Physics 201Professor P. Q. Hung

311B, Physics Building

Physics 201 – p. 1/34

Thermodynamics

Thermodynamics: Study of physical systemsinvolving the transfer of something calledHeat.

Heat: Energy transferred between objectsbecause of temperature difference.

Temperature: Property of a body to be mademore precise below.

Thermal contact: Objects are in thermalcontact when heat flows between them.

Thermal equilibrium: Objects are in thermalequilibrium with each other when they reachthe same temperature.

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Thermodynamics

Thermodynamics: Study of physical systemsinvolving the transfer of something calledHeat.

Heat: Energy transferred between objectsbecause of temperature difference.

Temperature: Property of a body to be mademore precise below.

Thermal contact: Objects are in thermalcontact when heat flows between them.

Thermal equilibrium: Objects are in thermalequilibrium with each other when they reachthe same temperature.

Physics 201 – p. 2/34

Thermodynamics

Thermodynamics: Study of physical systemsinvolving the transfer of something calledHeat.

Heat: Energy transferred between objectsbecause of temperature difference.

Temperature: Property of a body to be mademore precise below.

Thermal contact: Objects are in thermalcontact when heat flows between them.

Thermal equilibrium: Objects are in thermalequilibrium with each other when they reachthe same temperature.

Physics 201 – p. 2/34

Thermodynamics

Thermodynamics: Study of physical systemsinvolving the transfer of something calledHeat.

Heat: Energy transferred between objectsbecause of temperature difference.

Temperature: Property of a body to be mademore precise below.

Thermal contact: Objects are in thermalcontact when heat flows between them.

Thermal equilibrium: Objects are in thermalequilibrium with each other when they reachthe same temperature.

Physics 201 – p. 2/34

Thermodynamics

Thermodynamics: Study of physical systemsinvolving the transfer of something calledHeat.

Heat: Energy transferred between objectsbecause of temperature difference.

Temperature: Property of a body to be mademore precise below.

Thermal contact: Objects are in thermalcontact when heat flows between them.

Thermal equilibrium: Objects are in thermalequilibrium with each other when they reachthe same temperature. Physics 201 – p. 2/34

Thermodynamics

Zeroth Law of Thermodynamics

If bodies A and B are in thermal equilibrium witha third body C, then they are in thermalequilibrium with each other.

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Thermodynamics

The measurement of temperature andTemperature scales

Kelvin scale:International agreement for standard fixedpoint: point where liquid water, solid ice andwater vapor can coexist. This point is definedas having a temperatureT3 = 273.16KTriple fixed point.The size of the Kelvin is 1

273.16 of the differencebetween absolute zero and the triple-pointtemperature of water.

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Thermodynamics

The measurement of temperature andTemperature scales

Thermometer:Standard Thermometer: Constant VolumeGas Thermometer.Measure the pressure of the gas at the triplepoint. Call it P3. The pressure changes as thegas is heated up. Call that P . Thetemperature at that point will be proportionalto the ratio P/P3.

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Thermodynamics

The measurement of temperature andTemperature scales

Different gases give slightly different results atthe boiling point unless we put smaller andsmaller amount of gas until they converge toa single temperature. This givesT = (273.16 K)(limgas→0

PP3

)

Using this thermometer, the boiling point ismeasured to be at T = 373.15 K.

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Thermodynamics

The measurement of temperature andTemperature scales

Different gases give slightly different results atthe boiling point unless we put smaller andsmaller amount of gas until they converge toa single temperature. This givesT = (273.16 K)(limgas→0

PP3

)

Using this thermometer, the boiling point ismeasured to be at T = 373.15 K.

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Thermodynamics

The measurement of temperature andTemperature scales

Constant Volume Gas Thermometer

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Thermodynamics

The measurement of temperature andTemperature scales

Celsius scale: Defined to be 0 0C at thefreezing point and 100 0C at the boiling point.It is related to the scientific Kelvin scale byTC = T − 273.15 KSo the triple point in the Celsius scale isTC = 0.01 0C.

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Thermodynamics

The measurement of temperature andTemperature scales

Fahrenheit scale: The Fahrenheit scale isrelated to the Celsius scale byTF = 9

5TC + 320

At the freezing point, TC = 0 0C givingTF = 32 0F . At the boiling point, TC = 100 0Cgiving TF = 212 0F

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Thermodynamics

The measurement of temperature andTemperature scales

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Thermodynamics

Thermal expansion

Linear expansion:∆L = Lα ∆Tα is the linear coefficient of expansion givenin Table 16-1 for various material. It has theunit of K−1.

Area expansion:When α∆T � 1, one has∆A ≈ Aα ∆T

Volume expansion:∆V = V β ∆Twithβ = 3α

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Thermodynamics

Thermal expansion

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Thermodynamics

Thermal expansionPeculiar property of water (unlike other fluids):Above 4 0C, water expands as the temperaturerises so the density decreases. Between 0 0Cand 4 0C, water shrinks as the temperature risesso the density increases. This fact explains thebehavior of a frozen lake for example. See theexplanation on p. 550.

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Thermodynamics

Thermal expansion

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Thermodynamics

Thermal expansionOn a hot day in Las Vegas, an oil trucker loaded9785 gal of diesel fuel. He encountered coldweather on the way to Payso, Utah, where thetemperature was 41 0F lower than in Las Vegas,and where he delivered his load. How manygallons did he delivered? The coefficient ofvolume expansion for diesel fuel is 9.5 × 10−4/ 0Cand the coefficient of linear expansion for hissteel truck tank is 11 × 10−6/ 0C.

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Thermodynamics

Thermal expansion

∆V = βV ∆T = (9785gal)(9.5 ×

10−4/ 0C)(−41 0F )(5

9) = −212gal.

Amount delivered: Vdel = V + ∆V = 9600 gal.The steel tank is irrelevant for this problem.

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Thermodynamics

Thermal expansion

∆V = βV ∆T = (9785gal)(9.5 ×

10−4/ 0C)(−41 0F )(5

9) = −212gal.

Amount delivered: Vdel = V + ∆V = 9600 gal.The steel tank is irrelevant for this problem.

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Thermodynamics

Heat and mechanical work:

One kilocalorie (kcal) is defined as theamount of heat needed to raise 1 kg of waterfrom 14.5 0C to 15.5 0C.

Joule’s experiment showed that1 cal = 4.186 J

Heat is denoted byQ = energy transferred due to temperaturedifferences. Unit:J

Physics 201 – p. 17/34

Thermodynamics

Heat and mechanical work:

One kilocalorie (kcal) is defined as theamount of heat needed to raise 1 kg of waterfrom 14.5 0C to 15.5 0C.

Joule’s experiment showed that1 cal = 4.186 J

Heat is denoted byQ = energy transferred due to temperaturedifferences. Unit:J

Physics 201 – p. 17/34

Thermodynamics

Heat and mechanical work:

One kilocalorie (kcal) is defined as theamount of heat needed to raise 1 kg of waterfrom 14.5 0C to 15.5 0C.

Joule’s experiment showed that1 cal = 4.186 J

Heat is denoted byQ = energy transferred due to temperaturedifferences. Unit:J

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Thermodynamics

Heat Capacity and Specific HeatThe amount of heat energy Q needed to raisethe temperature of a substance is proportional tothe temperature change and to the mass of thesubstanceQ = C ∆T = cm ∆T

C: Heat capacity. Unit: J/K.

c: Specific Heat. It depends only on thesubstance. Unit: J/kg.K.

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Thermodynamics

Heat Capacity and Specific HeatThe amount of heat energy Q needed to raisethe temperature of a substance is proportional tothe temperature change and to the mass of thesubstanceQ = C ∆T = cm ∆T

C: Heat capacity. Unit: J/K.

c: Specific Heat. It depends only on thesubstance. Unit: J/kg.K.

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Thermodynamics

Heat Capacity and Specific Heat

Q > 0 ⇒ ∆T > 0: Heat added.

Q < 0 ⇒ ∆T < 0: Heat removed.cW = 4186 J/kg.K for water andcL = 128 J/kg.K for lead. More examples aregiven in Table 16-2.

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Thermodynamics

Heat Capacity and Specific Heat

Q > 0 ⇒ ∆T > 0: Heat added.

Q < 0 ⇒ ∆T < 0: Heat removed.cW = 4186 J/kg.K for water andcL = 128 J/kg.K for lead. More examples aregiven in Table 16-2.

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Thermodynamics

Specific Heat

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Thermodynamics

Heat Capacity and Specific HeatA 2 kg Aluminum block is originally at 10 0C. If36 kJ of heat energy are added to the block, whatis its final temperature?

For aluminum, c = 900 J/kg.K.

Q = cm ∆T = cm (Tf − Ti)

Tf = Ti + Qmc = 10 0C + 36kJ

2kg×0.9kJ/kg.0C = 30 0C.

Physics 201 – p. 21/34

Thermodynamics

Heat Capacity and Specific HeatA 2 kg Aluminum block is originally at 10 0C. If36 kJ of heat energy are added to the block, whatis its final temperature?

For aluminum, c = 900 J/kg.K.

Q = cm ∆T = cm (Tf − Ti)

Tf = Ti + Qmc = 10 0C + 36kJ

2kg×0.9kJ/kg.0C = 30 0C.

Physics 201 – p. 21/34

Thermodynamics

Heat Capacity and Specific HeatA 2 kg Aluminum block is originally at 10 0C. If36 kJ of heat energy are added to the block, whatis its final temperature?

For aluminum, c = 900 J/kg.K.

Q = cm ∆T = cm (Tf − Ti)

Tf = Ti + Qmc = 10 0C + 36kJ

2kg×0.9kJ/kg.0C = 30 0C.

Physics 201 – p. 21/34

Thermodynamics

CalorimetryLead shot of mass 600g is heated to 100 0C andplaced in an aluminum can of mass 200 g whichcontains 500 g of water initially at 17.3 0C. Thespecific heat of aluminum is 0.9kJ/kg.0C. Thefinal equilibrium temperature of the mixture is20.0 0C. What is the specific heat of lead?

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Thermodynamics

CalorimetryWe have assumed above that the aluminum cancontaining water is insulated, in the sense thatheat transfer only occurs between the containerand objects which are in thermal contact with it.The procedure is called calorimetry and thecontainer is called calorimeter. Since there is noheat transfer to the surroundings, the sum ofheat transfer between indidual componentsshould sum up to zero.∑

Qi = 0

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Thermodynamics

Calorimetry

From the above, one concludes:Qlead + Qaluminum + Qwater = 0.

Qlead = (0.6 kg)clead(20.00C − 100 0C) =

−(0.6 kg)clead 80.0 0C.

Qaluminum =(0.2 kg)(0.900 kJ/kg.0C)(20.0 0C − 17.3 0C) =(0.2 kg)(0.900 kJ/kg.0C)(2.7 0C) = 0.486 kJ .

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Thermodynamics

Calorimetry

From the above, one concludes:Qlead + Qaluminum + Qwater = 0.

Qlead = (0.6 kg)clead(20.00C − 100 0C) =

−(0.6 kg)clead 80.0 0C.

Qaluminum =(0.2 kg)(0.900 kJ/kg.0C)(20.0 0C − 17.3 0C) =(0.2 kg)(0.900 kJ/kg.0C)(2.7 0C) = 0.486 kJ .

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Thermodynamics

Calorimetry

From the above, one concludes:Qlead + Qaluminum + Qwater = 0.

Qlead = (0.6 kg)clead(20.00C − 100 0C) =

−(0.6 kg)clead 80.0 0C.

Qaluminum =(0.2 kg)(0.900 kJ/kg.0C)(20.0 0C − 17.3 0C) =(0.2 kg)(0.900 kJ/kg.0C)(2.7 0C) = 0.486 kJ .

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Thermodynamics

Calorimetry

Qwater =(0.5 kg)(4.18 kJ/kg.0C)(20.0 0C − 17.3 0C) =(0.5 kg)(4.18 kJ/kg.0C)(2.7 0C) = 5.64 kJ .

Using the equation in (1), one obtainsclead = 6.13 kJ

48.0kg.K = 0.128 kJ/kg.K.

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Thermodynamics

Calorimetry

Qwater =(0.5 kg)(4.18 kJ/kg.0C)(20.0 0C − 17.3 0C) =(0.5 kg)(4.18 kJ/kg.0C)(2.7 0C) = 5.64 kJ .

Using the equation in (1), one obtainsclead = 6.13 kJ

48.0kg.K = 0.128 kJ/kg.K.

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Thermodynamics

Conduction, Convection, RadiationConduction:Heat energy can be transferred from one point toanother point in a given material. This is calledthermal conduction. Experimentally, the rate ofconduction is found to be

Pcond = Qt = k A ∆T

L

A= area; L = length.

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Thermodynamics

Conduction, Convection, RadiationConduction:Heat energy can be transferred from one point toanother point in a given material. This is calledthermal conduction. Experimentally, the rate ofconduction is found to be

Pcond = Qt = k A ∆T

L

A= area; L = length.

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Thermodynamics

Conduction, Convection, Radiation

k = thermal conductivity. A constant whichdepends on the material. A good thermalconductor has high k, such as silver forexample. Air is a poor thermal conductor.Also woods, etc..

In building construction, one needs a poorthermal conductor i.e. a good thermalinsulator. Define a heat resistance R = L

k .This is also called the R-value. Low k ⇒ HighR. For buildings in cold climates, usually anR-value of around 30 is desirable.

Physics 201 – p. 27/34

Thermodynamics

Conduction, Convection, Radiation

k = thermal conductivity. A constant whichdepends on the material. A good thermalconductor has high k, such as silver forexample. Air is a poor thermal conductor.Also woods, etc..

In building construction, one needs a poorthermal conductor i.e. a good thermalinsulator. Define a heat resistance R = L

k .This is also called the R-value. Low k ⇒ HighR. For buildings in cold climates, usually anR-value of around 30 is desirable. Physics 201 – p. 27/34

Thermodynamics

Conduction

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Thermodynamics

Conduction

Physics 201 – p. 29/34

Thermodynamics

Conduction, Convection, Radiation

Convection:Take a heat source such as a candle. As airaround it heats up, it expands and its densitydecreases. Buoyant force makes it rise.Cooler air flows down to take its place. All ofthis creates a phenomenon called convectioncurrent.

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Thermodynamics

Conduction

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Thermodynamics

Conduction, Convection, Radiation

Radiation:Another form of heat transfer is by radiatingelectromagnetic waves. This form of energytransfer is called thermal radiation. Nomedium is required for heat transfer bythermal radiation because electromagneticwaves can propagate even in vacuum.

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Thermodynamics

Conduction, Convection, RadiationYou feel hot standing in front of a fire is becauseyou absorb thermal radiation coming from thefire.The rate at which an object emits energy viaelectromagnetic radiation is given by theStefan-Boltzman law

Prad = eσ AT 4

e: emissivity

σ = 5.6703 × 10−8W/m2.K4: Stefan-Boltzmanconstant

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Thermodynamics

Conduction, Convection, RadiationYou feel hot standing in front of a fire is becauseyou absorb thermal radiation coming from thefire.The rate at which an object emits energy viaelectromagnetic radiation is given by theStefan-Boltzman law

Prad = eσ AT 4

e: emissivity

σ = 5.6703 × 10−8W/m2.K4: Stefan-Boltzmanconstant

Physics 201 – p. 33/34

Thermodynamics

Conduction, Convection, RadiationYou feel hot standing in front of a fire is becauseyou absorb thermal radiation coming from thefire.The rate at which an object emits energy viaelectromagnetic radiation is given by theStefan-Boltzman law

Prad = eσ AT 4

e: emissivity

σ = 5.6703 × 10−8W/m2.K4: Stefan-Boltzmanconstant

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Thermodynamics

ConductionIdeal Reflector

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