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Thermodynamics
Chapter 17: Temperature and heatp p(a macroscopic study)Chapter 18: Thermal properties of matter p p p(a microscopic study)Chapter 19: The first law of thermodynamicsp yChapter 20: the second law of thermodynamics
Thermodynamics: study of energyThermodynamics: study of energy transformation involving heat, mechanical work, and other aspects of energy and how these transformations relate to the properties of matter.
Homework problems from thUniversity physics 12th edition
• 28-36-49-52-68-109-11928 36 49 52 68 109 119
“Hot” and “Cool”Hot and Cool
• How can we quantify concepts of “hot” andHow can we quantify concepts of hot and “cold”?
• What are the microscopic and macroscopic p pmanifestations of “hot” and “cold”?
• Temperature is a macroscopic measure for p p“hotness” and “coldness” of an object.
• Kinetic energy of the molecules is a microscopic measure for “hotness” and “coldness” of an object.
How can we quantify concepts of “h ” d “ ld”?“hot” and “cold”?
• We need to measure those properties of the matter thatWe need to measure those properties of the matter that change during a heating or cooling process. – Length of a metal changes by heat (L).– Volume of a liquid changes by heat (V).– Pressure of a constant-volume gas changes with
heat (p).– Resistance of a conducting wire changes with heat
(R)(R).– Emission spectrum of the material change by heat.
• How can we use these properties to make a• How can we use these properties to make a thermometer?
Temperature measurement:
• How is it guaranteed that these systems are actually measuring temperature of the systems under study?
Other kinds of thermometer
• Bimetallic strip: uses the pdifferences in thermal expansion of material.A th il i l t i d i• A thermopile is an electronic device composed of thermocouples connected in series or parallel that converts thermal energy into electrical energy. It uses the infrared radiation emitted by theinfrared radiation emitted by the objects and generates an electrical voltage (diode). This is a non-contact t h i tl d i di ltechnique mostly used in medical thermometers.
A US Patent: Microencapsulated cholesteric liquid crystal temperature measuring device for determining y p g g
the temperature of non-planar or planar surfaces
• A temperature measuring device engageable with a surface of any contour to determine the individual temperature of each of the several portions of the surface by providing a thermal map. The map is observable and may be photographed if a record thereof is desired. The temperature measuring device has a portion engageable with the surface and conformable to thedevice has a portion engageable with the surface and conformable to the surface. The portion which engages the surface includes a coating of cholesteric material, such as microencapsulated liquid crystal material which is temperature-sensitive and light-reflecting. The device also includes
t t fill t i l th ithi t t it th th l ia transparent fill material therewithin to transmit the thermal image or map therethrough. Thus, a thermal map or image of the surface is provided. The thermal map is observable and/or recordable through another portion of the device.
• cholesteric Material: A liquid crystal material in which the elongated molecules are parallel to each other within the plane of a layer, but the direction of orientation is twisted slightly from layer to layer to form a helix h h h lthrough the layers.
The zeroth law of thermodynamics• If two systems (A and B) are in thermal equilibrium with a• If two systems (A and B) are in thermal equilibrium with a
third system (C), then they are at thermal equilibrium with each other.
• Two systems are in thermal equilibrium only if they have• Two systems are in thermal equilibrium only if they have the same temperature.
• Zeroth law guarantees that thermometers measure temperature of the bodies in contact with themtemperature of the bodies in contact with them.
Temperature scale
Water boiling: 1000C
Celsius scaleFahrenheit scale
Water boiling: 2120F
100 degrees180 degrees
Water freezing: 00CWater freezing: 320F0
0 0 100100 180 1oC F F C⎛ ⎞⎜ ⎟
0 0
00 0 0
100 180 1180
51 ; and 0 occurs at 32
oC F F C
F C C F
⎛ ⎞= → = ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟1 ; and 0 occurs at 329
F C C F= ⎜ ⎟⎝ ⎠
0F C
9T = T +32
5F C
0C F
55
T = (T -32 )9
Example: Fahrenheit and Celsius lscales
• Find the temperature that Fahrenheit andFind the temperature that Fahrenheit and Celsius scales agree (-400).
Gas thermometers and Kelvin saleGas thermometers and Kelvin sale• Usually the properties that are used in
th t t li th tithermometers are not linear over the entire scale of measurable temperatures.
• We need a material-independent• We need a material-independent temperature scale.
• Gas thermometers offer such a property. p p y• Pressure of an ideal gas kept in constant
volume changes linearly with temperature (PV RT)(PV=nRT).
• Kelvin is a temperature scale defined based on pressure of an ideal gason pressure of an ideal gas.
• Name an ideal gas.
Finding the absolute temperature (Kelvin scale)(Kelvin scale)
• Using a constant-volume gas thermometer we can Diff tPgas thermometer we can find the “ultimate cold” or absolute zero -273.15K l i l i t bli h d
Different gasses1
23
P
• Kelvin scale is established based on absolute zero and the temperature rise
Liquid found by extrapolation Measured
temperaturespbetween two degrees is defined to match the Celsius scale (T0C)0 200100-100-200273 15
p
Celsius scale. • 10C=1K & TK=TC+273.15• What happens beyond
(T0C)
T(K)
0 200100100200
300 4002001000
273 15
-273.15
pp y-273.15? Is it achievable!
273.15
Pressure and temperature1 2
1 1 2 2
Let us have some ideal gas trapped in a container heated from to and
T TpV nRT p V nRT= =
2 2
1 1
T in Kelvin unknownunknown known
known
pT p T TT p p
= → =
We need to know temperature of at least one unique and reproducable point to use
this relation. If we had such a point then we could read the
d l l t th t tpressure and calculate the temperature.
HeatHeat
Triple point of water as a reference point for gas thermometerspoint for gas thermometers
Triple point of water is a combination of temperature and pressure
0
at which solid water (ice), liquid water, and vapor can all coexist. It occurs at 0.01 C. We call this temperature 273.16 kelvin.
triT 273.16 then we have
0.6117ple
triple
K
p KPa
=
=
unknownunknown triple
triple
pT Tp
=
00 273.16 when 0 T K C p= = − =When this can happen?
Some interesting temperatures in h lthree scales
All positive values for temperatureKelvin is a scale defined based onKelvin is a scale defined based on kinetic energy and negative kinetic energy does not mean much.
Thermal expansion• Most material expand when their temperature
increases Why?increases. Why?
212
KE kx=2
Linear thermal expansion
0
Objects expand as heated. &
hL T L L αΔ ∝ Δ Δ ∝ →
one - dimensional objects :
0ΔL= LΔT
where -f iT T TΔ =
0 is the original length of an object at temperatue is the change in length due to a change in
iL TL TΔ Δ temeperature
-1 0 1
0 0
is the coefficient of linear expansion. Units are K or (C )
Final length: (1 )L L L L T
α
α
−
= + Δ = + Δ
L0
L
LΔ
Thermal expansionObj d i h di i h d
0 & 3 V V Tβ β αΔ = Δ =Objects expand in three - dimensions as heated
0 0
0
Final volume: (1 )
V initial volume
V V V V Tβ= + Δ = + Δ
0
-1 0 1
change in temperatureis the coefficient of linear expansion Units are K or (C )
Tβ −
Δ
is the coefficient of linear expansion. Units are K or (C )β
and are not constant at all temperatures.
E i f t i l i t i Wh ?
α β
Expansion of some material is assymetric. Why?What happens if we keep heating up the material
k hi i ?or keep streatching a spring?
Which material would you choose as a mountyou choose as a mount for a glass lens? Which one for the Quartz?
How objects expand? Whi h i ?Which one is correct?
Unique behavior of h l i fthermal expansion of water
• Why ice floats on
Negative thermal
ywater?
• If volume decreases expansion coefficientVolume decreases as temperature increases
with temperature density of ice should b hi h th th t temperature increasesbe higher than that of water. Why it is not?not?
ExamplesExamples• What is the exact length of a steel measuring tape g g p
at 350C if its length is 50.000m at 200C? (50.009 m)If di i d b hi h• If a distance is measured by this tape on a hot (350C) summer day has a value of 35.794 m, what is the actual distance? The tape is calibrated foris the actual distance? The tape is calibrated for use at 200C.
• A glass flask with volume 200cm3 is filled with mercury at 200C. How much mercury overflows when the temperature is raised to 1000C? The coefficient of linear expansion for glass is 0 40x10-coefficient of linear expansion for glass is 0.40x105K-1 and for mercury is 6.0x10-5K-1. (2.7 cm3)
Summary I
Thermal stress IHeating/cooling an object without allowing it to expand/contract will cause thermal stressHeating/cooling an object without allowing it to expand/contract, will cause thermal stress.
Tensile stress Force per unit area /Recall: Young's modulusTensile strain Fractional change in length /
F AYL
= = → =Δ 0L
S i
0
a) the amount the rod would expand (or contract) had it been free: thermal
L TL
α⎡ ⎤Δ
= Δ⎢ ⎥⎣ ⎦
To calculate Stress in a clamped rod we calculate :
0
b) the amount of stress needed to compress (or stretch) thermal⎣ ⎦
it back to its original length.
/ /
F A L FYL L L AY
⎡ ⎤Δ= → =⎢ ⎥Δ ⎣ ⎦0 0/
c) If length is to be constant, the total fractional change in length must be zero:
L + 0
tensionL L L AY
L FT
⎢ ⎥Δ ⎣ ⎦
⎡ ⎤ ⎡ ⎤Δ ΔΔ +⎢ ⎥ ⎢ ⎥
0 0thermal
+ 0L
Thetension
TL AY
α= Δ + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
result is an equation for stress caused by temperature change or thermal stress:
FFA
Y Tα= − Δ
Thermal Stress IIF
0
F A
is the temperature change ( C)
Y T
T T T
α= − Δ
Δ = −2 12
0 -1
is the temperature change ( C),
Y is Young's modulus (N/m or Pa), is the linear thermal expansion coefficient ( )
T T T
Cα
Δ =
is the linear thermal expansion coefficient ( ),F is the compress
Cα
2
ive or tensile force (N), A is cross sectional area of the object (m )A is cross sectional area of the object (m ). When an object's temperature changes, to keep its length constant:
FFwhile heatinng 0 0 compressive forces are needed
hil
TA
Δ > → <
Fli 0 0 t il f d dTΔ < → >while cooling 0 0 tensile forces are neededTA
Δ < → >
Example 17 30Example 17.30
• A brass rod is 185 cm long and 1.6 cm inA brass rod is 185 cm long and 1.6 cm in diameter. What force must be applied to each end of the rod to prevent it from pcontracting when it is cooled from 1200C to 100C? Young’s Modulus for brass is 9.0 1010 P d li i ffi i tx1010 Pa and linear expansion coefficient
is 2x10-5 (C0) -1 . (answer: 40000N). If th d f t t t h l• If the rod was free to contract how long would it be at 100C?
Heat and mechanical energy•Joule 1818 1889 understood mechanical energy can be•Joule 1818-1889 understood mechanical energy can be converted to heat. Examples? •We can conclude: heat is a form of energy.
Water heats up as a result of the mechanical work of the paddle on it.
Water heats up as a result of the thermal energy trasfare to it .
Heat and temperatureHeat and temperature• Heat and temperature are not the same
Temperature depends on the physical state of a material– Temperature depends on the physical state of a material.– Heat is the amount of thermal energy needed to bring a
system to a certain temperature. Heat is a relative quantity. W ’t d t i th b l t t f th h tWe can’t determine the absolute amount of the heat a system has. But we can determine the amount of heat needed to bring a system from one state (T1) to another t t (T )state (T2).
• If we add heat to a system we can change its temperaturetemperature.
• If we divide mass of a system to half, the amount of heat divides to half but its temperature stays the ysame.
Quantity of heat• One Calorie (cal) is the amount of heat that raises
the temperature of one gram of water from 14.5 0C to 15 5 0Cto 15.5 0C
• One British Thermal Unit (BTU) is the amount of energy that required to raise temperature of oneenergy that required to raise temperature of one pound water by 1 0F from 63 0F to 64 0F
• Conversion of thermal and mechanical energy:– 1 cal=4.18 J (Joule or N.m or kg.m2/s2 or Pa. m3)– 1 kcal= 1000 cal =4180J
1 BTU=778 ft lb=252 cal=1055 J– 1 BTU=778 ft.lb=252 cal=1055 J• Calorie or Btu are not SI units. • The fundamental SI unit for all forms of energy is• The fundamental SI unit for all forms of energy is
Joule (N.m)
Specific heat and Molar heat capacityAmount of heat, required to increase temperature of a mass is proportional to amount of temperature change - and mass:f i
Q mT T TΔ =
For infinitesimal change
Q m T Q mc T dQ mcdT∝ Δ → = Δ → = in temperature is proportional to
i h fdQ dT
, is the amount of enrgy needed to change temperature of the unit mass of a material by
1
c
dQ
Specific heat of the material,
1one temperature unit:
We can als
dQcm dT
=
o write m nM=23M is the molar weight or weight of 6.022 x 10 molecules of a substance
n is number of moles. Using we define : 1
dQ mcdTdQ
= molar heat capacity C1 and then dQC Mcn dT
= = Q nC T= Δ
Specific heat is not constant over all temperature rangesall temperature ranges.
• Example: specific heat of water as a function of temperatureS ifi h t f t t t t 4190 J/(k K)• Specific heat of water at room temperature: 4190 J/(kg.K) or 1 cal/g. 0Cor 1 Btu/lb. 0F
• Calculate molar heat capacityCalculate molar heat capacity of water at room temperature.C=McC=(0 0180 kg/mol)(4190 J/kg K)C=(0.0180 kg/mol)(4190 J/kg.K)C=75.4 J/mol.KCaution: heat is energy in transit It does not reside in atransit. It does not reside in a matter so heat capacity does not mean energy is contained in the matter It is the energy needed tomatter. It is the energy needed to change state of the matter
Example: 17 7Example: 17.7
• You are designing an electronic circuitYou are designing an electronic circuit made of 23 mg silicon. The electric current through it adds energy at the rate of 7 4through it adds energy at the rate of 7.4 mW (how many J/s?). If your circuit does not allow any heat transfer out of thenot allow any heat transfer out of the element, at what rate does its temperature increase? The specific heat of Si is 705increase? The specific heat of Si is 705 J/kg.K. (dT/dt=0.46)
Calorimetry (measuring heat) and Phase changesand Phase changes
• Heat transfer in and out of a system (containing material) can causematerial) can cause– temperature change– phase change of the materialphase change of the material
• Phase is specific state of the matter for example solid, liquid, etc.solid, liquid, etc.
• For any given pressure, phase change or phase transition occur at a specific temperature p p
• Phase transition involves absorption or emission of energy followed by change in state of the matter, its gy y gvolume, and its density.
Heating/cooling phase transition and temperature changephase transition and temperature change
an experiment
Heat of fusion and vaporizationQ tit f h t t f d i h h iQuantity of heat transfered in a phase change is:
Q mL= ±
The amount of heat reuired
to change state of the nit mass of a material fromfLHeat of fusion or :
to change state of the unit mass of a material from solid to lqiuid.
vLHeat of vaporization or : The amount of heat reuired to change state of the unit mass of a material
g
from lqiuid to vapor.
L LB th d d d thv fL LBoth and depend on the pressure.
Phase transition and reversibilityreversibility
is a condition during a pahse Pahse equilibrium g ptransition that more than one state of the matter can coexist at that constant temperature and pressure
q
can coexist at that constant temperature and pressure. If a phase transition costs Q amount of heat, the revrse
For example liquid to vapor transition costs the same transition releases the same Q amount of heat.
p q p
as vapor to liquid transition releases.
QuestionsQuestions
• Why boiling temperature of water is different atWhy boiling temperature of water is different at different elevations?
• Why temperature of water/ice stays the same while y p ymelting or freezing? For water at atmospheric pressure:
5
p p3.34 10 / ; fL J kg= ×
6
5
2.25 10 / ;
4 19 10 /vL J kg
Q J k
= ×
05
0 1004.19 10 /
CQ J kg
→= ×
Example 17.8:Temperature change i h h hwith no phase change
• An aluminium cup has a mass of 0 120 kgAn aluminium cup has a mass of 0.120 kg and it is initially at 200C. You pour 0.300kg of water at 700C into the cup What is theof water at 70 C into the cup. What is the final temperature after the equilibrium is attained? Assume no heat exchange withattained? Assume no heat exchange with outside world. (T=66.00C)
Example 17.9: changes in both d htemperature and phase
• You want to cool 0 25 kg of diet colaYou want to cool 0.25 kg of diet cola (mostly water) initially at 250C, by adding ice initially at -200C How much ice shouldice initially at 20 C. How much ice should you add so that the final temperature is 00C with all the ice melted Neglect heat0 C with all the ice melted. Neglect heat capacity of the container.
Specific heat and molar heat capacitiescapacities
Heat of fusion and vaporizationHeat of fusion and vaporization
Summary IISummary II
Mechanisms of heat transferMechanisms of heat transfer
• How heat is transferred from one point toHow heat is transferred from one point to another.
• Conduction (bodies in contact)• Conduction (bodies in contact)• Convection (some material mediates the
t f )energy transfer)• Radiation (photons or waves carry the
energy)• Do we always need material for heat y
transfer?
How conduction happens?
• Why it feels colder sitting on a steel bench on a ld d i k d t itti dcold day in a park compared to sitting on a wooden
bench? Do they have different temperatures?
• Is there any difference between the mechanisms of heat conduction in a metal (steel) and anof heat conduction in a metal (steel) and an insulator (wood)?
Heat conductionHeat conductionH CT TdQ dTH kA kA−
= = = − H kA kAdt L dx
= = =
How conduction happens?
High Low
Direction of heat flow in an object with its one end at T and other end at T .
A is the cross sectional area L is the length. Rate of heat flow or heat current, dQ/dt is proportional to:
dQH ∝1& &A T TQH
dt= ∝ & &
or
H CA T TL
dQ TH
−
Δ= ∝ the temperature gradientor
H C
Hdt x
T TdQ dTH kA kA
= ∝Δ
−= = =
the temperature gradient
H kA kAdt L dx
= = = −
How conduction happens?
H CT TdQ dTH kA kAdt L dx
−= = = −
(unit: Watt or J/s or Btu/h) Negative sign is to assure heat always flows from hot to cold. H is heat current in conduction
g g yonstant of proportionalk is thermal conductivity (c ity,
unit: W/m.K). u t: W/ . ).Engineers difine :
( )A T TL
thermal resistance
( ) and H becomes: H CA T TLR Hk R
−= =
Problem solvingProblem solving
• Identify heat transfer mechanisms.Identify heat transfer mechanisms. • If there is more than one mechanism, calculate
them separately and then add them up.p y p• Identify direction of heat flow, L, and A.• Identify target variablesIdentify target variables• Pay attention to the signs. Heat gain is positive
and heat loss is negative for each system.a d eat oss s egat e o eac syste• Be consistent with units.• Evaluate the resultsEvaluate the results.
Conduction through two barsll l d iparallel and series
• A steel bar 10.0 cm long is welded to a copper bar g pp20.0cm long. Both bars are insulated perfectly on their sides. Each bar has a square cross section,
f f2.00 cm on a side. The free end of the steel bar is maintained at 1000C and the other end of copper bar is maintained at 00C Find the temperature at thebar is maintained at 00C. Find the temperature at the junction of the two bars and the total rate of heat flow. (T=20.7 00C, Hsteel=15.9 W, Hcopper=15.9 W)flow. (T 20.7 0 C, Hsteel 15.9 W, Hcopper 15.9 W)
• Suppose the two bars are separated. One end of each bar is maintained at 1000C and the other end at 00C. What is the total rate of heat flow in the two bars? (Htotal=91.7 W)
Conduction through two barsll l d iparallel and series
Compare and discuss the results of the two problemsCompare and discuss the results of the two problems.
Convection
Most important for heat transfer in fluids, gasses and liquids.Examples: flow of blood in the body, cooling fluid in engine, heating system of most buildingsheating system of most buildings. Forced convection: circulation forced by a pump (blood flow)Natural convection: or free convection (atmosphere)
Convection & Fluid mechanics• Heat current is directly proportional to the surface
area. • Viscosity slows the natural convection near• Viscosity slows the natural convection near
stationary surfaces with formation of a stationary insulating thin film. F d ti kill th t ti fil l di• Forced convection kills the stationary film leading to faster heat exchange (wind-chill factor)
• Heat current due to convection is proportional toHeat current due to convection is proportional to the ΔT5/4 where ΔT is the temperature difference between the main body of the fluid and its surface.
Hot bodyT1
Cold bodyT2T1
RadiationRadiated heat current from a body at temperature T:
H=dQ/dt=AeσT4
– A is surface area from which radiation happensA is surface area from which radiation happens
– T is absolute temperature in Kelvin scale
– e is emissivity, a property of the material, that is a measure of the b b d di ti itt d b i t i l 0 1 1 f d ll bl kabsorbed radiation emitted by a given material. 0<e<1, e~1 for dull black
material, and e~0 for absolute reflectors.
– σ=5.670400 x10-8 W/m2.K4 is the Stephan Boltzmann constant
• Radiation flow is not directional like conduction or convection.
• Everybody at any temperature radiates heat to its surroundings in all direction, some less and some more.
• The net heat exchanged between a body at temperature T and its surrounding at temperature T is:its surrounding at temperature Ts is:
4 4 4 4( )net s sH Ae T Ae T Ae T Tσ σ σ= − = −
Radiation
Example: radiationExample: radiation• Total surface area of a human body is ~1.20m2
d th f t t i 300C Fi d thand the surface temperature is 300C. Find the total rate of radiation of energy from the body If th di t t t f 200C• If the surroundings are at temperature of 200C, what is the net rate of heat loss by the body by radiation? Emissivity of the human body is veryradiation? Emissivity of the human body is very close to unity.
• How many calories we need to consume per• How many calories we need to consume per day to keep up with this rate of energy loss?
• Do we loose heat from our body due to other• Do we loose heat from our body due to other mechanisms? What are those mechanisms?
Summary IIISummary III
