑢𝑙𝑒𝑠
1.00
or 4.18
𝑔°𝐶
𝑔°𝐶
Since Joules are the official SI unit, we will use
◦ cwater = 4.18
𝐽𝑜𝑢𝑙𝑒𝑠
𝑔°𝐶
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Specific Heat Capacities (C) of Some Substances
(I’ll give you these, you only have to know water)
Substance
Aluminum
Bismuth
Copper
Brass
Gold
Lead
Silver
Tungsten
Zinc
Mercury
Alcohol(ethyl)
Water
Ice (-10 C)
Granite
Glass
c in J/g·ºC
(most common)
0.900
0.123
0.386
0.380
0.126
0.128
0.233
0.134
0.387
0.140
2.4
4.186
2.05
.790
.84
C in cal/g·ºC
0.215
0.0294
0.0923
0.092
0.0301
0.0305
0.0558
0.0321
0.0925
0.033
0.58
1.00
0.49
0.19
0.20
Molar ºC
J/mol·ºC
24.3
25.7
24.5
...
25.6
26.4
24.9
24.8
25.2
28.3
111
75.2
36.9
...
11
...






Specific heat capacity is a measure of the amount
of heat a substance can hold.
𝐽𝑜𝑢𝑙𝑒𝑠
Aluminum has a specific heat of 0.900 𝑔°𝐶 and
𝐽𝑜𝑢𝑙𝑒𝑠
water has a specific heat of 4.18 𝑔°𝐶 .
This means that water can “hold” over 4 times as
much heat as aluminum (for the same mass of
each).
Ex: why soil freezes before water
Ex: why baked potatoes stay hot, but the Al foil
does not.
Calculations tomorrow.
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



Read chapter 10.1-10.4
Complete worksheet 10.1
There is a reading guide posted online if you
need help getting through the chapter’s
readings.
You could do the reading guide as the
chapter outline
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How is the transferred heat calculated?
Dq = mcDT
Dq = change in heat
m = mass
c = specific heat
DT = change in Temp.
Change in Heat = Δq depends on:
•
temperature of an object is changed by DT
• ΔT= Tf -Ti
•
object with a mass of “m”
•
object with a specific heat capacity “c”
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Heat calculations
q = mcΔT
1. You take one liter of water (1000 g) out of the fridge (0°C)
and want to boil (100ºC) it. How much energy will it take to
do that?
2. You do the same with a 1000 g block of copper (0ºC
100ºC). How much energy will it take to do that?
The Joules required are different because water
and copper have different specific heat capacities. Water can
“hold” more heat than copper can.
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

1. How much energy is needed to raise the
𝐽𝑜𝑢𝑙𝑒𝑠
temperature of 150. g of water (c=4.18 𝑔°𝐶 )
by 75°C?
2. A sample of gold (c = 0.126 𝑔°𝐶 ) is
heated from 15°C to 135°C upon the addition
of 14500 J of heat. What is the mass of the
gold?
𝐽𝑜𝑢𝑙𝑒𝑠
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

3. A 75.3 g sample of mercury (c = 0.140
𝐽𝑜𝑢𝑙𝑒𝑠
) at 22.5°C absorbs 3756 J of heat. What
𝑔°𝐶
is the final T of the mercury?
4. How much heat, in kilojoules, is needed to
raise the temperature of 225 g of water
(about an 8oz glass) from 0°C to 100°C?
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
We will be doing an experiment in which we
determine the specific heat of a piece of
metal.
◦ Take a piece of metal, determine its mass and heat
it in boiling water. This will give you its initial T.
◦ Add the hot metal to a determined mass of water at
room T.
◦ The heat from the metal will transfer to the water.
◦ You may assume that the heat gained by the water
is equal to the heat lost by the metal.
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
A 135 g sample of tin is heated to 94.6°C and
placed into 75.2 g of water (c = 4.18
𝐽𝑜𝑢𝑙𝑒𝑠
𝑔°𝐶
)
initially at 22.5°C. After the mixture reaches
thermal equilibrium, the temperature is
31.9°C. Determine the specific heat of the tin.
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
A sample of brass (c = 0.380
) at a
temperature of 93.5°C is added to 81.2 g of
water at 22.5°C. The final temperature of the
mixture is 37.2°C. What is the mass of the
brass sample?
𝐽𝑜𝑢𝑙𝑒𝑠
𝑔°𝐶
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
A 135 g sample of zinc (c = 0.387
) at a
temperature of 97.2°C is added to 37.5 g of
water at 25.3°C. What is the final temperature
of the mixture?
𝐽𝑜𝑢𝑙𝑒𝑠
𝑔°𝐶
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