Herriman High Honors Physics Chapter 9 Temperature and Heat.

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Herriman High Honors Physics Chapter 9 Temperature and Heat

Transcript of Herriman High Honors Physics Chapter 9 Temperature and Heat.

Page 1: Herriman High Honors Physics Chapter 9 Temperature and Heat.

Herriman High Honors Physics

Chapter 9

Temperature and Heat

Page 2: Herriman High Honors Physics Chapter 9 Temperature and Heat.

Herriman High Honors Physics

Temperature Scales Temperature is defined as a

measure of the average kinetic energy of the molecules.

Temperature scales were developed using the freezing and boiling points of water at sea level as the standard reference points.

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Herriman High Honors Physics

Temperature Scales There are three Temperature

Scales used in Science Fahrenheit – Used primarily in the

United States Celsius – the standard for the Metric

System Kelvin – Also know as the “Absolute

Zero” Scale.

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Herriman High Honors Physics

Standard Reference Points

Scale Boiling Point

Freezing Point

Fahrenheit 212 °F 32 °F

Centigrade 100 °C 0 °C

Kelvin 373 273

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Herriman High Honors Physics

Conversion Equations Fahrenheit to Centigrade

C = 5/9 (F – 32)Ex: C = 5/9 (212-32) = 100

Centigrade to FahrenheitF = (9/5 C) + 32Ex: F = (9/5)(100) + 32 = 212

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Herriman High Honors Physics

Conversion Equations Centigrade to Kelvin

K = C +273Ex: K = 100 + 273 = 373

Kelvin to Centigrade C = K - 273Ex: K = 373 - 273 = 100

You TryPractice

Set AP. 303

#1,3,& 5

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Herriman High Honors Physics

Heat Heat is a form of Energy Transfer Heat flows from areas of high

energy to areas of lower energy Heat is transferred three way

Conduction – requires contact Convection – mass movement of

molecules Radiation – transfer over a distance

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Heat Heat has the units of Energy

English System – calories – the amount of energy required to raise the temperature of 1 gram of water 1°C.

Metric System – Joule – 4.18 Joules in one calorie.

Herriman High Honors Physics

Page 9: Herriman High Honors Physics Chapter 9 Temperature and Heat.

Conservation of Energy ΔPE + ΔKE + ΔU = 0

The change in potential energy + the change in kinetic energy + the change in internal energy = 0

Energy is not created or destroyed, it just changes form.

Herriman High Honors Physics

Practice BP. 311

Problems 2 & 4

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Herriman High Honors Physics

Thermal Expansion Most objects tend to expand when

their temperature rises and to contract when the temperature drops.

Do you know what the one notable exception is?

Page 11: Herriman High Honors Physics Chapter 9 Temperature and Heat.

Herriman High Honors Physics

Thermal Expansion Objects can expand linearly whereas

liquids expand volumetrically! Each substance has a constant for

which describes its ability to expand. Linear constants are denoted by the

Greek letter alpha, α and the volumetric constant is denoted by the Greek letter beta, β.

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Herriman High Honors Physics

Conduction

Each material has an innate ability to absorb or give off heat – specific heat

The amount of heat an object can transfer depends upon three things: The mass of the object, m, (in Kg) The specific heat of the object, Cp, ( in J/g°C) The temperature change of the object, t, (in

ºC)

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Herriman High Honors Physics

Calorimetry The symbol for heat is a capitol Q Law of Conservation of Energy

says thatQlost= Qgained

Mathematically Q = mCpΔt

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Herriman High Honors Physics

Sample Problem How much energy is required to

raise the temperature of 5 Kg of water from 0°C to 100°C ? (Cp = 4186 J/kg°C)

Q= m CpΔt =

(5 kg)(Cp = 4186 J/kg°C)(100°C)

= 2093000 Joules

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Herriman High Honors Physics

Sample Problem A 5 Kg copper ball is heated to 180

ºC and dropped into a container of water at 100 °C. When the temperature of the ball/water system equalizes the final temperature is found to be 110 ºC. How much water is in the container? (Cp= 390 J/Kg°C)

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Herriman High Honors Physics

Solution

Qlost=Qgained

Q copper = Q water

mC copperΔt = mC water Δt

(5 kg)(390 J/Kg°C)(70°C)=m(4186 J/Kg°C)(10°C)

m = 3.26 kg Practice Cp. 316

Problems 1 & 3

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Herriman High Honors Physics

Phases Changes Phase changes require that a

substance absorb energy or release energy to occur.

There is NO Change in Temperature associated with a phase change!

Different words are used to denote direction when dealing with a phase change.

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Herriman High Honors Physics

Vocabulary of a Phase Change Freezing – change from liquid to solid.

Energy is released! Melting – change from solid to liquid

Energy is absorbed!

Both of these changes happen at the same point; 0° for water!

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Herriman High Honors Physics

Vocabulary of a Phase Change Condensing – change from gas or

vapor to a liquid. Energy is released!

Boiling – change from liquid to gas or vapor Energy is absorbed!

Vapor is the gas phase of any substance which is normally a liquid at room temperature!

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Herriman High Honors Physics

Energy Required for a Phase Change

Just like specific heat is a set amount of energy for each substance – the amount of energy required for a phase change is also substance specific.

Heat of Fusion – ΔHf , is the amount of energy absorbed or released when a substance melts or freezes!

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Herriman High Honors Physics

Energy Required for a Phase Change

Heat of Vaporization – ΔHv , is the amount of energy absorbed or released when a substance boils or condenses!

Mathematically:Q = mΔHf for Fusion

orQ = mΔHv for Vaporization

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Herriman High Honors Physics

Sample Problem

How much heat is required to melt 5 kg of ice at 0° C?

SolutionQ = mΔHf = (5 kg)(3.33 x 105 J/kg) =

1.665 x 106 J

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Herriman High Honors Physics

Phase Diagram

Ice

Ice - Water

Water

Water-Steam

Steam

-5

0

100

120

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Herriman High Honors Physics

Sample ProblemHow much energy is required to

convert 5 kg of ice at -5 °C to Steam at 120 °C?

StepsRaise temperature of ice to melting point

Melt IceRaise temperature of Water to boiling point

Vaporize WaterRaise Temperature of Steam to 120 °C

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Herriman High Honors Physics

Solution

Step 1: Raise temperature of IceQ = mCΔt = (5 Kg)(2100 J/kg°C)(5 °C)= 52500 J

Step 2: Melt IceQ = mΔHf = (5 Kg)(3.33 x 105 J/kg) = 1665000 JStep 3: Raise temperature of water to boiling

Q = mCΔt = (5 Kg)(4186 J/kg°C)(100 °C)= 2093000 JStep 4: Vaporize Water

Q = mΔHv = (5 Kg)(22.6 x 105 J/kg) = 11300000 JStep 5: Raise temperature of Steam to 120°C

Q = mCΔt = (5 Kg)(2010 J/kg°C)(20 °C)= 201000 JStep 6: Get Total

Qtotal= 52500 + 1665000 + 2093000 + 11300000 + 201000 = 15311500J

Page 26: Herriman High Honors Physics Chapter 9 Temperature and Heat.

Herriman High Honors Physics

Convection & Radiation Convection – transfer of energy by

mass movement of molecules Most common form is Wind

Radiation – transfer of energy through waves – most common form is light or electromagnetic waves

Will discuss both in more detail in later chapters

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Herriman High Honors Physics

Problem Types Temperature Scales and

Conversions Thermal Expansion Conduction

Calorimetry Latent Heat

Fusion Vaporization