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AnnouncementsAnnouncements & Agenda& Agenda (01/15/07)(01/15/07)

Office Hours (SC 3063)Office Hours (SC 3063) Permanent Change: Thurs 9-11am replaces Mon 3-5pmPermanent 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 WedQuiz 1 (the CD quizzes) will be accepted until WedFirst in-class quiz (Quiz 2) will be Wed. (Ch 1 & 2)First in-class quiz (Quiz 2) will be Wed. (Ch 1 & 2)See overhead for your clicker numberSee overhead for your clicker numberYou should currently be reading Ch 3 You should currently be reading Ch 3

Today: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)

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. l. l22. . . . l . . . . l. . . . l . . . . l33 . . . . l . . . . l . . . . l . . . . l44. . cm. . cm

• The markings on the meter stick at the end of the The markings on the meter stick at the end of the orange line are read asorange line are read as

the first digit the first digit 2 2

plus the second digit plus the second digit 2.7 2.7

• The last digit is obtained by The last digit is obtained by estimatingestimating..

• The end of the line might beThe end of the line might be estimated between 2.7–estimated between 2.7–2.8 as half-way (0.5) or a little more (0.6), which gives 2.8 as half-way (0.5) or a little more (0.6), which gives a reported length of 2.7a reported length of 2.755 cm or 2.7 cm or 2.766 cm. cm.

Last time: Making MeasurementsLast time: Making Measurements

Measured #s always need units!Measured #s always need units!

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LAST TIME: EnergyLAST TIME: Energy

EnergyEnergy – the ability to do work – the ability to do work1.1. Kinetic Energy: energy of motionKinetic Energy: energy of motion

2.2. Potential Energy: stored energyPotential Energy: stored energy IMPORTANT EXAMPLES:IMPORTANT EXAMPLES:– ENERGY OF BONDSENERGY OF BONDS– ENERGY OF FOOD MOLECULESENERGY OF FOOD MOLECULES

3.3. Thermal EnergyThermal Energy

4.4. Radiant EnergyRadiant Energy

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• Write the given and needed units.Write the given and needed units.• Write a unit plan to convert the given unit to the needed Write a unit plan to convert the given unit to the needed

unit.unit.• Write equalities and conversion factors that connect the Write equalities and conversion factors that connect the

units.units.• Use conversion factors to cancel the given unit and provide Use conversion factors to cancel the given unit and provide

the needed unit.the needed unit.

Unit 1 x Unit 1 x Unit 2Unit 2 = Unit 2 = Unit 2Unit 1Unit 1

Given Given x x ConversionConversion = Needed= Needed unit unit factor factor unit unit

Last Time: Last Time: Conversion Factors in Problem SolvingConversion Factors in Problem Solving

PRACTICE!PRACTICE!

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Heat & TemperatureHeat & Temperature

Particles are always moving.Particles are always moving.

When you heat water, the water molecules move faster.When you heat water, the water molecules move faster.

When molecules move faster, the substance gets hotter.When molecules move faster, the substance gets hotter.

When a substance gets hotter, its temperature and total When a substance gets hotter, its temperature and total

energy content increases.energy content increases. measure temperature changes with a thermometermeasure temperature changes with a thermometer

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Three Important Temperature ScalesThree Important Temperature Scales

FahrenheitFahrenheit Celsius Celsius KelvinKelvin

Water boils 212°F 100°CWater boils 212°F 100°C 373 K 373 K

180°180° 100°C 100K 100°C 100K

Water freezes 32°F 0°CWater freezes 32°F 0°C 273 K 273 K

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• On the On the Fahrenheit scaleFahrenheit scale, there are 180, there are 180°F °F between the between the freezing and boiling pointsfreezing and boiling points and on the and on the Celsius scaleCelsius scale, , there are there are 100100°C. °C.

180°F180°F = = 9°F 9°F == 1.8°F 1.8°F 100°C 100°C 5°C 5°C 1°C1°C

• In the In the formula for the formula for the Fahrenheit temperatureFahrenheit temperature, adding, adding 32 32 adjusts the zero point of water from 0adjusts the zero point of water from 0°C to °C to 3232°F.°F.

TTFF = 9/5 T = 9/5 TCC + 32 + 32

oror

TTFF = 1.8 T = 1.8 TCC + 32 + 32

Fahrenheit FormulaFahrenheit Formula

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• TTC C is obtained by rearranging the equation for Tis obtained by rearranging the equation for TF.F.

TTFF = 1.8T= 1.8TCC + 32 + 32

• Subtract 32 from both sides.Subtract 32 from both sides.

TTFF - 32 - 32 = = 1.8T1.8TCC ( +32 - 32) ( +32 - 32)

TTFF - 32 - 32 = = 1.8T1.8TCC

• Divide by 1.8 =Divide by 1.8 = °F - 32°F - 32 = = 1.8 T1.8 TCC

1.8 1.81.8 1.8

TTFF - 32 - 32 = = TTCC

1.81.8

Celsius FormulaCelsius Formula

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Solving A Temperature ProblemSolving A Temperature Problem

A person with hypothermia has aA person with hypothermia has abody temperature of 34.8°C. body temperature of 34.8°C. What is that temperature in °F? What is that temperature in °F?

TTFF = 1.8 T = 1.8 TCC + 32 + 32

TTFF = 1.8 (34.8°C) + 32° = 1.8 (34.8°C) + 32° exact tenth's exactexact tenth's exact

= 62.6 + 32°= 62.6 + 32° = 94.6°F = 94.6°F tenth’stenth’s Copyright © 2005 by Pearson Education, Inc.

Publishing as Benjamin Cummings3 sig. figs. based on 3 sig. figs. based on measured numbers!measured numbers!

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Specific Heat (2.4)Specific Heat (2.4)

Different substances have different capacities for Different substances have different capacities for storing energystoring energyIt may take 20 minutes to heat water to 75°C. It may take 20 minutes to heat water to 75°C. However, the same mass of aluminum might require However, the same mass of aluminum might require 5 minutes and the same amount of copper may take 5 minutes and the same amount of copper may take only 2 minutes to reach the same temperature.only 2 minutes to reach the same temperature.

THINK ABOUT BOB MAKING MAC & CHEESETHINK ABOUT BOB MAKING MAC & CHEESE

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Examples of Specific HeatsExamples of Specific Heats

TABLE 2.6TABLE 2.6

cal/g°Ccal/g°C

0.2140.214

0.09200.0920

0.03080.0308

0.1080.108

0.05620.0562

0.1250.125

0.4880.488

0.5880.588

0.2070.207

0.1000.100

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Specific heat (SH)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 is the amount of heat that raises the temperature of 1 g of a substance by 1°C of a substance by 1°C

• Heat of a process = (Specific Heat) Heat of a process = (Specific Heat) xx (mass) (mass) xx ( (T)T)• in the SI system has units of J/gin the SI system has units of J/gC.C.

• in the metric system has units of cal/gin the metric system has units of cal/gC.C.

Specific Heat: Mathematical Description Specific Heat: Mathematical Description

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Heat EquationHeat Equation

Rearranging the specific heat expression gives theRearranging the specific heat expression gives theheat equationheat equation..

Heat = g x Heat = g x °C x °C x J J = J = J gg°C°C

The amount of heat lost or gained by a substance isThe amount of heat lost or gained by a substance iscalculated from thecalculated from the

• mass of substance (g).mass of substance (g).• temperature change (temperature change (T).T).• specific heat of the substance (J/gspecific heat of the substance (J/g°C)°C)..

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A. When ocean water cools, the surrounding air A. When ocean water cools, the surrounding air 1) cools. 1) cools. 2) warms.2) warms. 3) stays the same.3) stays the same.

B. Sand in the desert is hot in the day, and coolB. Sand in the desert is hot in the day, and cool at night. Sand must have aat night. Sand must have a

1) high specific heat. 1) high specific heat. 2) low specific heat.2) low specific heat.

Learning CheckLearning Check

Key Point: If one substance “heats up” by a certain amount, another substance must exactly lose that same amount of heat!

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How many kilojoules are needed to raise the How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C temperature of 325 g of water from 15.0°C to 77.0°C (SH of water = 4.184 J/(gto 77.0°C (SH of water = 4.184 J/(gooC)?C)?

0%

0%

0% 1) 20.4 kJ1) 20.4 kJ

2) 77.7 kJ2) 77.7 kJ

3) 84.3 kJ3) 84.3 kJ

11 22 33 44 55

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How many kilojoules are needed to raise the How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to temperature of 325 g of water from 15.5°C to 77.5°C?77.5°C?

3) 84.3 kJ 3) 84.3 kJ

77.0°C – 15.0°C = 62.0°C77.0°C – 15.0°C = 62.0°C

325 g x 62.0°C x 325 g x 62.0°C x 4.184 J4.184 J x x 1 kJ1 kJ

g °C 1000 Jg °C 1000 J

= 84.3 kJ= 84.3 kJ

SolutionSolution

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Another ExampleAnother Example

What is the specific heat of a metal if 24.8 g absorbsWhat is the specific heat of a metal if 24.8 g absorbs275 J of energy and the temperature rises from 20.2275 J of energy and the temperature rises from 20.2C toC to24.524.5C?C?

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SolutionSolution

What is the specific heat of a metal if 24.8 g absorbsWhat is the specific heat of a metal if 24.8 g absorbs275 J of energy and the temperature rises from 20.2275 J of energy and the temperature rises from 20.2C toC to24.524.5C?C?

Given:Given: 24.8 g metal, 275 J of energy, 20.2 24.8 g metal, 275 J of energy, 20.2C to 24.5C to 24.5C C Need:Need: J/g J/gCC Plan:Plan: SH = Heat/g SH = Heat/gCC

ΔΔT = 24.5T = 24.5CC – 20.2 – 20.2CC = 4.3 = 4.3 CC

Set Up:Set Up: 275 J 275 J = 2.6 J/g = 2.6 J/gCC (24.8 g)(4.3(24.8 g)(4.3C)C)

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Why are we so interested in heat?Why are we so interested in heat?

Chemical reactions Chemical reactions that produce heatthat produce heat

Chemical reactions Chemical reactions that absorb heatthat absorb heat

““Exothermic”Exothermic”

““Endothermic”Endothermic”

Heat is related to whether a chemical or Heat is related to whether a chemical or biological process will happen!biological process will happen!

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States of MatterStates of Matter

SolidSolid LiquidLiquid Gas Gas

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States of MatterStates of Matter

SolidSolid LiquidLiquid Gas Gas

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Phases ChangesPhases Changes

meltingmelting

vaporizationvaporizationcondensationcondensation

freezingfreezing

depositiondeposition

sublimationsublimation(e.g. freeze-drying)(e.g. freeze-drying)

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The Heat of FusionThe Heat of Fusion

The The heat of fusionheat of fusion • is the amount of heat is the amount of heat releasedreleased when 1 gram of when 1 gram of

liquid freezes (at its freezing point). liquid freezes (at its freezing point). • is the amount of heat is the amount of heat neededneeded to melt 1 gram of a to melt 1 gram of a

solid (at its melting point).solid (at its melting point).• for water (at 0for water (at 0°C)°C) is is

80. cal 80. cal 1 g water1 g water

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The heat needed to freeze (or melt) a specific mass of The heat needed to freeze (or melt) a specific mass of water (or ice) is calculated using the heat of fusion.water (or ice) is calculated using the heat of fusion.

Heat = g water x Heat = g water x 80. cal 80. cal 1 g water1 g water

ExampleExample: How much heat in cal is needed to melt 15.0 : How much heat in cal is needed to melt 15.0 g of water?g of water?

15.0 g water x 15.0 g water x 80. cal 80. cal = 1200 cal = 1200 cal

1 g water1 g water

Calculations Using Heat of FusionCalculations Using Heat of Fusion

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Heat of VaporizationHeat of Vaporization

The The heat of vaporizationheat of vaporization is the amount of heatis the amount of heat• absorbedabsorbed to vaporize 1 g of a liquid to gas at the to vaporize 1 g of a liquid to gas at the

boiling point.boiling point.• releasedreleased when 1 g of a gas condenses to liquid at the when 1 g of a gas condenses to liquid at the

boiling point.boiling point.

Boiling Point of Water = 100°CBoiling Point of Water = 100°C

Heat of Vaporization (water) Heat of Vaporization (water) = = 540 cal 540 cal

1 g water1 g water

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How many kilocalories (kcal) are released How many kilocalories (kcal) are released when 50.0 g of steam from a volcano when 50.0 g of steam from a volcano condenses at 100condenses at 100°°C? (Heat of Vaporization C? (Heat of Vaporization = 540 cal/g)= 540 cal/g)

0%

0%

0%

0% 1) 1) 27 kcal27 kcal

2) 2) 540 kcal540 kcal

3)3) 54 kcal 54 kcal

4) 4) 2700 kcal2700 kcal

11 22 33 44 55

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Summary of PhaseSummary of PhaseChanges: Heating CurvesChanges: Heating Curves

A A heating curveheating curve • illustrates the illustrates the

changes of state as changes of state as a solid is heated.a solid is heated.

• uses sloped lines to uses sloped lines to show an increase in show an increase in temperature.temperature.

• uses plateaus (flat uses plateaus (flat lines) to indicate a lines) to indicate a change of state.change of state.

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Cooling CurveCooling Curve

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

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To reduce a fever, an infant is packed in 250. g of ice. If 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 the ice (at 0°C) melts and warms to body temperature (37.0°C), how many calories are removed from the body?(37.0°C), how many calories are removed from the body?Step 1: Diagram the changesStep 1: Diagram the changes

37°C 37°C T = 37.0°C - 0°C = 37.0°C T = 37.0°C - 0°C = 37.0°C

temperature increasetemperature increase

0°C 0°C solid liquidsolid liquid

meltingmelting

Combined Heat CalculationsCombined Heat Calculations

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Combined Heat Calculations Combined Heat Calculations (continued.)(continued.)

Step 2: Calculate the heat to melt ice (fusion)Step 2: Calculate the heat to melt ice (fusion) 250. g ice x 250. g ice x 80. cal80. cal = 2.000 = 2.000 10 1044 cal cal

1 g ice1 g ice

Step 3: Calculate the heat to warm the water fromStep 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 250. g x 37.0°C x 1.00 cal1.00 cal = 9 250 cal = 9 250 cal g °C g °C

Total: Step 2 + Step 3Total: Step 2 + Step 3 = 29 200 cal = 29 200 cal(rounded to 3 SF)(rounded to 3 SF)