1 CHAPTER 4. Energy Energy is the capacity to do work. Potential energy is stored energy. Kinetic...
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Transcript of 1 CHAPTER 4. Energy Energy is the capacity to do work. Potential energy is stored energy. Kinetic...
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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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EnergyThe 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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Cal/g cal/g
Protein 4 4,000
Carbohydrate 4 4,000
Fat 9 9,000
Focus on The Human BodyEnergy 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.
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Identify the original quantity and the desired quantity.3 g protein23 g carbohydrates ? Caloriginal quantities desired quantity
Step [1]
Step [2]
Write out conversion factors.4 Cal 4 Cal .1 g protein 1 g carbohydrate
Focus on The Human BodyEnergy 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.
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Multiply the original quantity by the conversion factor for both protein and carbohydrates and add up the results.
Total Cal = Cal due to protein + Cal due to carbohydrate
= 3 g × 4 Cal + 23 g × 4 Cal . 1 g protein 1 g carbohydrate
Step [3]
grams cancel
Total Cal = 12 Cal + 92 Cal= 104 Cal, rounded to 100 Cal
Focus on The Human BodyEnergy and Nutrition
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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
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Intermolecular Forces, Boiling Point,and Melting Point
Intermolecular forces are the attractive forces thatexist between molecules.
In order of increasing strength, these are:
• London dispersion forces
• Dipole–dipole interactions
• Hydrogen bonding
The strength of the intermolecular forces determineswhether a compound has a high or low melting pointand boiling point, and thus whether it is a solid,liquid, or gas at a given temperature.
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Intermolecular ForcesLondon Dispersion Forces
London dispersion forces are very weak interactionsdue to the momentary changes in electron densityin a molecule.
• The change in electron density creates a temporary dipole.
• All covalent compounds exhibit London dispersion forces.
• The weak interaction between these temporary dipoles constitutes London dispersion forces.
• The larger the molecule, the larger the attractive force, and the stronger the intermolecular forces.
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Intermolecular ForcesLondon Dispersion Forces
More e− densityin one region creates a partialnegative charge (δ−).
Less e− densityin one regioncreates a partialpositive charge (δ+).
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Intermolecular ForcesDipole–Dipole Interactions
Dipole–dipole interactions are the attractive forcesbetween the permanent dipoles of two polar molecules.
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Intermolecular ForcesHydrogen Bonding
Hydrogen bonding occurs when a hydrogen atombonded 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 typesof intermolecular forces.
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Intermolecular ForcesHydrogen Bonding in DNA
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Intermolecular Forces
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Intermolecular ForcesBoiling 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 ForcesBoiling Point and Melting Point
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Intermolecular ForcesBoiling 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.
• The larger the specific heat of a substance, the less its temperature will change when it absorbs a particular amount of heat energy.
specific heat = heat = cal (or J) mass x ΔT g • °C
specific heat = heat = cal (or J) mass x ΔT g • °C
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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
Identify the known and desired quantities.mass = 1,600g ? CalT1 = 25°C desired quantityT2 = 100°CSp. heat of water = 1.00 cal/g°C known quantities
Step [1]
Determine the change in temperature.ΔT = T2 - T1 = 100°C - 25°C = 75°C
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?
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Specific Heat
Step [2]
Write the equation.
heat = mass x ΔT x specific heat
cal = g x °C x cal g•°C
Step [3]
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 ChangesConverting 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 ChangesConverting 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 ChangesConverting a Solid to a Gas
solid CO2gaseous 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.
The plateau B C occurs at the melting point, while the plateau D E occurs at the boiling point.
A
B
C
D
E
melting
boiling
solid
liquid
gas
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condensation
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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.
The plateau W X occurs at the boiling point, while the plateau Y Z occurs at the freezing point.
V
XW
freezingsolid
liquid
gas
Y
Z
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Heating and Cooling Curves
Identify the original and desired quantities.mass = 25.0g ? CalT1 = 25°C desired quantityT2 = 100°Cknown quantities
Step [1]
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.
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.
specific heat heat of vaporization
Conversion factors are needed for both the specificheat and the heat of vaporization.
1.00 cal 1 g • 1 °C
1 g • 1 °C 1.00 cal
or 540 cal 1 g
1 g 540 cal
or
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.
cal = 25.0 g x 75°C x 1,000 cal = 1,900 cal 1 g• 1°C
Answer = 16,000 cal
heat = mass x ΔT x specific heat
Calculate the heat needed to change the temperature of water.
Calculate the heat needed for the phase change.
cal = 25.0 g x 540 cal = 14,000 cal 1 g
Add the two together: 1,900 cal + 14,000 cal =