Transcript CHAPTER 4. Energy

CHAPTER 4.
Energy
• Energy is the capacity to do work.
• Potential energy is stored energy.
• Kinetic energy is the energy of motion.
• The law of conservation of energy states that the
total energy in a system does not change. Energy
cannot be created or destroyed.
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Energy
The Units of Energy
• A calorie (cal) is the amount of energy needed to
raise the temperature of 1 g of water by 1o C.
• A joule (J) is another unit of energy.
1 cal = 4.184 J
• Both joules and calories can be reported in the
larger units kilojoules (kJ) and kilocalories (kcal).
1,000 J = 1 kJ
1,000 cal = 1 kcal
1 kcal = 4.184 kJ
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Focus on The Human Body
Energy and Nutrition
• The amount of stored energy in food is measured
using nutritional Calories (upper case C), where
1 Cal = 1,000 cal.
• Upon metabolism, proteins, carbohydrates, and fat
each release a predictable amount of energy, the
caloric value of the substance.
Cal/g
cal/g
Protein
4
4,000
Carbohydrate
4
4,000
Fat
9
9,000
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Focus on The Human Body
Energy and Nutrition
Sample Problem 4.2
If a baked potato contains 3 g of protein, a trace
of fat, and 23 g of carbohydrates, estimate its
number of Calories.
Step [1] Identify the original quantity and the
desired quantity.
3 g protein
23 g carbohydrates
? Cal
original quantities
desired quantity
Step [2]
Write out conversion factors.
4 Cal
4 Cal
.
1 g protein
1 g carbohydrate
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Focus on The Human Body
Energy and Nutrition
Step [3]
Total Cal
Multiply the original quantity by the
conversion factor for both protein and
carbohydrates and add up the results.
= Cal due to protein + Cal due to carbohydrate
=3g×
4 Cal
+
1 g protein
23 g ×
4 Cal
.
1 g carbohydrate
grams cancel
Total Cal
=
12 Cal
+
92 Cal
= 104 Cal, rounded to 100 Cal
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The Three States of Matter
Whether a substance exists as a gas, liquid, or solid
depends on the balance between the kinetic energy
of its particles and the strength of the interactions
between the particles.
Gas: kinetic energy is high and particles are far apart.
The attractive forces between molecules are negligible
allowing them to move freely.
Liquid: attractive forces hold the molecules much
more closely together. The distance between
molecules and the kinetic energy is much less.
Solid: attractive forces between molecules are even
stronger. The distance between particles is small and
there is little freedom of motion.
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The Three States of Matter
Intermolecular Forces, Boiling Point,
and Melting Point
Intermolecular forces are the attractive forces that
exist between molecules.
In order of increasing strength, these are:
• London dispersion forces
• Dipole–dipole interactions
• Hydrogen bonding
The strength of the intermolecular forces determines
whether a compound has a high or low melting point
and boiling point, and thus whether it is a solid,
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liquid, or gas at a given temperature.
Intermolecular Forces
London Dispersion Forces
London dispersion forces are very weak interactions
due to the momentary changes in electron density
in a molecule.
• The change in electron density creates a
temporary dipole.
• The weak interaction between these temporary
dipoles constitutes London dispersion forces.
• All covalent compounds exhibit London
dispersion forces.
• The larger the molecule, the larger the attractive
force, and the stronger the intermolecular forces.
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Intermolecular Forces
London Dispersion Forces
More e− density
in one region
creates a partial
negative charge (δ−).
Less e− density
in one region
creates a partial
positive charge (δ+).
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Intermolecular Forces
Dipole–Dipole Interactions
Dipole–dipole interactions are the attractive forces
between the permanent dipoles of two polar
molecules.
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Intermolecular Forces
Hydrogen Bonding
Hydrogen bonding occurs when a hydrogen atom
bonded to O, N, or F is electrostatically attracted
to an O, N, or F atom in another molecule.
Hydrogen bonds are the strongest of the three types
of intermolecular forces.
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Intermolecular Forces
Hydrogen Bonding in DNA
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Intermolecular Forces
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Intermolecular Forces
Boiling Point and Melting Point
• The boiling point is the temperature at which a
liquid is converted to the gas phase.
• The melting point is the temperature at which a
solid is converted to the liquid phase.
• The stronger the intermolecular forces, the higher
the boiling point and melting point.
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Intermolecular Forces
Boiling Point and Melting Point
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Intermolecular Forces
Boiling Point and Melting Point
• Both propane and butane have London
dispersion forces and nonpolar bonds.
• In this case, the larger molecule will have stronger
attractive forces.
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Specific Heat
• The specific heat is the amount of heat energy (in
calories or joules) needed to raise the temperature
of 1 g of a substance by 1°C.
specific heat =
heat
mass x ΔT
= cal (or J)
g • °C
• The larger the specific heat of a substance, the
less its temperature will change when it absorbs a
particular amount of heat energy.
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Specific Heat
• The specific heat of water is 1.00 cal/(g∙°C), meaning
that 1.00 cal of heat must be added to increase the
temperature of 1.00 g of water by 1.00 °C.
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Specific Heat
Sample Problem 4.6
How many calories are needed to heat a pot of
1,600 g of water from 25°C to 100.°C?
Step [1]
Identify the known and desired quantities.
mass = 1,600g
? Cal
T1 = 25°C
desired quantity
T2 = 100°C
Sp. heat of water = 1.00 cal/g°C
known quantities
Determine the change in temperature.
ΔT = T2 - T1 = 100°C - 25°C = 75°C
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Specific Heat
Step [2]
Write the equation.
heat = mass x ΔT x specific heat
cal =
Step [3]
g
x °C x
cal
g•°C
Solve the equation.
cal = 1,600 g x 75°C x 1,000 cal
1 g• 1 °C
Answer =
1.2 x 105 cal
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Energy and Phase Changes
• When energy is absorbed, a process is said to be
endothermic.
• When energy is released, a process is said to be
exothermic.
• In a phase change, the physical state of a
substance is altered without changing its
composition.
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Converting a Solid to a Liquid
• Converting a solid to a liquid is called melting.
• Melting is endothermic—it absorbs heat from
the surroundings.
• Freezing converts a liquid to a solid.
• Freezing is exothermic—it gives off heat to
the surroundings.
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Energy and Phase Changes
Converting a Solid to a Liquid
solid water
liquid water
The amount of energy needed to melt 1 gram of a
substance is called its heat of fusion.
The heat of fusion of water = 79.7 cal/g
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Converting a Liquid to a Gas
• Vaporization is the conversion of liquids into the
gas phase.
• Vaporization is endothermic—it absorbs heat from
the surroundings.
• Condensation is the conversion of gases into the
liquid phase.
• Condensation is exothermic—it gives off heat to
the surroundings.
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Energy and Phase Changes
Converting a Liquid to a Gas
liquid water
gaseous water
The amount of energy needed to vaporize 1 gram of
a substance is called its heat of vaporization.
The heat of vaporization of water = 540 cal/g
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Converting a Solid to a Gas
• Sublimation is the conversion of solids directly
into the gas phase.
• Sublimation is endothermic—it absorbs heat from
the surroundings.
• Deposition is the conversion of gases into the
solid phase.
• Deposition is exothermic—it gives off heat to
the surroundings.
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Energy and Phase Changes
Converting a Solid to a Gas
solid CO2
gaseous CO2
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Heating and Cooling Curves
A heating curve shows how the temperature of a
substance (plotted on the vertical axis) changes as
heat is added.
D
gas
boiling
E
liquid
B melting
C
solid
A
The plateau B  C occurs at the melting point, while
the plateau D  E occurs at the boiling point.
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Heating and Cooling Curves
A cooling curve illustrates how the temperature of
a substance (plotted on the vertical axis)
changes as heat is removed.
V
gas
X
W condensation
liquid
Z
Y
freezing
solid
The plateau W  X occurs at the boiling point, while
the plateau Y  Z occurs at the freezing point.
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Heating and Cooling Curves
Sample Problem 4.11
How much energy is required to heat 25.0 g of water
from 25°C to a gas at its boiling point of 100.°C? The
specific heat of water is 1.00 cal/(g∙°C), and the heat
of vaporization of water is 540 cal/g.
Step [1]
Identify the original and desired quantities.
mass = 25.0g
? Cal
T1 = 25°C
desired quantity
T2 = 100°C
known quantities
Determine the change in temperature.
ΔT = T2 - T1 = 100°C - 25°C = 75°C
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Heating and Cooling Curves
Step [2]
Write out the conversion factors.
Conversion factors are needed for both the specific
heat and the heat of vaporization.
specific heat
1.00 cal or
1 g • 1 °C
1 g • 1 °C
1.00 cal
heat of vaporization
540 cal or
1g
1g
540 cal
Choose the conversion factors with
the unwanted units– (g • °C) and g–
in the denominator.
.
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Heating and Cooling Curves
Step [3]
Solve the problem.
Calculate the heat needed to change the
temperature of water.
heat = mass x ΔT x specific heat
cal = 25.0 g x 75°C x 1,000 cal = 1,900 cal
1 g• 1°C
Calculate the heat needed for the phase change.
cal = 25.0 g x 540 cal = 14,000 cal
1g
Add the two together: 1,900 cal + 14,000 cal =
Answer =
16,000 cal
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