1. 1 Announcements & Agenda (01/15/07) Office Hours (SC 3063) Permanent Change: Thurs 9-11am replaces Mon 3-5pm Permanent Change: Thurs 9-11am replaces Mon 3-5pm This Week: must cancel Wed 3-5pm OH (will leave @ 2pm) This Week: must cancel Wed 3-5pm OH (will leave @ 2pm) Quiz 1 (the CD quizzes) will be accepted until Wed First in-class quiz (Quiz 2) will be Wed. (Ch 1 & 2) See overhead for your clicker number You should currently be reading Ch 3 You should currently be reading Ch 3 Today: Heat and temperature (2.3, 2.4) Heat and temperature (2.3, 2.4) Phase changes (2.5, 2.6) Phase changes (2.5, 2.6)

2. 2. l 2.... l.... l 3.... l.... l 4.. cm The markings on the meter stick at the end of the orange line are read as The markings on the meter stick at the end of the orange line are read as the first digit 2 plus the second digit 2.7 The last digit is obtained by estimating. The last digit is obtained by estimating. The end of the line might be estimated between 2.7– 2.8 as half-way (0.5) or a little more (0.6), which gives a reported length of 2.75 cm or 2.76 cm. The end of the line might be estimated between 2.7– 2.8 as half-way (0.5) or a little more (0.6), which gives a reported length of 2.75 cm or 2.76 cm. Last time: Making Measurements Measured #s always need units!

3. 3 LAST TIME: Energy Energy – the ability to do work 1.Kinetic Energy: energy of motion 2.Potential Energy: stored energy IMPORTANT EXAMPLES: IMPORTANT EXAMPLES: –ENERGY OF BONDS –ENERGY OF FOOD MOLECULES 3.Thermal Energy 4.Radiant Energy

4. 4 Write the given and needed units.Write the given and needed units. Write a unit plan to convert the given unit to the needed unit.Write a unit plan to convert the given unit to the needed unit. Write equalities and conversion factors that connect the units.Write equalities and conversion factors that connect the units. Use conversion factors to cancel the given unit and provide the needed unit.Use conversion factors to cancel the given unit and provide the needed unit. Unit 1 x Unit 2 = Unit 2 Unit 1 Given x Conversion= Needed unit factor unit unit factor unit Last Time: Conversion Factors in Problem Solving PRACTICE!

5. 5 Heat & Temperature Particles are always moving. When you heat water, the water molecules move faster. When molecules move faster, the substance gets hotter. When a substance gets hotter, its temperature and total energy content increases. measure temperature changes with a thermometer measure temperature changes with a thermometer

6. 6 Three Important Temperature Scales Fahrenheit Celsius Kelvin Fahrenheit Celsius Kelvin Water boils 212°F 100°C 373 K 180° 100°C 100K 180° 100°C 100K Water freezes 32°F 0°C 273 K

7. 7 On the Fahrenheit scale, there are 180°F between the freezing and boiling points and on the Celsius scale, there are 100°C.On the Fahrenheit scale, there are 180°F between the freezing and boiling points and on the Celsius scale, there are 100°C. 180°F = 9°F =1.8°F 100°C 5°C 1°C In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0°C to 32°F.In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0°C to 32°F. T F = 9/5 T C + 32  or or T F = 1.8 T C + 32  Fahrenheit Formula

8. 8 T C is obtained by rearranging the equation for T F.T C is obtained by rearranging the equation for T F. T F = 1.8T C + 32 Subtract 32 from both sides.Subtract 32 from both sides. T F - 32 = 1.8T C ( +32 - 32) T F - 32 = 1.8T C Divide by 1.8 =°F - 32 = 1.8 T CDivide by 1.8 =°F - 32 = 1.8 T C 1.8 1.8 1.8 1.8 T F - 32 = T C T F - 32 = T C 1.8 1.8 Celsius Formula

9. 9 Solving A Temperature Problem A person with hypothermia has a body temperature of 34.8°C. What is that temperature in °F? T F = 1.8 T C + 32  T F = 1.8 (34.8°C) + 32° exact tenth's exact exact tenth's exact = 62.6 + 32° = 62.6 + 32° = 94.6°F = 94.6°F tenth’s tenth’s Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 3 sig. figs. based on measured numbers!

10. 10 Specific Heat (2.4) Different substances have different capacities for storing energy It may take 20 minutes to heat water to 75°C. However, the same mass of aluminum might require 5 minutes and the same amount of copper may take only 2 minutes to reach the same temperature. THINK ABOUT BOB MAKING MAC & CHEESE

11. 11 Examples of Specific Heats TABLE 2.6 cal/g°C0.2140.09200.03080.1080.05620.1250.4880.5880.2070.100

12. 12 Specific heat (SH) is different for different substances.is different for different substances. is the amount of heat that raises the temperature of 1 g of a substance by 1°Cis the amount of heat that raises the temperature of 1 g of a substance by 1°C Heat of a process = (Specific Heat) x (mass) x (  T)Heat of a process = (Specific Heat) x (mass) x (  T) in the SI system has units of J/g  C.in the SI system has units of J/g  C. in the metric system has units of cal/g  C.in the metric system has units of cal/g  C. Specific Heat: Mathematical Description

13. 13 Heat Equation Rearranging the specific heat expression gives the heat equation. Heat = g x °C x J = J g°C g°C The amount of heat lost or gained by a substance is calculated from the mass of substance (g).mass of substance (g). temperature change (  T).temperature change (  T). specific heat of the substance (J/g°C).specific heat of the substance (J/g°C).

14. 14 A. When ocean water cools, the surrounding air 1) cools. 2) warms.3) stays the same. 1) cools. 2) warms.3) stays the same. B. Sand in the desert is hot in the day, and cool at night. Sand must have a at night. Sand must have a 1) high specific heat. 2) low specific heat. 1) high specific heat. 2) low specific heat. Learning Check Key Point: If one substance “heats up” by a certain amount, another substance must exactly lose that same amount of heat!

15. 15 How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C (SH of water = 4.184 J/(g o C)? 1) 20.4 kJ 2) 77.7 kJ 3) 84.3 kJ 12345

16. 16 How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 3) 84.3 kJ 77.0°C – 15.0°C = 62.0°C 325 g x 62.0°C x 4.184 J x 1 kJ g °C 1000 J g °C 1000 J = 84.3 kJ = 84.3 kJ Solution

17. 17 Another Example What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C?

18. 18 Solution What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C? Given: 24.8 g metal, 275 J of energy, 20.2  C to 24.5  C Need: J/g  C Plan: SH = Heat/g  C ΔT = 24.5  C – 20.2  C = 4.3  C ΔT = 24.5  C – 20.2  C = 4.3  C Set Up: 275 J = 2.6 J/g  C (24.8 g)(4.3  C)

19. 19 Why are we so interested in heat? Chemical reactions that produce heat Chemical reactions that absorb heat “Exothermic” “Exothermic” “Endothermic” “Endothermic” Heat is related to whether a chemical or biological process will happen!

20. 20 States of Matter Solid Liquid Gas Solid Liquid Gas

21. 21 States of Matter Solid Liquid Gas Solid Liquid Gas

22. 22 Phases Changes melting vaporization condensation freezing deposition sublimation (e.g. freeze-drying)

23. 23 The Heat of Fusion The heat of fusion is the amount of heat released when 1 gram of liquid freezes (at its freezing point).is the amount of heat released when 1 gram of liquid freezes (at its freezing point). is the amount of heat needed to melt 1 gram of a solid (at its melting point).is the amount of heat needed to melt 1 gram of a solid (at its melting point). for water (at 0°C) isfor water (at 0°C) is 80. cal 80. cal 1 g water

24. 24 The heat needed to freeze (or melt) a specific mass of water (or ice) is calculated using the heat of fusion. Heat = g water x 80. cal 1 g water 1 g water Example: How much heat in cal is needed to melt 15.0 g of water? 15.0 g water x 80. cal = 1200 cal 1 g water Calculations Using Heat of Fusion

25. 25 Heat of Vaporization The heat of vaporization is the amount of heat absorbed to vaporize 1 g of a liquid to gas at the boiling point.absorbed to vaporize 1 g of a liquid to gas at the boiling point. released when 1 g of a gas condenses to liquid at the boiling point.released when 1 g of a gas condenses to liquid at the boiling point. Boiling Point of Water = 100°C Heat of Vaporization (water) = 540 cal 1 g water

26. 26 How many kilocalories (kcal) are released when 50.0 g of steam from a volcano condenses at 100°C? (Heat of Vaporization = 540 cal/g) 1) 27 kcal 2) 540 kcal 3)54 kcal 4) 2700 kcal 12345

27. 27 Summary of Phase Changes: Heating Curves A heating curve illustrates the changes of state as a solid is heated.illustrates the changes of state as a solid is heated. uses sloped lines to show an increase in temperature.uses sloped lines to show an increase in temperature. uses plateaus (flat lines) to indicate a change of state.uses plateaus (flat lines) to indicate a change of state.

28. 28 Cooling Curve Using the heating curve of water as a guide, draw a cooling curve for water beginning with steam at 110°C and ending at -20°C.

29. 29 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 37°C  T = 37.0°C - 0°C = 37.0°C temperature increase temperature increase 0°C solid liquid 0°C solid liquid melting melting Combined Heat Calculations

30. 30 Combined Heat Calculations (continued.) Step 2: Calculate the heat to melt ice (fusion) 250. g ice x 80. cal= 2.000  10 4 cal 1 g ice 1 g ice Step 3: Calculate the heat to warm the water from 0°C to 37.0°C (SH of water = 1 cal/g) 0°C to 37.0°C (SH of water = 1 cal/g) 250. g x 37.0°C x 1.00 cal = 9 250 cal g °C 250. g x 37.0°C x 1.00 cal = 9 250 cal g °C Total: Step 2 + Step 3 = 29 200 cal (rounded to 3 SF)