Chapters 13 & 14

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# Chapters 13 & 14 - PowerPoint PPT Presentation

Chapters 13 & 14. Temperature and Heat. 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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### Chapters 13 & 14

Temperature and Heat

Alta 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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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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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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Conversion Equations

C = 5/9 (F – 32)

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

F = (9/5 C) + 32

Ex: F = (9/5)(100) + 32 = 212

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Conversion Equations

K = C +273

Ex: K = 100 + 273 = 373

C = K - 273

Ex: K = 373 - 273 = 100

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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?

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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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Thermal Expansion
• Linear Expansion

Δl = (α)(l)(Δt)

l’ = l + Δl

• Volumetric Expansion

Δl = (β)(V)(Δt)

V’ = V + ΔV

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Sample Problem
• A Brass strip is 3 cm long at 0° C. How long will it be at 100° C if the coefficient of linear expansion for Brass is 19 x 10 -6?

Δl = (α)(l)(Δt) = (19 x 10 -6)(0.03 m)(100 °C)=

0.000057 m

l’ = Δl + l = 0.000057 m + 0.03 m =

3.0057 cm

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

Qlost= Qgained

• Mathematically

Q = mCpΔt

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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)

= 493000 Joules

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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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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 = 3.02 kg

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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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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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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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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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Energy Required for a Phase Change
• Heat of Vaporization – ΔHf , is the amount of energy absorbed or released when a substance boils or condenses!
• Mathematically:

Q = mΔHf for Fusion

or

Q = mΔHv for Vaporization

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Sample Problem

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

Solution

Q = mΔHf = (5 kg)(3.33 x 105 J/kg °C) =

1.665 x 106 J

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Phase Diagram

Water-Steam

120

Steam

100

Water

Ice - Water

0

-5

Ice

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Sample Problem

How much energy is required to convert 5 kg of ice at -5 °C to Steam at 120 °C?

Steps

Raise temperature of ice to melting point

Melt Ice

Raise temperature of Water to boiling point

Vaporize Water

Raise Temperature of Steam to 120 °C

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Solution

Step 1: Raise temperature of Ice

Q = mCΔt = (5 Kg)(2100 J/kg°C)(5 °C)= 52500 J

Step 2: Melt Ice

Q = mΔHf = (5 Kg)(3.33 x 105 J/kg) = 1665000 J

Step 3: Raise temperature of water to boiling

Q = mCΔt = (5 Kg)(4186 J/kg°C)(100 °C)= 2093000 J

Step 4: Vaporize Water

Q = mΔHv = (5 Kg)(22.6 x 105 J/kg) = 11300000 J

Step 5: Raise temperature of Steam to 120°C

Q = mCΔt = (5 Kg)(2010 J/kg°C)(20 °C)= 201000 J

Step 6: Get Total

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

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• 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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Problem Types
• Temperature Scales and Conversions
• Thermal Expansion
• Conduction
• Calorimetry
• Latent Heat
• Fusion
• Vaporization

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