Chapter 20 •Heat •Heat Transfer •Phase Changes...
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Chapter 20 •Heat •Heat Transfer •Phase Changes •Specific Heat •Calorimetry •First Law of Thermo •Work
Transcript of Chapter 20 •Heat •Heat Transfer •Phase Changes...
Chapter 20•Heat•Heat Transfer•Phase Changes•Specific Heat•Calorimetry•First Law of Thermo•Work
•Heat Energy is a flow of energy from hotter to colder because of a difference in temperature. Objects do not have heat. [Heat] = Joule
•Heat Energy entering or leaving a system will cause either a Temperature Change: Q = mcΔT or a Phase Change: Q = mL
Any two systems placed in thermal contact will have an exchange of heat
energy until they reach the same temperature.
If the systems are in thermal equilibrium then no net changes take place.
•Heat Energy is a flow of energy from hotter to colder because of a difference in temperature. Objects do not have heat. [Heat] = Joule
• Internal Energy of a system is a measure of the total Energy due to ALL random molecular motions INTERNAL of the system (Translations KE, Rotational KE, Vibrational KE) and internal POTENTIAL energies due to interactive forces (electromagnetic, strong, weak, gravitational) Objects have energy.
•Mechanical Energy is due to the kinetic and potential energies of the system itself in an external reference frame.
•Mechancial Equivalent of Heat: mechanical energy converted to heat energy by doing work on the system: 1.000 kcal = 4186J
• The change of internal energy of a system due to a temperature or phase change is given by:
Temperature Change: Q = mcΔTPhase Change: Q = mL
• Q is positive when the system GAINS heat and negative when it LOSES heat.
Specific Heat: Thermal InertiaThe Specific Heat of a substance is the amount of Energy it requires to raise the temperature of 1 kg, 1 degree Celsius.
Q mc T= Δ 0
Q Jcm T kg C
= =Δ ⋅
•The higher the specific heat, the more energy it takes and the longer it takes to heat up and to cool off.
•The lower the specific heat, the less energy it takes and the quicker it takes to heat up and cool off.
•Substances with HIGH specific heat STORE heat energy and make good thermal moderators. (Ex: Water, Oceans)
Some Specific Heat Values
More Specific Heat Values
Specific Heat
Why does water have such a highspecific heat?
Heat goes into other modes of energy so that temperature changes slowly.
0
0
0
4186
2410
452
water
glycerin
iron
Jckg C
Jckg C
Jckg C
=⋅
=⋅
=⋅
A 10,000 kg truck applies the brakes and descends 75.0 m at a constant speed, causing the brakes to smoke as shown. If the brakes have a mass of 100.00 kg and a specific heat of 800 J/kg°C, calculate the temperature increase of the brakes.
truck brakesm gh m c T= Δ
92.0truck
brakes
m ghT Cm c
Δ = = o
Q mc T= ΔHow much heat is required to raise the temperature of a 0.750kg aluminum pot containing 2.50kg of water at 30ºC to the boiling point?
Al Al w wQ m c T m c T= Δ + Δ
.75 (900 / ) 2.5 (4186 / ) (70 )kg J kg C kg J kg C C⎡ ⎤= +⎣ ⎦o o o
( )Al Al w wm c m c T= + Δ
57.798 10Q x J=
Phase Change Q mL=•A change from one phase to another•A phase change always occurs with an exchange of energy!•A phase change always occurs at constant temperature!
Sample Latent Heat Values
Phase ChangeEnergy goes into the system and breaks molecular bonds..
Energy is given up by the system by forming molecular bonds
Melting: Energy goes into the system and breaks molecular bonds..
Freezing: Energy is given up by the system by forming molecular bonds
Phase Change: Melting & Freezing
Phase Change: Melting & Freezing
Phase Change: Melting & Freezing•Melting: Solid to Liquid @ the melting temperature•Melting is a cooling process•Freezing: Liquid to Solid @ the melting temperature•Freezing is a warming process.
Why do farmers spray peaches with water to save them from frost?
Freezing is a warming process!
If you were in an igloo on a freezing night. You would be warmed more by
a) a bucket of ice melting. b) a bucket of water freezing c) the same either way. d) neither - are you nuts?
Phase Change: Evaporation•Takes place at the surface of a liquid due to escaping molecules.•Occurs at all temperatures•Evaporation occurs when water vapor pressure in the liquid exceeds the pressure of water vapor in the surrounding air. •Evaporation is a cooling process.
Evaporation is a Cooling Process
Phase Change: Boiling•Boiling is evaporation under the surface of the liquid.•Liquid boils at the temperature for which its vapor pressure exceeds the external pressure (mostly atmospheric pressure.) •Boiling point depends on temperature AND pressure:•@ 1 atm, bp of water is 100ºC, @ 5atm, bp of water is 374 ºC•Boiling is a cooling process.•At low pressures, liquids are boiled (‘freeze-dried’) into solids.
Increasing the PressureDoes increasing the external pressure increase or
decrease the boiling temperature of water?Increases! Boiling happens when vapor pressure in the liquid exceeds the external vapor pressure - now greater due to the
increased pressure – so the boiling temperature increases!
Phase Change: Condensation•Gas molecules condense to form a liquid.•Condensation is a warming process•Why is a rainy day warmer than a cloudy or clear day in winter?•Why do we feel uncomfortable on a muggy day?
Condensation is a Warming Process
Phase Change: Humidity•Vapor is the gas phase of a substance below its boiling temperature. •Air can ‘hold’ only so much water vapor before it becomes saturated and condensation occurs. Humidity is a measure of vapor density.•Warm air can hold more water vapor. More condensation occurs at cooler temperatures because the molecules are moving slower.
Slow moving water molecules coalesce upon collision.
Windward: WetLeeward: Dry
WarmHumidAir PushedUp
Cools and condenses at Top
WarmDryAirFallsDown
Stormy Weather
When warm air rises, it expands and cools. The water vapor in the air soon condenses into
water droplets, which form clouds and eventually these droplets fall from the sky as rain.
Phase Change:Sublimation
The conversion of a solid directly to a gas & visa versaExamples: snowflakes, Moth Balls, dry ice
Phase Change: Triple PointA temperature and pressure at which all three phases
exist in equilibrium.
Freezing-Melting Evaporation-Condensation
Sublimation
Lines ofequilibrium
Phase ChangePhase change occurs at a Constant Temperature!Latent Heats of: Fusion & Evaporation Lf, Lv
Q mL=
( )( )
334 / solid-liquid
2256 / liquid-gasf
v
L kJ kg
L kJ kg
=
=Water:
Phase Change: WaterHow much steam @ 100 °C does it take to melt 1kg of ice at -30 °C?
Q mL=
0
0
334 /2256 /
2090 /
4186 /
f
v
ice
water
L kJ kgL kJ kg
c J kg C
c J kg C
=
=
= ⋅
= ⋅
•How much energy is needed to raise the ices to 0 °C •How much energy is needed to melt 1kg of ice?•How much energy is given up by the steam? •What happens to the steam that is melting the ice?
Phase Change: WaterHow much steam @ 100 °C does it take to melt 1kg of ice at -30 °C?
Q mL=
0
0
334 /2256 /
2090 /
4186 /
f
v
ice
water
L kJ kgL kJ kg
c J kg C
c J kg C
=
=
= ⋅
= ⋅
How much energy is needed to raise the ices to 0 °C 0 0
1 1 (2090 / )(30 )Q kg J kg C C= ⋅
62700J=
Phase Change: WaterHow much steam @ 100 °C does it take to melt 1kg of ice at -30 °C?
Q mL=
0
0
334 /2256 /
2090 /
4186 /
f
v
ice
water
L kJ kgL kJ kg
c J kg C
c J kg C
=
=
= ⋅
= ⋅
How much energy is needed to melt 1kg of ice?
2Q mL=
2 334Q kJ=1 (334 / )kg kJ kg=
1 62700Q J=
2 334Q kJ=
Phase Change: WaterHow much steam @ 100 °C does it take to melt 1kg of ice at -30 °C?
Q mL=
0
0
334 /2256 /
2090 /
4186 /
f
v
ice
water
L kJ kgL kJ kg
c J kg C
c J kg C
=
=
= ⋅
= ⋅
(397 ) /(2256 / )m kJ kJ kg=
1 62700Q J=
2 334Q kJ=
•How much energy is given up by the steam? •What happens to the steam that is melting the ice?
/totalm Q L=.18kg=
397totalQ kJ=
•Heat flows from HOT to COLD•Conduction (solids)•Convection (liquids & gases)•Radiation (solids, gases, plasma)
Energy transferred via molecular collisions
•Good Conductors: Most Metals (free electrons!)
•Bad Conductors: Organic & Inert Materials•Good Insulators: Air, Water, Wood•Good Conductors are BAD Insulators •& Visa Versa
Heat energy is transferred in solids by collisions collisions between free electrons and
vibrating atoms.
The heat Q conducted during a time t through a material witha thermal conductivity k. dT/dx is the Temperature Gradient.
dTP kAdx
=
Some Thermal Conductivities
Temperature Gradient
h cdT T Tdx L
−=
The quantity |dT / dx| is called the temperature gradient
Q dTkAt dx
℘= =Δ
Compound Slab: R values
( )( )
h c
i ii
A T TL k−
℘=∑
• For a compound slab containing several materials of various thicknesses (L1, L2, …) and various thermal conductivities (k1, k2, …) the rate of energy transfer depends on the materials and the temperatures at the outer edges:
• Substances are rated by their R values– R = L / k and the rate becomes
– For multiple layers, the total R value is the sum of the R values of each layer
• Wind increases the energy loss by conduction in a home
( )h c
ii
A T TR−
℘=∑
Conduction Problem
A bar of gold is in thermal contact with a bar of silver of the same length and area as shown. One end of the compound bar is maintained at 80.0°C while the opposite end is at 30.0°C. When the energy transfer reaches steady state, what is the temperature at the junction? Ignore thermal expansion of the metals.
h cT TkAL−⎛ ⎞℘= ⎜ ⎟
⎝ ⎠
In the same room, at the same temperature, the tile floor feels
cooler than wood floor.How can they be the same
temperature?
kA TtELΔ
=
HW13.45 A fire walker runs across a bed of hot coals without sustaining burns. Calculate the energy transferred by conduction into the sole of one foot of a a fire walker given that the bottom of the foot is a 3.00mm thick callous with a conductivity of 0.08 J/smC and its density is 300kg/m3. The area of contact is 25.0 cm2, the temperature of the coals is 700C, and the time in contact is 1.00s. Assume his foot has an initial temperature of 37C (98.6F)
4 2(0.08 / )(25.0 10 )(700 37 )(1 ).003
kA Tt J smC x m C C sEL m
−Δ −= =
44.2E J=
What temperature increase is produced on his foot?
dT EP kAdx t
= =
A fire walker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a a fire walker given that the bottom of the foot is a 3.00mm thick callous with a conductivity of 0.08 J/smC and its density is 300kg/m3. The area of contact is 25.0 cm2, the temperature of the coals is 700C, and the time in contact is 1.00s. Assume his foot has an initial temperature of 37C (98.6F)
44.2E J=What temperature increase is produced on his foot?
Q mc T= Δ Vc Tρ= ΔQTVcρ
=> Δ = 1.68 C= o
(~35F)
kA TtQLΔ
=dT EP kAdx t
= =
Electromagnetic Radiation is emitted and absorbed via atomic excitations. All objects absorb and emit EM waves.
Electromagnetic Radiation is emitted and absorbed via atomic excitations. All objects absorb and emit EM waves.
When an object it heated it will glow first in the infrared, then the visible. Most solid materials break down before they emit UV and higher frequency EM waves.
Frequency ~ Temperature
Long
Short
Radiation
• Radiation does not require physical contact• All objects radiate energy continuously in
the form of electromagnetic waves due to thermal vibrations of their molecules
• Rate of radiation is given by Stefan’s law
Stefan’s Law
• P = σAeT 4
– P is the rate of energy transfer, in Watts– σ = 5.6696 x 10-8 W/m2 . K4
– A is the surface area of the object– e is a constant called the emissivity
• e varies from 0 to 1• The emissivity is also equal to the absorptivity
– T is the temperature in Kelvins
A good absorber reflects little and appears BlackA good absorber is also a good emitter.
Ideal Absorbers
• An ideal absorber is defined as an object that absorbs all of the energy incident on it– e = 1
• This type of object is called a black body• An ideal absorber is also an ideal radiator of
energy
Ideal Reflector
• An ideal reflector absorbs none of the energy incident on it: e = 0
4P e T Aσ=
Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m2 of 1200C fresh lava into 30.0C surroundings, assuming lava’s emissivity is 1.
The net heat transfer by radiation is: 4 4
2 1( )P e A T Tσ= −4 4
2 1( )P e A T Tσ= −8 4 2 4 41(5.67 10 / )1 ((303.15 ) (1473.15 ) )x J smK m K K−= −
266P kW= −
The heating effect of a medium such as glass or the Earth’s atmosphere that is transparent to short wavelengths but opaqueto longer wavelengths: Short get in, longer are trapped!
CO2 : Greenhouse Gas
Unless we change our direction, we are likely to end up where we are headed.
Hot Air rises, expands and cools, and then sinks back downcausing convection currents that transport heat energy.
Hot air rises because fast moving molecules tend to migrate towardregions of least obstruction - UP - into regions of lesser density!
Rising air cools because a decrease in densityreduces number of collisions & speeds decrease. As the air cools, it becomes denser, sinking down,producing a convection current.
Uneven heating on the earth and over water cause convection currents in the atmosphere, resulting in WINDS.
Global wind patterns (Trade Winds, Jet Streams) are due to convection current from warmer regions (equator) to cooler regions (poles) plus rotation of Earth.
Convection Currents in the Ocean (Gulf Stream) transport energy throughout the oceans.
Air & Ocean Convection causesthe WEATHER.
Fur is filled with air. Convection currents are slow because the convection loops are so small.
How do fur coats keep you warm?
Convection between water and land causes the Winds.
Sea Breeze
High PressureDry Warm Weather
Low PressureStormy Weather
Happy Earth Day!
“The organic and inorganic components of Planet Earth have evolved together as a single living, self-regulating systemLife maintains conditions suitable for its own survival.”
- James Lovelock
“It is much too late for sustainable development; what we need is a sustainable retreat.”
-James Lovelock, The Revenge of Gaia
“...we’re all astronauts aboard a little spaceship called Earth”- Bucky Fuller
One island in one ocean...from space
Our Spaceship Earth
"We are on a spaceship; a beautiful one. It took billions of years to develop.
We're not going to get another.”- Bucky Fuller,
Operating Manual for Spaceship Earth
500,000 miles/hr67,000 miles/hr
Space Ecology
Art by Ray Troll
Work in Thermodynamics• Work can be done on a deformable
system, such as a gas• Consider a cylinder with a moveable
piston• A force is applied to slowly compress the
gas– The compression is slow enough for
all the system to remain essentially in thermal equilibrium
– This is said to occur quasi-statically
ˆ ˆ dW d F dy Fdy PA dy PdV= ⋅ = − ⋅ = − = − = −F r j j
dW PdV= −
Work
• Interpreting dW = - P dV– If the gas is compressed, dV is negative and the
work done on the gas is positive– If the gas expands, dV is positive and the work
done on the gas is negative– If the volume remains constant, the work done
is zero• The total work done is: f
i
V
VW P dV= −∫
PV Diagrams
• Used when the pressure and volume are known at each step of the process
• The state of the gas at each step can be plotted on a graph called a PV diagram– This allows us to visualize the
process through which the gas is progressing
• The curve is called the path
f
i
V
VW P dV= −∫
PV Diagrams
• The work done on a gas in a quasi-static process that takes the gas from an initial state to a final state is the negative of the area under the curve on the PV diagram, evaluated between the initial and final states– This is true whether or not the pressure stays
constant– The work done does depend on the path taken
The First Law of Thermodynamics
• The First Law of Thermodynamics is a special case of the Law of Conservation of Energy– It takes into account changes in internal energy and energy
transfers by heat and work• Although Q and W each are dependent on the path, Q + W is
independent of the path
intE Q WΔ = +
Work Done By Various Paths
( )f f iW P V V= − −
f
i
V
VW P dV= −∫
( )i f iW P V V= − − ( )W P V dV= −∫
The work done depends on the path taken!
Not necessarily an isotherm!
f
i
V
VW P dV= −∫
HW Problem 21
Adiabatic Process
• An adiabatic process is one during which no energy enters or leaves the system by heat: Q = 0
– This is achieved by:• Thermally insulating the walls of the system• Having the process proceed so quickly that no heat can be exchanged
• Since Q = 0, ΔEint = W• If the gas is compressed adiabatically, W is positive so ΔEint is positive and the
temperature of the gas increases• If the gas expands adiabatically, the temperature of the gas decreases• Examples of adiabatic processes related to engineering are:
– The expansion of hot gases in an internal combustion engine– The liquefaction of gases in a cooling system– The compression stroke in a diesel engine– Adiabatic free expansion of a gas
• The gas expands into a vacuum, no piston: W = 0• Since Q = 0 and W = 0, ΔEint = 0 : initial and final states are the same,
no change in temperature is expected.
intE WΔ =
intE Q WΔ = + f
i
V
VW P dV= −∫
Special Case: Adiabatic Free Expansion
• This is an example of adiabatic free expansion
• The process is adiabatic because it takes place in an insulated container
• Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0
• Since Q = 0 and W = 0, ΔEint = 0 and the initial and final states are the same and no change in temperature is expected. – No change in temperature is expected
Isothermal Process• At right is a PV diagram of an isothermal
expansion• The curve is a hyperbola• The curve is called an isotherm
• The curve of the PV diagram indicates PV = constant– The equation of a hyperbola
• Because it is an ideal gas and the process is quasi-static, PV = nRT and
f f f
i i i
V V V
V V V
nRT dVW P dV dV nRTV V
= − = − = −∫ ∫ ∫
ln i
f
VW nRTV⎛ ⎞
= ⎜ ⎟⎜ ⎟⎝ ⎠
Isobaric Processes• An isobaric process is one that occurs at a
constant pressure• The values of the heat and the work are
generally both nonzero• The work done is W = P (Vf – Vi) where P
is the constant pressure
intE Q WΔ = + f
i
V
VW P dV= −∫PV nRT=
Isovolumetric Processes• An isovolumetric process is one in which there is
no change in the volume• Since the volume does not change, W = 0• From the first law, ΔEint = Q• If energy is added by heat to a system kept at
constant volume, all of the transferred energy remains in the system as an increase in its internal energy
intE Q WΔ = + f
i
V
VW P dV= −∫PV nRT=
Isothermal Process• An isothermal process is one that occurs at
a constant temperature• Since there is no change in temperature, ΔEint = 0
• Therefore, Q = - W• Any energy that enters the system by heat
must leave the system by work
intE Q WΔ = + f
i
V
VW P dV= −∫PV nRT=
Thermo Processes• Adiabatic
– No heat exchanged– Q = 0 and ΔEint = W
• Isobaric– Constant pressure– W = P (Vf – Vi) and ΔEint = Q + W
• Isovolumetric– Constant Volume– W = 0 and ΔEint = Q
• Isothermal– Constant temperatureΔEint = 0 and Q = -W
intE Q WΔ = +
ln i
f
VW nRTV⎛ ⎞
= ⎜ ⎟⎜ ⎟⎝ ⎠
Was ist das?
Fig. 20-9, p. 569
HW Problem 22
Fig. P20-22, p. 582
Cyclic Processes
• A cyclic process is one that starts and ends in the same state– On a PV diagram, a cyclic process appears as a
closed curve• If ΔEint = 0, Q = -W• In a cyclic process, the net work done on the
system per cycle equals the area enclosed by the path representing the process on a PVdiagram
intE Q WΔ = +
A gas is taken through the cyclic process as shown. (a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed—that is, the process follows the path ACBA—what is the net energy input per cycle by heat?
intE Q WΔ = +f
i
V
VW P dV= −∫
A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from Bto C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy E(B) – E(A).
intE Q WΔ = + f
i
V
VW P dV= −∫PV nRT=
HW Problem 28
Fig. P20-34, p. 582
HW Problem 34
