Transcript Heat and Energy - Gordon State College

Chapter 6
Energy and States of Matter
6.1 Energy
1
Energy
Energy

Makes objects move.

Makes things stop.

Is needed to “do work.”
2
Work
Work is done when
 You go up stairs.
 You play soccer.
 You lift a bag of groceries.
 You ride a bicycle.
 You breathe.
 Your heart pumps blood.
 Water goes over a dam.
3
Potential Energy



Potential energy is
energy that is stored
for use at a later
time. Examples are:
Water behind a dam
A compressed spring
Chemical bonds in
gasoline, coal, or food
4
Kinetic Energy
Kinetic energy is the energy of motion. Examples
are:
 Hammering a nail
 Water flowing over a dam
 Working out
 Burning gasoline
5
Learning Check
Identify the energy as 1) potential or 2) kinetic
A. Roller blading.
B. A peanut butter and jelly sandwich.
C. Mowing the lawn.
D. Gasoline in the gas tank.
6
Solution
Identify the energy as 1) potential or 2) kinetic
A. Roller blading. (2 kinetic)
B. A peanut butter and jelly sandwich.
(1 potential)
C. Mowing the lawn. (2 kinetic)
D. Gasoline in the gas tank. (1 potential)
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Forms of Energy
Energy has many forms:
 Mechanical
 Electrical
 Thermal (heat)
 Chemical
 Solar (light)
 Nuclear
8
Heat
 Heat energy flows from a warmer object to
a colder object.
 The colder object gains kinetic energy when
it is heated.
 During heat flow, the loss of heat by a
warmer object is equal to the heat gained by
the colder object.
9
Some Equalities for Heat
 Heat is measured in calories or joules.
1 kilocalorie (kcal) = 1000 calories (cal)
1 calorie = 4.18 Joules (J)
1 kJ = 1000 J
10
6.2 Energy in Chemical
Reactions
11
Reaction Conditions



A chemical reaction occurs when the reacting
molecules collide.
Collisions between molecules must have
sufficient energy to break the bonds in the
reactants.
Once the bonds between atoms of the
reactants are broken, new bonds can form to
give the product.
12
Chemical Reactions

In the reaction H2 + I2
2 HI, the bonds of
H2 and I2 must break, and bonds for HI must
form.
H2 + I2
collision
bonds break
2HI
13
Activation Energy


The activation energy
is the minimum
energy needed for a
reaction to take place.
When a collision has
the energy that is
equal to or greater
than the activation
energy, reaction can
occur.
14
Exothermic Reactions


The heat of reaction is
the difference in the
energy of the
reactants and the
products.
An exothermic
reaction releases heat
because the energy of
the products is less
that the reactants.
15
Endothermic Reactions

In an
endothermic
reaction, heat is
absorbed
because the
energy of the
products is
greater that that
of the reactants.
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Learning Check
Identify each reaction as
1) exothermic or 2) endothermic
A. N2 + 3H2
2NH3 + 22 kcal
B. CaCO3 + 133 kcal
CaO + CO2
C. 2SO2 + O2
2SO3 + heat
17
Solution
Identify each reaction as
1) exothermic or
2) endothermic
1 A. N2 + 3H2
2NH3 + 22 kcal
2 B. CaCO3 + 133 kcal
CaO + CO2
1 C. 2SO2 + O2
2SO3 + heat
18
6.3 Specific Heat
Specific heat is
the amount of
heat (calories or
Joules) that
raises the
temperature of 1
g of a substance
by 1°C.
19
Learning Check
A. A substance with a large specific heat
1) heats up quickly
2) heats up slowly
B. When ocean water cools, the surrounding air
1) cools
2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
1) high specific heat
2) low specific heat
20
Solution
A. A substance with a large specific heat
2) heats up slowly
B. When ocean water cools, the surrounding air
2) warms
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
2) low specific heat
21
Learning Check
When 200 g of water are heated, the water
temperature rises from 10°C to 18°C.
200 g
400 g
If 400 g of water at 10°C are heated with the
same amount of heat, the final temperature
would be
1) 10 °C
2) 14°C
3) 18°C
22
Solution
When 200 g of water are heated, the water
temperature rises from 10°C to 18°C.
200 g
400 g
If 400 g of water at 10°C are heated with the
same amount of heat, the final temperature
would be
2) 14°C
23
Calculation with Specific Heat
 To calculate the amount of heat lost or gained
by a substance, we need the
grams of substance,
temperature change T, and the
specific heat of the substance.
 Heat = g x °C x cal (or J) = cal ( or J)
g °C
24
Sample Calculation for Heat
A hot-water bottle contains 750 g of water at
65°C. If the water cools to body temperature
(37°C), how many calories of heat could be
transferred to sore muscles?
The temperature change is 65°C - 37°C = 28°C.
heat (cal) =
g x T x Sp. Ht. (H2O)
750 g x 28°C x 1.00 cal
g°C
= 21 000 cal
25
Learning Check
How many kcal are needed to raise the
temperature of 120 g of water from 15°C to
75°C?
1) 1.8 kcal
2) 7.2 kcal
3) 9.0 kcal
26
Solution
How many kcal are needed to raise the
temperature of 120 g of water from 15°C to
75°C?
2) 7.2 kcal
75°C - 15°C = 60 °C
120 g x (60°C) x 1.00 cal x 1 kcal
g °C
1000 cal
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6.4 Energy and Nutrition


On nutrition and food
labels, the nutritional
Calorie, written with a
capital C, is used.
1 Cal is actually 1000
calories.
1 Calorie = 1 kcal
1 Cal = 1000 cal
28
Caloric Food Values

The caloric values for foods indicate the
number of kcal provided by 1 g of each
type of food.
29
Calories in Some Foods
30
Energy Requirements

The amount of energy needed each day
depends on age, sex, and physical activity.
31
Loss and Gain of Weight

If food intake
exceeds energy
use, a person
gains weight. If
food intake is
less than energy
use, a person
loses weight.
32
Learning Check
A cup of whole milk contains 12 g of
carbohydrates, 9.0 g of fat, and 9.0 g of
protein. How many kcal (Cal) does a
cup of milk contain?
1) 48 kcal
2) 81 kcal
3) 165 kcal
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Solution
3) 165 kcal
12 g carb x 4 kcal/g
=
48 kcal
x 9 kcal/g =
81 kcal
9.0 g protein x 4 kcal/g =
36 kcal
Total kcal =
165 kcal
9.0 g fat
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6.5 States of Matter
35
Solids
Solids have
 A definite shape.
 A definite volume.
 Particles that are close together in a fixed
arrangement.
 Particles that move very slowly.
36
Liquids
Liquids have
 An indefinite shape, but
a definite volume.
 The same shape as their
container.
 Particles that are close
together, but mobile.
 Particles that move
slowly.
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Gases
Gases have
 An indefinite shape.
 An indefinite volume.
 The same shape and volume
as their container.
 Particles that are far apart.
 Particles that move fast.
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Summary of the States of Matter
39
Learning Check
Identify each as: 1) solid 2) liquid or 3) gas.
___ A. It has a definite volume, but takes the
shape of the container.
__ B. Its particles are moving rapidly.
__C. It fills the volume of a container.
__D. It has particles in a fixed arrangement.
__E. It has particles close together that are
mobile.
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Solution
Identify each as: 1) solid 2) liquid or 3) gas.
2 A. It has a definite volume, but takes the
shape of the container.
3 B. Its particles are moving rapidly.
3 C. It fills the volume of a container.
1 D. It has particles in a fixed arrangement.
2 E. It has particles close together that are
mobile.
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Attractive Forces between
Particles

In ionic compounds, ionic bonds are strong
attractive forces that hold positive and
negative ions together.
42
Attractive Forces between
Particles


In covalent compounds, polar molecules exert
attractive forces called dipole-dipole
attractions.
Hydrogen bonds are strong dipole attractions
between hydrogen atoms and atoms of F, O,
or N, which are very electronegative.
43
Attractive Forces between
Particles


Nonpolar molecules form liquids or solids
through weak attractions called dispersion
forces.
Dispersion forces are caused by temporary
dipoles that develop when electrons are not
distributed equally.
44
Melting Points and Attractive
Forces



Ionic compounds require large amounts of
energy to break apart ionic bonds. Thus,
they have high melting points.
Hydrogen bonds are the strongest type of
dipole-dipole attractions. They require more
energy to break than other dipole
attractions.
Dispersion forces are weak interactions and
very little energy is needed to change state.
45
Melting Points and Attractive
Forces of Some Typical Substances
46
Learning Check
Identify the type of attractive forces for each:
1) ionic
2) dipole-dipole
3) hydrogen bonds
4) dispersion
A. NCl3
B. H2O
C. Br-Br
D. KCl
E. NH3
47
Solution
Identify the type of attractive forces for each:
1) ionic
2) dipole-dipole
3) hydrogen bonds
4) dispersion
2
A. NCl3
3
B. H2O
4
C. Br-Br
1
D. KCl
3
E. NH3
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6.6 Changes of State
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Melting and Freezing



A substance is melting
while it changes from a
solid to a liquid.
A substance is freezing
while it changes from a
liquid to a solid.
The freezing (melting)
point of water is 0°C.
50
Calculations Using Heat of Fusion



The heat of fusion is the amount of heat
released when 1 gram of liquid freezes at its
freezing point.
The heat of fusion is the amount of heat
needed to melt 1 gram of a solid at its melting
point.
For water the heat of fusion (at 0°C) is
80. cal
1 g water
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Calculation Using Heat of Fusion

The heat involved in the freezing (or melting)
a specific mass of water (or ice) is calculated
using the heat of fusion.
Heat = g water x
80. cal
g water
Problem: How much heat in calories is needed
to melt 15.0 g of water?
15.0 g water x
80. cal
= 1200 cal
1 g water
52
Learning Check
A. How many calories are needed to melt 5.0 g
of ice of 0°C?
1) 80. cal
2) 400 cal
3) 0 cal
B. How many calories are released when 25 g
of water at 0°C freezes?
1) 80. cal
2) 0 cal
3) 2000 cal
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Solution
A. How many calories are needed to melt 5.0 g
of ice of 0°C?
2) 400 cal
5.0 g x 80. cal
1g
B. How many calories are released when 25 g
of water at 0°C freezes?
3) 2000 cal
25 g x 80. cal
1g
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Boiling & Condensation


Water evaporates
when molecules
on the surface
gain enough
energy to form a
gas.
At boiling, all the
water molecules
acquire enough
energy to form a
gas.
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Heat of Vaporization
The heat of vaporization
 Is the amount of heat needed to change 1 g of
liquid to gas at the boiling point.
 Is the amount of heat released when 1 g of a
gas changes to liquid at the boiling point.
Boiling (Condensing) Point of Water = 100°C
Heat of Vaporization (water) = 540 cal
1 g water
56
Learning Check
How many kilocalories (kcal) are released
when 50.0 g of steam in a volcano condenses
at 100°C?
1) 27 kcal
2) 540 kcal
3) 27 000 kcal
57
Solution
How many kilocalories (kcal) are released
when 50.0 g of steam in a volcano condenses at
100°C?
1) 27 kcal
50.0 g steam x 540 cal
x 1 kcal = 27 kcal
1 g steam 1000 cal
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Heating and Cooling Curves
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Heating Curve



A heating curve
illustrates the changes
of state as a solid is
heated.
Sloped lines indicate an
increase in
temperature.
Plateaus (flat lines)
indicate a change of
state.
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Learning Check
A. A flat line on a heating curve represents
1) a temperature change
2) a constant temperature
3) a change of state
B. A sloped line on a heating curve represents
1) a temperature change
2) a constant temperature
3) a change of state
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Solution
A. A flat line on a heating curve represents
2) a constant temperature
3) a change of state
B. A sloped line on a heating curve represents
1) a temperature change
62
Cooling Curve
 A cooling curve illustrates
the changes of state as a
gas is cooled.
 Sloped lines indicate a
decrease in temperature.
 This cooling curve for
water begins at 140°C
and ends at -30°C.
63
Learning Check
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
1) 0°C
2) 50°C
3) 100°C
B. At a temperature of 0°C, water
1) freezes
2) melts
3) changes to a gas
C. At 40 °C, water is a
1) solid
2) liquid
3) gas
D. When water freezes, heat is
1) removed 2) added
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Solution
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
3) 100°C
B. At a temperature of 0°C, water
1) freezes
C. At 40 °C, water is a
2) liquid
D. When water freezes, heat is
1) removed
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Combined Heat Calculations
To reduce a fever, an infant is packed in 250 g
of ice. If the ice at 0°C melts and warms to
body temperature (37.0°C), how many
calories are removed from the body?
Step 1: Diagram the changes.
37°C
T = 37.0°C - 0°C =
37.0°C
0°C
S
L
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Combined Heat Calculations
(continued)
Step 2: Calculate the heat to melt ice (fusion)
250 g ice x 80. cal
= 20 000 cal
1 g ice
Step 3: Calculate the heat to warm the water from
0°C to 37.0°C
250 g x 37.0°C x 1.00 cal = 9250 cal
g °C
Total: Step 2 + Step 3
= 29 250 cal
Final answer
= 29 000 cal
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Learning Check
150 g of steam at 100°C is released from a
boiler. How many kilocalories are lost when
the steam condenses and cools to 15°C?
1) 81 kcal
2) 13 kcal
3) 94 kcal
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Solution
3) 94 kcal
Condense: 150 g x 540 cal x 1 kcal = 81 kcal
1g
1000 cal
Cool: 150 g x 85°C x 1 cal x 1 kcal = 13 kcal
g °C
1000 cal
Total: 81 kcal + 13 kcal = 94 kcal
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