1. 1 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.

2. 2 Energy The Units of Energy A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 o 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 kJ1,000 cal = 1 kcal 1 kcal = 4.184 kJ

3. 3 Cal/gcal/g Protein44,000 Carbohydrate44,000 Fat99,000 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.

4. 4 Identify the original quantity and the desired quantity. 3 g protein 23 g carbohydrates ? Cal original 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 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.

5. 5 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 Body Energy and Nutrition

6. 6 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.

7. The Three States of Matter

8. 8 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, liquid, or gas at a given temperature.

9. 9 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. 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.

10. 10 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+).

11. 11 Intermolecular Forces Dipole–Dipole Interactions Dipole–dipole interactions are the attractive forces between the permanent dipoles of two polar molecules.

12. 12 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.

13. 13 Intermolecular Forces Hydrogen Bonding in DNA

14. 14 Intermolecular Forces

15. 15 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.

16. 16 Intermolecular Forces Boiling Point and Melting Point

17. 17 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.

18. 18 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

19. 19 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.

20. 20 Specific Heat Identify the known and desired quantities. mass = 1,600g ? Cal T 1 = 25°C desired quantity T 2 = 100°C Sp. heat of water = 1.00 cal/g°C known quantities Step [1] Determine the change in temperature. ΔT = T 2 - T 1 = 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?

21. 21 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 10 5 cal

22. 22 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.

23. 23 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.

24. 24 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

25. 25 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.

26. 26 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

27. 27 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.

28. 28 Energy and Phase Changes Converting a Solid to a Gas solid CO 2 gaseous CO 2

29. 29 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

30. condensation 30 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 X W freezing solid liquid gas Y Z

31. 31 Heating and Cooling Curves Identify the original and desired quantities. mass = 25.0g ? Cal T 1 = 25°C desired quantity T 2 = 100°C known 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 = T 2 - T 1 = 100°C - 25°C = 75°C

32. 32 Heating and Cooling Curves Step [2] Write out the conversion factors. specific heatheat of vaporization Conversion factors are needed for both the specific heat and the heat of vaporization. 1.00 cal 1 g 1 °C 1.00 cal or 540 cal 1 g 540 cal or Choose the conversion factors with the unwanted units– (g °C) and g– in the denominator..

33. 33 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 =