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


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


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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.

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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• 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

condensation

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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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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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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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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 =

1 chapter 4. energy energy is the capacity to do work. potential energy is stored energy. kinetic...

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