Chapter 6
Energy
Thermodynamics
1
Calorimetry
2
Big Ideas



The laws of thermodynamics describe the
essential role of energy and explain and
predict the direction of changes in matter
Any bond or intermolecular attraction that
can be formed can be broken. These two
processes are in a dynamic competition,
sensitive to initial conditions and external
perturbations
Energy is neither created nor destroyed, but
only transformed from one form to another.
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1
Essential Questions


Energy is transferred between systems either
through heat transfer or through one system
doing work on the other system.
When two systems are in contact with each
other and are otherwise isolated, the energy
that comes out of one system is equal to the
energy that goes into the other system. The
combined energy of the two systems remains
fixed. Energy transfer can occur through
either heat exchange or work.
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Calorimetry
Measuring heat.
Use a calorimeter.
 Two kinds


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Calorimetry

Measuring heat.
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2
Specific Heat

DEF The amount of heat (in J, cal)
required to raise the temperature of 1
gram of substance 1̊C at constant pressure
heat = specific heat x mass x change
∆

Specific heat constant of H2O
1
4.184
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Materials and Their Specific Heat
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Materials and Their Specific Heats


Old Book, page 239
Table 6.1 left column
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3
Specific Heat

Why is it a good thing that water has a
very high specific heat constant value?

Why are pots and pans manufactured
out of metals such as “cast iron” or
copper?
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Calorimetry
How will we use calorimetry?
 Determine the amount of heat associated
with chemical or physical changes.

– Changing the temperature of an amount of
substance i.e. Making hot cup of tea, boiling
water to make spaghetti
– Placing a hot object in cool water and measuring
the heat transferred
– Dissolving a salt in water and measuring the
temperature change i.e. Hot/cold packs
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Example Problem
Calculate the heat absorbed by 15.0 g of
water to raise the temperature from
20.0 to 50.0 .

∆

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4
First Law of Thermodynamics

“Law of Conservation of Energy”
– Energy cannot be created or destroyed, only
changed

Heat
– Flow of energy from High Temperature to Low
Temperature
– Quantity of heat gained by one substance
equals the Quantity of heat lost by another
substance
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First Law of Thermodynamics
“Law of Conservation of Energy”
 Heat Gained = - Heat Lost

DEMO: Placing hot water into cold water
What happens to the temperature of the
hot water? Why?
What happens to the temperature of the
cold water? Why?
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First Law of Thermodynamics

“Law of Conservation of Energy”
Heat Gained = - Heat Lost
DEMO: Placing hot metal into cold water
What happens to the temperature of the metal?
Why?
What happens to the temperature of the water?
Why?
What is the final temp of water? Of metal?
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5
First Law of Thermodynamics
“Law of Conservation of Energy”
 Heat Gained
= - Heat Lost
Heat gained = - Heat Lost
by water
by hot metal

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Example

A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and
the copper is 21.8ºC. What is the specific
heat of copper?
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Calorimetry


How will we use calorimetry?
Determine the amount of heat associated
with chemical or physical changes.
– Dissolving a salt in water and measuring the
temperature change i.e. Hot/cold packs
– Things to remember:
• The Dissociation of the compound is the SYSTEM
• The water is the surroundings
– The thermometer measures the temperature of the
water!!!!!!!!
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6
Calorimetry

Calorimeter
– DEF a device used to measure the heat
absorbed or evolved during a chemical or
physical change
– Dissolving a salt in water is a physical change
– What happens to a “salt” (aka a strong
electrolyte) when its dissolved in water?
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Example

The mass of a sample of potassium bromide in a beaker is
24.89 g. (The beaker itself weighs 23.14 g.) The potassium
bromide was then dissolved in 100.00 g of distilled water,
raising the temperature from 20.05˚C to 25.40˚C. Assume
all of the heat was absorbed by the water, calculate the
amount of energy associated with this process.
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Example

Using the calculations from the previous problem,
calculate the amount of energy associated with
dissolving 1.0 mole of potassium bromide in 100.00 g
of water.

Write a thermochemical equation that represents this
process.
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7
Heat Capacity C



DEF quantity of heat needed to raise the
temperature of a sample of substance or a
calorimeter 1̊C or 1 Kelvin, K
– Units: J/ ̊C
Sometimes called “Molar Heat Capacity” if
talking about a specific amount of substance
– Very similar to specific heat constant
Units: J/mol ̊C
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Molar Heat Capacity C
Old Book, pg 239
 Table 6.1, right column
 “Molar Heat Capacity”
 Units: J/mol ̊C

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Heat Capacity C

How do we use it?

Formula: q = C ΔT




q, heat
ΔT, change in temp
C, Heat capacity or
molar heat capacity
units: J, KJ, cal
units: ̊C
units: KJ/ ̊C
units: J/ mol ̊C
WATCH UNITS!!!!
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8
Example

It takes 6.70 J of heat energy to raise the
temperature of a piece of iron one
degree celsius. What is the heat
capacity of that piece of iron?

How much heat energy is required to
raise the temperature of that piece of
iron from 25.0 C to 35.0 C?
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Example


Heat gained = - Heat Lost problem
A 18.3 g piece of aluminum (which has a
molar heat capacity of 24.03 J/°C·mol) is
heated to 82.4°C and dropped into a
calorimeter containing water (specific heat
capacity of water is 4.18 J/g°C) initially at
22.3°C. The final temperature of the water is
25.8°C. Calculate the mass of water in the
calorimeter.
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Heat Capacity of a Calorimeter

How does a calorimeter work
– heat (q) released by a reaction or process is
absorbed by the calorimeter and any substances in
the calorimeter
Heat lost by the reaction or process = heat gained by the calorimeter
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9
Heat Capacity of a Calorimeter

How do we calculate the “Heat Gained
By the Calorimeter”?
qcal = Ccal ΔT
– qcal , heat gained by the calorimeter
• Units: KJ, J,cal
–
Ccal, heat capacity of the calorimeter
• Constant
• Units: KJ/ ̊C
– ΔT, temp change of the calorimeter
• Units: ̊C
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Example
qcal = Ccal ΔT
Suppose 0.562 g of graphite is placed in a
calorimeter with excess oxygen at 25.00C and 1 atm
pressure. During the reaction, the calorimeter
temperature rises from 25.00 C to 25.89 C. The heat
capacity of the calorimeter was determined to be
20.7 KJ/C. What is the heat of reaction under these
conditions?
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