Dr. Joseph W. Howard ©Summer 2006 Kinetic & Potential Energy on the Nanoscale.

Dr. Joseph W. Howard

©Summer 2006

Kinetic & Potential EnergyKinetic & Potential Energy on the Nanoscaleon the Nanoscale

Kinetic & Potential EnergyKinetic & Potential Energy on the Nanoscaleon the Nanoscale


©Summer 2006

TemperatureTemperature

Cool Block

Warm Block


©Summer 2006

Thermal EnergyThermal Energy

Heat vs. Temperature.

Heat is a form of Energy!

Temperature is the measure of “degrees of hotness”


©Summer 2006

Energy RemindersEnergy Reminders

Hmmmm… What kind of energy(ies) have we already talked about?

Ah Ha!

Potential energy (gravitational)

Kinetic Energy (energy of motion)


©Summer 2006

TemperatureTemperature

All matter is composed of continuously jiggling atoms and molecules (solids, liquids, gases)

The average jiggling (avg. KE) of all the individual particles of an object is related to “How Hot.”


©Summer 2006

Kinetic Energy on the NanoscaleKinetic Energy on the NanoscaleKinetic Energy on the NanoscaleKinetic Energy on the Nanoscale

thermal energy

BaseballLooking at a tiny piece within the baseball.


©Summer 2006

Temperature ScalesTemperature ScalesTemperature ScalesTemperature Scales

Based ONLY on atomic and molecular motion (Avg. KE)

Fast jiggling Hot!No upper

limit

Slow jiggling Cooler

NO jiggling Coldest Possible

A lower limit

Shocking!


©Summer 2006

Fahrenheit Temperature ScaleFahrenheit Temperature ScaleFahrenheit Temperature ScaleFahrenheit Temperature Scale

Weatherman?

32oF Water Freezes/Melts

212oF Water Boils / Water Vapor Condenses


©Summer 2006

Celsius Temperature ScaleCelsius Temperature ScaleCelsius Temperature ScaleCelsius Temperature Scale

0oC Water Freezes/Melts

100oCWater Boils / Water Vapor Condenses


©Summer 2006

273oK Water Freezes/Melts

373oKWater Boils / Water Vapor Condenses

Kelvin Temperature ScaleKelvin Temperature ScaleKelvin Temperature ScaleKelvin Temperature Scale

0oKAbsolute Zero

No KE anywhere

32oF

0oC

100oC

212oF

-273oC

-460oF


©Summer 2006

TemperatureTemperatureTemperatureTemperature

Fahrenheit

212oF

32oF

Celsius

100oC

0oC

Kelvin

373K

273K

oF = (9/5 × oC) + 32oC = 5/9 × (oF – 32)

oF = (9/5 × oC) + 32oC = 5/9 × (oF – 32)


©Summer 2006

Time to move Heat Time to move Heat (energy) around.(energy) around.

Time to move Heat Time to move Heat (energy) around.(energy) around.

Reminder: Total Amount of Energy is always conserved. Reminder: Total Amount of Energy is always conserved.

We can only transform types of energyWe can only transform types of energy

PEPE KEKE HeatHeat


©Summer 2006

How to “Heat” something?How to “Heat” something?How to “Heat” something?How to “Heat” something?

To change its temperatureTo change its temperature

To change its phase!

To change its phase!


©Summer 2006

Specific HeatSpecific HeatSpecific HeatSpecific HeatSpecific Heat – “heat-ability” is different for

different substances. (Specific Heat Capacity)Specific Heat – “heat-ability” is different for

different substances. (Specific Heat Capacity)

The amount of heat (energy) required to increase the temperature of 1 gram of a

substance by one degree Celsius.

The amount of heat (energy) required to increase the temperature of 1 gram of a

substance by one degree Celsius.

Anyone heard of calories?Anyone heard of calories?


©Summer 2006

The “calorie”The “calorie”The “calorie”The “calorie”

One calorie is the amount of heat (energy) to raise the temperature of 1 gram of water by 1oC.

One calorie is the amount of heat (energy) to raise the temperature of 1 gram of water by 1oC.

Food Calorie? = 1 Calorie = 1000 caloriesFood Calorie? = 1 Calorie = 1000 calories

1 calorie = 4.184 Joules1 calorie = 4.184 Joules


©Summer 2006

Heat: Situation #1Heat: Situation #1

Using heat to ONLY change the Using heat to ONLY change the temperature of a material.temperature of a material.

Using heat to ONLY change the Using heat to ONLY change the temperature of a material.temperature of a material.

“Heating” something? “Heating” something?

Add Heat!Add Heat! Result?Result?

Change the temperatureChange the temperature


©Summer 2006

Conceptual PitfallConceptual PitfallConceptual PitfallConceptual PitfallKinetic Energy on the Nanoscale

Heat: Situation #1Temperature

Kinetic Energy on the NanoscaleHeat: Situation #1

Temperature


©Summer 2006

Heat to change the temperatureHeat to change the temperatureHeat to change the temperatureHeat to change the temperature

Heat = (mass)Heat = (mass)××(specific heat) (specific heat) ××(change in temperature)(change in temperature)

How much stuff is How much stuff is there to “heat”there to “heat”

““Heat ability” of Heat ability” of substancesubstance Change in Change in

temperature temperature (T(Tfinalfinal – T – Tinitialinitial))

Q = m × c ×(Q = m × c ×(t)t)Q = m × c ×(Q = m × c ×(t)t)


©Summer 2006

ExampleExampleExampleExample

If I add 418.4 Joules of heat to 20 grams of liquid HIf I add 418.4 Joules of heat to 20 grams of liquid H220 0 at 20at 20ooC, how much will the temperature of the HC, how much will the temperature of the H22O O change? What is the final temperature of the Hchange? What is the final temperature of the H22O?O?

If I add 418.4 Joules of heat to 20 grams of liquid HIf I add 418.4 Joules of heat to 20 grams of liquid H220 0 at 20at 20ooC, how much will the temperature of the HC, how much will the temperature of the H22O O change? What is the final temperature of the Hchange? What is the final temperature of the H22O?O?

Q = m c T

418.4 J = (20 grams) (4.184 J/goC) T

T = (418.4 J) / (20g)(4.184 J/goC)

T = 5oC

T = Tfinal – Tinitial

5oC = Tfinal – 20oC

Tfinal = 25oC


©Summer 2006

Example Con’tExample Con’tExample Con’tExample Con’t

What if the situation was the same, except the What if the situation was the same, except the heat of 418.4 J were added to heat of 418.4 J were added to IronIron and not water? and not water?What if the situation was the same, except the What if the situation was the same, except the heat of 418.4 J were added to heat of 418.4 J were added to IronIron and not water? and not water?

First, iron is a different material and has a different Specific heat of 0.451 J/goC

Q = m c T

418.4 J = (20 grams) (0.451 J/goC) T

T = (418.4 J) / (20g)(0.451 J/goC)

T = 46oC Much easier to change the temp of iron.

Much easier to change the temp of iron.


©Summer 2006

Using “Heat”Using “Heat”Using “Heat”Using “Heat”

Hmmmm…. What will happen to our 20 g water sample Hmmmm…. What will happen to our 20 g water sample if we keep adding “heat”?if we keep adding “heat”?

Another 418.4 JAnother 418.4 J 3030ooCC




Temp Temp keeps keeps going up!going up!

Water Water boilsboils


©Summer 2006

Boiling?Boiling?Boiling?Boiling?

What is boiling? What is boiling?

A liquid begins to change into a gas!!A liquid begins to change into a gas!!

Phase ChangePhase Change


©Summer 2006

Heat: Situation #2Heat: Situation #2Heat: Situation #2Heat: Situation #2

Heat ONLY to change the phase of a materialHeat ONLY to change the phase of a material..Heat ONLY to change the phase of a materialHeat ONLY to change the phase of a material..

Add heat to liquid (water) temperature goes up, but when it reaches the “boiling point”, add heat and NO CHANGE IN TEMPERATURE until all the liquid converts to gas!

Add heat to liquid (water) temperature goes up, but when it reaches the “boiling point”, add heat and NO CHANGE IN TEMPERATURE until all the liquid converts to gas!

It takes some energy (heat) just to convert phase!It takes some energy (heat) just to convert phase!

This is true for any change of This is true for any change of phasephase!!

This is true for any change of This is true for any change of phasephase!!


©Summer 2006

Conceptual PitfallConceptual PitfallConceptual PitfallConceptual Pitfall

Potential Energy on the NanoscaleHeat: Situation #2

Phase Change

Potential Energy on the NanoscaleHeat: Situation #2

Phase Change

Case 1

Case 2

Different position means different potential energy & different phase


©Summer 2006

Phase changes!Phase changes!Phase changes!Phase changes!

Heat of vaporizationHeat of vaporization – how much “heat” (energy) is – how much “heat” (energy) is removed/added when a gas condenses / when liquid vaporizes.removed/added when a gas condenses / when liquid vaporizes.Heat of vaporizationHeat of vaporization – how much “heat” (energy) is – how much “heat” (energy) is removed/added when a gas condenses / when liquid vaporizes.removed/added when a gas condenses / when liquid vaporizes.

Heat of fusionHeat of fusion – how much “heat” (energy) is – how much “heat” (energy) is removed/added when a liquid freezes / when solid melts.removed/added when a liquid freezes / when solid melts.Heat of fusionHeat of fusion – how much “heat” (energy) is – how much “heat” (energy) is removed/added when a liquid freezes / when solid melts.removed/added when a liquid freezes / when solid melts.

Q=m×LQ=m×L

Q = (mass)Q = (mass)× × (Heat needed for phase change)(Heat needed for phase change)Q = (mass)Q = (mass)× × (Heat needed for phase change)(Heat needed for phase change)


©Summer 2006

Phase ChangesPhase ChangesPhase ChangesPhase Changes

Description of Phase Change

Name of Change

Solid Solid Liquid Liquid MeltingMelting or or FusionFusion

Liquid Liquid Solid Solid FreezingFreezing

Liquid Liquid Gas Gas VaporizationVaporization

Solid Solid Gas Gas (directly (directly without changing to liquid without changing to liquid

first, dry ice does this)first, dry ice does this)

SublimationSublimation

Gas Gas Solid Solid (directly (directly without changing to liquid without changing to liquid

first)first)

CondensationCondensation or or DepositionDeposition


©Summer 2006

Ice Bath Activity

• Was there a temperature change?• Was there a flow of heat?• Was energy conserved?


©Summer 2006

ExampleExample

Calculate how much heat should be removed to Calculate how much heat should be removed to change 100 g of steam at 100change 100 g of steam at 100ooC to water at 100C to water at 100ooC.C.Calculate how much heat should be removed to Calculate how much heat should be removed to change 100 g of steam at 100change 100 g of steam at 100ooC to water at 100C to water at 100ooC.C.

1 g steam 1 g water 2260 J/g

Q=m×L =(2260J/g)(100g) = 226000J

Notice!! Heat was removed! Only the phase changed! Not the temperature!


©Summer 2006

Example Con’tExample Con’tExample Con’tExample Con’t

Suppose this heat removed from the steam is Suppose this heat removed from the steam is used to heat a 2000 g rock from 25used to heat a 2000 g rock from 25ooC to 79C to 79ooC. C. What is the specific heat of the rock?What is the specific heat of the rock?

Suppose this heat removed from the steam is Suppose this heat removed from the steam is used to heat a 2000 g rock from 25used to heat a 2000 g rock from 25ooC to 79C to 79ooC. C. What is the specific heat of the rock?What is the specific heat of the rock?

Q = m c T

226000 J = (2000 g) c (79oC-25oC)

c = (226000 J) / (2000g)(54oC)

c = 2.09 J/goC


©Summer 2006

Heat & EnergyHeat & EnergyHeat & EnergyHeat & Energy

• Heat can change the kinetic energy of a particles in which case the temperature will change.

• Heat can change the potential energy of particles (as during a phase change)

• Heat can change the kinetic energy of a particles in which case the temperature will change.

• Heat can change the potential energy of particles (as during a phase change)


©Summer 2006

Follow the HeatFollow the HeatFollow the HeatFollow the HeatA 50.0g piece of copper at 90oC is placed into 100.0g of water at 15oC. The final temperature of the water and copper is 18oC. The specific heat of water is 4.184 J/(goC).

How much did the temperature of the copper change?

How much did the temperature of the water change?

Copper? T = Tfinal – Tinitial T = 18oC – 90oC = - 72oC

How much heat was gained by the water?

Water? T = Tfinal – Tinitial T = 18oC – 15oC = + 3oC

Q = m c T Q = (100g)(4.184 J/goC)(3oC) = 1255 Joules


©Summer 2006

Follow the HeatFollow the HeatFollow the HeatFollow the HeatA 50.0g piece of copper at 90oC is placed into 100.0g of water at 15oC. The final temperature of the water and copper is 18oC. The specific heat of water is 4.184 J/(goC).

Calculate the specific heat of copper.

If the mass of the copper were doubled how much would the specific heat of the copper change?

How much heat was lost by the copper?

-1255 Joules !!

Q = m c T c = Qm T

c = (-1255 J)

(50g)(-72oC)= 0.35 J/goC

It would not change!

Dr. Joseph W. Howard ©Summer 2006 Kinetic & Potential Energy on the Nanoscale.

Documents

heat energy

energy heat

massspecific heat change

joules of heat

different specific heat

liquid water temperature

change of phase

specific heat capacitythe