TEMPERATURE INTERNAL ENERGY PER UNIT MOLECULE NOT A MEASURE OF THE TOTAL KINETIC ENERGY OF A...
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Transcript of TEMPERATURE INTERNAL ENERGY PER UNIT MOLECULE NOT A MEASURE OF THE TOTAL KINETIC ENERGY OF A...
TEMPERATURE
• INTERNAL ENERGY PER UNIT MOLECULE
• NOT A MEASURE OF THE TOTAL KINETIC ENERGY OF A SUBSTANCE
• NOT THE SAME AS HEAT
HEAT
• The energy that transfers from one object to another because of a temperature difference
• Energy in transit• Always from higher
temperature object to a lower temp. object
Thermal Contact ant Thermal equilibrium
• When heat flows from one substance to another it is in contact with it is in
Thermal Contact• When objects in Thermal Contact come to
the same temperature they are in
Thermal Equilibrium
Internal EnergyKinds of Energy in a substance
• Translational Kinetic Energy of moving molecules
• Rotational Kinetic Energy
• Potential energy due to forces between molecules
ExampleICE
Quantity of Heat
• The unit of heat is defined as the heat necessary to produce some standard, agreed upon temperature change for a specified amount of material
• Calorie - the amount of heat necessary to raise 1 gram of water by 1 degree Celsius
• 1 kilocalorie = 1000 calories = 1 Calorie• In SI System 1 calorie = 4.187 J
Question
• A woman with an average diet consumes and expends about 2000 Calories per day. The energy used by her body is eventually given off as heat. How many joules per second does her body give off? In other words, what is her average thermal output?
12.2 The Kelvin Temperature Scale
15.273+= cTT
Kelvin temperature
12.4 Linear Thermal Expansion
NORMAL SOLIDS
12.4 Linear Thermal Expansion
oLL ∝Δ
12.4 Linear Thermal Expansion
LINEAR THERMAL EXPANSION OF A SOLID
The length of an object changes when its temperature changes:
TLL oΔ=Δ α
coefficient of linear expansion
Common Unit for the Coefficient of Linear Expansion: ( )1C
C
1 −= o
o
12.4 Linear Thermal Expansion
12.4 Linear Thermal Expansion
Example 3 The Buckling of a Sidewalk
A concrete sidewalk is constructed between two buildings on a day when the temperature is 25oC. As the temperature rises to 38oC, the slabs expand, but no space is provided forthermal expansion. Determine the distance yin part (b) of the drawing.
12.4 Linear Thermal Expansion
( )[ ]( )( ) m 00047.0C 13m 0.3C101216 =×=
Δ=Δ−− oo
TLL oα
( ) ( ) m 053.0m 00000.3m 00047.3 22 =−=y
12.4 Linear Thermal Expansion
THE BIMETALLIC STRIP
12.4 Linear Thermal Expansion
12.4 Linear Thermal Expansion
THE EXPANSION OF HOLES
Conceptual Example 5 The Expansion of Holes
The figure shows eight square tiles that are arranged to form a square patternwith a hold in the center. If the tiled are heated, what happens to the size of the hole?
12.4 Linear Thermal Expansion
A hole in a piece of solid material expands when heated and contracts whencooled, just as if it were filled with the material that surrounds it.
12.4 Linear Thermal Expansion
Conceptual Example 7 Expanding Cylinders
Each cylinder is made from a different material. All three have the same temperature and theybarely fit inside each other.
As the cylinders are heated to the same,but higher, temperature, cylinder C fallsoff, while cylinder A becomes tightly wedgedto cylinder B.
Which cylinder is made from which material?
12.5 Volume Thermal Expansion
VOLUME THERMAL EXPANSION
The volume of an object changes when its temperature changes:
TVV oΔ=Δ β
coefficient of volume expansion
Common Unit for the Coefficient of Volume Expansion: ( )1C
C
1 −= o
o
12.5 Volume Thermal Expansion
Example 8 An Automobile Radiator
A small plastic container, called the coolant reservoir, catchesthe radiator fluid that overflows when an automobile enginebecomes hot. The radiator is made of copper and the coolant has an expansion coefficient of 4.0x10-4 (Co)-1. If the radiator is filled to its 15-quart capacitywhen the engine is cold (6oC),how much overflow will spill into the reservoir when the coolant reaches its operating temperature (92oC)?
12.5 Volume Thermal Expansion
( )( )( )( ) quarts 53.0C 86quarts 15C1010.414
coolant =×=Δ−− ooV
( )( )( )( ) quarts 066.0C 86quarts 15C105116
radiator =×=Δ−− ooV
quarts 0.46quarts 066.0quarts 53.0spill =−=ΔV
12.5 Volume Thermal Expansion
Expansion of water.
12.7 Heat and Temperature Change: Specific Heat Capacity
SOLIDS AND LIQUIDS
HEAT SUPPLIED OR REMOVED IN CHANGING THE TEMPERATUREOF A SUBSTANCE
The heat that must be supplied or removed to change the temperature ofa substance is
TmcQ Δ=
specific heatcapacity
Common Unit for Specific Heat Capacity: J/(kg·Co)
12.7 Heat and Temperature Change: Specific Heat Capacity
12.7 Heat and Temperature Change: Specific Heat Capacity
Example 9 A Hot Jogger
In a half-hour, a 65-kg jogger can generate 8.0x105J of heat. This heatis removed from the body by a variety of means, including the body’s owntemperature-regulating mechanisms. If the heat were not removed, how much would the body temperature increase?
TmcQ Δ=
( ) ( )[ ]o
o C 5.3CkgJ3500kg 65
J100.8 5
=⋅
×==Δ
mc
QT
12.7 Heat and Temperature Change: Specific Heat Capacity
GASES
The value of the specific heat of a gas depends on whether the pressure orvolume is held constant.
This distinction is not important for solids.
OTHER UNITS
1 kcal = 4186 joules
1 cal = 4.186 joules
12.7 Heat and Temperature Change: Specific Heat Capacity
CALORIMETRY
If there is no heat loss to the surroundings,the heat lost by the hotter object equals theheat gained by the cooler ones.
12.7 Heat and Temperature Change: Specific Heat Capacity
Example 12 Measuring the Specific Heat Capacity
The calorimeter is made of 0.15 kg of aliminumand contains 0.20 kg of water. Initially, thewater and cup have the same temperatureof 18.0oC. A 0.040 kg mass of unknown material is heated to a temperature of 97.0oC and then added to the water.
After thermal equilibrium is reached, thetemperature of the water, the cup, and the material is 22.0oC. Ignoring the small amountof heat gained by the thermometer, find the specific heat capacity of theunknown material.
12.7 Heat and Temperature Change: Specific Heat Capacity
( ) ( ) ( )unknownwaterAl TmcTmcTmc Δ=Δ+Δ
( ) ( )( )
( )[ ]( )( ) ( )[ ]( )( )( )( )
( )o
o
oooo
CkgJ1300
C 0.75kg 040.0
C 0.4kg 20.0CkgJ4186C 0.4kg 15.0CkgJ1000.9 2
unknown
waterAlunknown
⋅=
⋅+⋅×=
Δ
Δ+Δ=
Tm
TmcTmcc
12.8 Heat and Phase Change: Latent Heat
THE PHASES OF MATTER
12.8 Heat and Phase Change: Latent Heat
During a phase change, the temperature of the mixture does not change (provided the system is in thermal equilibrium).
12.8 Heat and Phase Change: Latent Heat
Conceptual Example 13 Saving Energy
Suppose you are cooking spaghetti for dinner, and the instructionssay “boil pasta in water for 10 minutes.” To cook spaghetti in an openpot with the least amount of energy, should you turn up the burnerto its fullest so the water vigorously boils, or should you turn downthe burner so the water barely boils?
12.8 Heat and Phase Change: Latent Heat
HEAT SUPPLIED OR REMOVED IN CHANGING THE PHASEOF A SUBSTANCE
The heat that must be supplied or removed to change the phaseof a mass m of a substance is
mLQ =
latent heat
SI Units of Latent Heat: J/kg
12.8 Heat and Phase Change: Latent Heat
12.8 Heat and Phase Change: Latent Heat
Example 14 Ice-cold Lemonade
Ice at 0oC is placed in a Styrofoam cup containing 0.32 kg of lemonadeat 27oC. The specific heat capacity of lemonade is virtually the same asthat of water. After the ice and lemonade reach and equilibriumtemperature, some ice still remains. Assume that mass of the cup isso small that it absorbs a negligible amount of heat.
( ) ( ) 44 344 2143421lemonade
bylost Heat
lemonade
iceby gainedHeat
iceTcmmL f Δ=
12.8 Heat and Phase Change: Latent Heat
( ) ( ) 44 344 2143421lemonade
bylost Heat
lemonade
iceby gainedHeat
iceTcmmL f Δ=
( )
( )[ ]( )( )kg 11.0
kgJ103.35
C0C27kg 32.0CkgJ4186
L
5
f
lemonadeice
=×
−⋅=
Δ=
ooo
Tcmm