chapter 10 heat thermal equilibrium bring two objects into thermal contact. –they can exchange...
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Chapter 10
Heat
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Thermal Equilibrium
• Bring two objects into thermal contact.
– They can exchange energy.
• When the flow of energy stops, the objects are in thermal equilibrium.
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Temperature• The quantity that tells how hot or
cold something is compared to the standard.
• Measure of average KE of particles
• Increase temp……matter expands
• decrease temp….matter shrinks
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Temperature• A method of assigning numbers to objects
that tells us about their thermal equilibrium.
– If the two objects are in thermal equilibrium, they are assigned the same numbers.
– If not, energy flows from the one with the higher temperature to the one with the lower temperature.
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Temperature
• Measure of the average KE of any substance.
• Thermal energy is proportional to the amount of matter, temp. is not.
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Temperature
• Newton proposed a method in 1701– Used an alcohol-in-glass
thermometer– Mixture of ice and water ---- 0– Human body ----------------- 12– Mark off 12 equal divisions
0
12
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Thermometer• Device used to measure
temp.
• Celsius, Kelvin and Fahrenheit
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Temperature
• International scale…..Celsius
• 0oC…………..water freezes
• 100oC…………water boils
• United States……..Fahrenheit Scale
• Scientific research is the SI scale…..Kelvin
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Temperature
• Gabriel Fahrenheit suggested zero correspond to mixture of ice and salt– Was lowest lab temperature– Avoided use of negative temperatures– Each of 12 degrees divided into 8– Human body -- 96 0
96
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Fahrenheit Temperature• Neither of fixed points was
reproducible– fp of pure water at standard
atmospheric pressure = 32– bp of pure water at standard
atmospheric pressure = 212
• Best overall agreement
• Body temperature = 98.6oF
32
212
98.6
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Celsius Temperature• Fixed points assigned new values
– 0oC -- freezing point
– 100oC -- boiling point
• Will leave conversions for you to learn.
0
100
37
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Kelvin Temperature
• All gas thermometers give a lowest temperature of 273.15oC
• Choose this as the zero point.
• Use better fixed point triple point of water 0.01oC
• 0 K absolute zero273
373
310
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Thermal Energy
• The total energy of particles in a material.
• KE and PE
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Thermal Energy Units
• calorie (cal) - amount of heat required to raise the temperature of 1 g of water by 1 c.
• Calorie (food) = kilocalorie
• British thermal unit (Btu) - amount of heat required to raise the temperature of 1 lb of water by 1oF.
• joule - same as in mechanics
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Mechanical Equivalent of Heat
• Can increase the internal energy of a body by – adding thermal energy– doing work on the body
• Therefore, there must be an equivalence– 1 cal = 4.186 J
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Example
• A drop of water falls a distance of 100 m. How much will its temperature rise when it hits the bottom?
• Any assumptions?
mghW
ghm
W 222 /m 980m 100m/s 8.9 s J/kg
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Heat• The energy transfer from one
object to another because of a temperature difference between them
• Matter doesn’t contain heat……..• Contains energy…..Heat is energy
in transit
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Heat • Thermal Contact…..
• ……..When heat flows from one object to another that it contacts
• Energy that flows from areas that are higher in temp to areas that are lower in temp
• There is NO such thing as cold
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Measurement of Heat• To quantify heat, we must specify
the mass and the kind of substance affected
• calorie….The amount of heat required to raise the temperature of 1 gram of water by 1oC
• Calorie………….1000 calories food
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Specific Heat (c)• The amount of heat required
to raise the temp of 1mass unit of a substance by 1 temp. unit.
• Thermal inertia…… signifies the resistance of a substance to change in temp.
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Computing thermal expansion
•Q = mCT• Q = heat gained or lost (J)
• m = mass (kg)T = change in temp (K)
• C = specific heat (J/kg • K)
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Calorimeter
• Device used to measure changes in thermal energy
• it is a closed system to measure energy transfer
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Heat of Fusion Hf
• Amount of energy needed to melt one kg of a substance
• ice= 3.34 x 105 J/kg
• ice= 80 cal
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Melting Point
• Temp at which a substance changes from a solid to a liquid
• same temp. as freezing point
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Heat of Vaporization Hv
• Amount of energy needed to vaporize a kg of liquid
• water = 2.26 x 106 J/kg
• water= 540 cal
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Boiling Point
• Temp at which a substance changes from a liquid to a gas
• Same as temp for condensation to occur
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• Q=mHf or Q=mHv
• Q….heat required to melt a solid or vaporize a liquid
• m….mass
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Thermal Energy
• The total energy of particles in a material.
• KE and PE
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Thermal Energy Transfer
• Occurs in 3 ways…
• CONDUCTION
• CONVECTION
• RADIATION
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Conduction• Transfer of energy from molecule to
molecule
• Most common in solids
• KE is transferred when particles collide
• MUST have direct contact
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Conductors
• Materials that conduct heat well
• Metals……….Silver
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Insulators• A material that is a poor conductor
• Delays the transfer of heat
• Examples… wood, wool, straw, paper
• Poor conductor is a good insulator
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Convection• Movement of all fluids (gas or liquid)
• Caused by substances having diff. Densities at diff. temps.
• Movement occurs in currents
• Example…….water, air
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Radiation• Thermal energy transferred
through space in the form of electromagnetic waves.
• Examples ….Solar energy,
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Example
• What does this mean?
• Convert to grams because we know what one cal does to one gram.
J 4.186
cal 1
kg
J980
kg
cal234
g 1000
kg 1
kg
cal234
g
cal234.0 C234.0 o T
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Specific Heat• Heat Capacity = amount of heat required to
raise T by 1oC– Not very useful because it is not a property of the
material
• Specific Heat = Heat Capacity per unit mass– Intrinsic property of the material
Tm
Qc
TmcQ
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Calorimetry
• Conservation of Energy
• Assume that no energy is lost to the surroundings
• Heat Gained = Heat Lost
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Calorimetry
• Assume that we drop a lump of gold (m = 20 g) at 100oC into 50 g of water at 20oC. What is the final equilibrium temperature?
• Heat lost = mgcg(Tg Te)
• Heat gained = mwcw(Te Tw)
– note that both are positive
– equate and solve for Te
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Calorimetry
• Do we need to use Kelvin temperatures?
wwgg
wwwggge cmcm
TcmTcmT
o o20 g 129 J/kg K 100 C 50 g 4186 J/kg K 20 C
20 g 129 J/kg K 50 g 4186 J/kg KeT
C21oeT
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Calorimetry• Let’s check to see if this is correct.
• It checks! The small difference is due to roundoff.
o
Heat lost
0.02 kg 129 J/kg K 79 C 204 J
g g gm c T
o
Heat gained
0.05 kg 4186 J/kg K 1 C 209 J
w w wm c T
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Latent Heat
• It requires heat to change the phase of a substance.
• Latent Heat is an intrinsic property– per unit mass
• Heat of fusion of water = 80 cal/g
• Heat of vaporization of water = 540 cal/g
• Q = mL
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Latent Heat• How much heat released in converting 1 g
of steam at 120oC to ice at 30oC?
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Cool steam, condense, cool water
TmcQ s
o1 g 0.5 cal/g K 20 C 10 cal
vmLQ cal 540cal/g 540g 1
TmcQ w
o1 g 1 cal/g K 100 C 100 cal
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Freeze and cool ice
fmLQ
cal 80cal/g 80g 1
TmcQ i
o1 g 0.5 cal/g K 30 C 15 cal
cal 745total Q J 3120cal 1
J 186.4
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Example
• A glass contains 120 g of ice at 0oC. What is the equilibrium temperature if you add 400 g of water at 20oC?
• Heat gained = heat lost
ewwwewifi TTcmTcmLm 0
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Example
• What ?!?• Check for mistakes.
– There are no mistakes.
• We made the assumption that all of the ice melted. Actually only 100 g melted!
wwi
fiwwwe cmm
LmTcmT
C67.2 o