Physics 1710—Warm-up Quiz In Antarctica the thermometer reads –20 C; what is this temperature on a Fahrenheit thermometer? A.-36 o F B.-11 o F C.-4 o

Physics 1710Physics 1710—Warm-up Quiz—Warm-up QuizIn Antarctica the thermometer reads In Antarctica the thermometer reads –20 C; what is this temperature on a Fahrenheit –20 C; what is this temperature on a Fahrenheit thermometer? thermometer?

A B C D E

6%

18%

0%

18%

57%A.A. -36 -36 ooFFB.B. -11 -11 ooFFC.C. -4 -4 ooFFD.D. -43 -43 ooFFE.E. -20 -20 ooFF

121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140

Physics 1710Physics 1710 C Chapter 20 Heat & 1hapter 20 Heat & 1stst Law of Law of ThermoThermo

Solution:Solution:

ooF = (180 F = (180 ooF/100 F/100 ooC) C) ooC +32C +32ooFF

=(1.8 =(1.8 ooF/F/ooC)(-20 C)(-20 ooC) +32C) +32ooF F = (-36 +32) = (-36 +32) ooF F = -4= -4ooFF

No Talking!No Talking!Think!Think! Confer!Confer!

Peer Instruction Peer Instruction TimeTime


What will happen to a heated ring? The What will happen to a heated ring? The hole willhole will

(1) get larger; (2) get smaller; (3) stay the (1) get larger; (2) get smaller; (3) stay the same.same.

What will happen to a heated ring? The hole What will happen to a heated ring? The hole willwill

Physics 1710 — Physics 1710 — e-Quize-Quiz

Answer Now !

1 2 3

71%

6%

24%

1.1. Get largerGet larger

2.2. Get smallerGet smaller

3.3. Stay the sameStay the same

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Time for Real Physics!Time for Real Physics!


Why did it happen?Why did it happen?


ΔL = α LΔT

ΔL

ΔL

Physics 1710Physics 1710Chapter 19 TemperatureChapter 19 Temperature

Ideal Gas LawIdeal Gas Law• Idealizations: Idealizations:

no interaction between atomsno interaction between atomsno volume occupied by atomsno volume occupied by atoms

• Number of atoms n = m/M, Number of atoms n = m/M, m = mass; M = molar massm = mass; M = molar mass

• PV = n R TPV = n R TR = 8.315 J/ mol K R = 8.315 J/ mol K = 0.08214 L= 0.08214 L‧‧atm/mol Katm/mol K = 22.4 L= 22.4 L‧‧atm /273.16 mol Katm /273.16 mol K


Boltzmann’ ConstantBoltzmann’ Constant• k = R/Nk = R/NAA

• k = 1.38 x 10 k = 1.38 x 10 -23-23 J/K = 1 yJ/ 7.25 K J/K = 1 yJ/ 7.25 K

= 1 eV / 11,600 K= 1 eV / 11,600 K

PV = N kTPV = N kT


Summary:Summary:•Temperature is a measure of the average Temperature is a measure of the average kinetic energy of a system of particles.kinetic energy of a system of particles.

• Thermal Equilibrium means that two bodies Thermal Equilibrium means that two bodies are at the same temperature.are at the same temperature.

• The “Zeroth Law of Thermodynamics” states The “Zeroth Law of Thermodynamics” states that if system A and B are n thermal equilibrium that if system A and B are n thermal equilibrium with system C, then A and B are in thermal with system C, then A and B are in thermal Equilibrium with each other.Equilibrium with each other.


•Kelvin is a unit of temperature where Kelvin is a unit of temperature where one degree K is 1/279.16 of the one degree K is 1/279.16 of the temperature of the triple point of water temperature of the triple point of water (near freezing).(near freezing).

TTC C = (100/180) (T= (100/180) (TFF – 32 – 32 ⁰F⁰F))

TTF F = (180/100) T= (180/100) TC C + 32 + 32 ⁰F⁰F• ∆∆L/L = L/L = αα∆T∆T• PV = n R T = N kTPV = n R T = N kT

11′ Lecture:′ Lecture:

• The internal energy is the total average energy of The internal energy is the total average energy of the atoms of an object, average kinetic plus the atoms of an object, average kinetic plus average potential.average potential.

• Heat is the change in internal energy.Heat is the change in internal energy.

• The change in temperature is proportional to the The change in temperature is proportional to the change in internal energy (heat flow) when there is change in internal energy (heat flow) when there is no change of phase and the system does no work.no change of phase and the system does no work.

• The first law of thermodynamics states The first law of thermodynamics states ∆∆E = ∆Q - WE = ∆Q - W


11′ Lecture:′ Lecture:

• Conduction is the flow of kinetic energy from Conduction is the flow of kinetic energy from atom to atom.atom to atom.

•Convection is the transport of energy by bulk Convection is the transport of energy by bulk motion of atoms.motion of atoms.

• Radiation is the transfer of energy by Radiation is the transfer of energy by electromagnetic waves.electromagnetic waves.


HeatHeat• Heat is the macroscopic manifestation Heat is the macroscopic manifestation

of microscopic internal energy.of microscopic internal energy.

• Heat Heat ∆Q in calories (or BTU)∆Q in calories (or BTU)1 calorie (cal) is the amount of energy 1 calorie (cal) is the amount of energy

required to raise the temperature of required to raise the temperature of 0.001 kg0.001 kg of of water from 14.5 C to 15.5 C water from 14.5 C to 15.5 C ((∆T = 1.00 C).∆T = 1.00 C).

1 BTU is the heat to raise 1 Lb by 1 ⁰F.1 BTU is the heat to raise 1 Lb by 1 ⁰F.

• James Prescott Joule (1818-1889) showed James Prescott Joule (1818-1889) showed 1 calorie of heat = 4.186 Joule1 calorie of heat = 4.186 Joule

1 J = 1 N1 J = 1 N‧‧mm


1 Calorie = 1000 calorie = 1 kcal1 Calorie = 1000 calorie = 1 kcal


Heat Capacity Heat Capacity ∆∆Q = C ∆TQ = C ∆TC ≡ dQ/dTC ≡ dQ/dT

Specific HeatSpecific Heatc =c = C /m C /m

C ≡ (1/m)dQ/dTC ≡ (1/m)dQ/dT

∆∆Q = (mc)∆TQ = (mc)∆T


How much heat is required to raise the How much heat is required to raise the temperature of 1 kg of water (1 liter) from temperature of 1 kg of water (1 liter) from 20 C to 100 C? Recall c = 1.00 kcal/kgC?20 C to 100 C? Recall c = 1.00 kcal/kgC?

Physics 1710 — Physics 1710 — e-Quize-Quiz

Answer Now !

1 2 3 4 5

4%

21%

4%0%

71%

1.1. 1.0 kcal1.0 kcal

2.2. 20 kcal 20 kcal

3.3. 80 kcal80 kcal

4.4. 100 kcal100 kcal

5.5. None of the above.None of the above.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Guided practice:Guided practice:∆∆Q = (mc)∆TQ = (mc)∆T

How much heat is required to raise the How much heat is required to raise the temperature of 1 kg of water (1 liter) from temperature of 1 kg of water (1 liter) from 20 C to 100 C? Recall c = 1.00 kcal/kgC?20 C to 100 C? Recall c = 1.00 kcal/kgC?

∆∆Q = (mc)∆TQ = (mc)∆T

∆∆Q Q = (1.00 kg)(1.00 kcal/kg C) (100. C- = (1.00 kg)(1.00 kcal/kg C) (100. C- 20. C)20. C)

∆∆Q = 80. kcal = 80.kcal‧ 4.186 J/cal = Q = 80. kcal = 80.kcal‧ 4.186 J/cal = 396. kJ396. kJ


Change of Phase and Latent HeatChange of Phase and Latent Heat

• It requires an energy “investment” to It requires an energy “investment” to change the phase from solid to liquid to change the phase from solid to liquid to gas.gas.

• By breaking the bonds that hold atoms, By breaking the bonds that hold atoms, they can have the same kinetic energy but they can have the same kinetic energy but different total energies.different total energies.

• The energy to change the phase is The energy to change the phase is “hidden” and therefore called “latent “hidden” and therefore called “latent heat.”heat.”


Latent heatLatent heat

∆∆Q = mLQ = mLFor Water:For Water: Fusion (and melting)Fusion (and melting)

LLf f = 333 kJ/kg = 79.4 kcal/kg= 333 kJ/kg = 79.4 kcal/kg

VaporizationVaporization

LLv v = 2260 kJ/kg = 540 kcal/kg= 2260 kJ/kg = 540 kcal/kg


Work from a Heat ReservoirWork from a Heat Reservoir

• The work done by a system is equal to The work done by a system is equal to the loss of the internal energythe loss of the internal energy..

• For an ideal gasFor an ideal gas

W = W = ∫∫V1V1V2 V2 PdVPdV


First Law of ThermodynamicsFirst Law of Thermodynamics

∆∆E = ∆Q -WE = ∆Q -W


Applications:Applications:Adiabatic Adiabatic (∆Q =0)(∆Q =0)

∆∆E = -WE = -W

Isovolumetric (∆V =0)Isovolumetric (∆V =0)∆∆E = ∆QE = ∆Q

Isothermal Isothermal (∆T =0)(∆T =0)

W =W =∫∫V1V1V2 V2 PdV = ∫PdV = ∫V iV i

V f (V f (nRT/V) dVnRT/V) dV

W = n R T ln(VW = n R T ln(Vf f /V/Vi i ) )


Mid Chapter SummaryMid Chapter Summary•The internal energy is the total average The internal energy is the total average energy of the atoms of an object.energy of the atoms of an object.

• Heat is the change in internal energy.Heat is the change in internal energy.

• The change in temperature is proportional to The change in temperature is proportional to the change in internal energy (heat flow) when the change in internal energy (heat flow) when there is no change of phase and the system there is no change of phase and the system does no work.does no work.

• The first law of thermodynamics states The first law of thermodynamics states

∆∆E = ∆Q - WE = ∆Q - W


Conduction:Conduction:

• PP = kA = kA ||dT/dx dT/dx ||• Examples:Examples:

– Thermos bottlesThermos bottles– BlanketsBlankets– Double pane windowsDouble pane windows– Newton’s law of cooling Newton’s law of cooling PP = h A(T = h A(T 2 2 – T– T11))– PansPans– R factor or R value R factor or R value

PP = A(T = A(T 2 2 – T– T11)/)/∑∑i i RRii


Convection:Convection:• Heat transfer by material transferHeat transfer by material transfer

• Forced convection (fluids)Forced convection (fluids)– External force produces material transferExternal force produces material transfer

• Natural ConvectionNatural Convection– Buoyancy-driven flowBuoyancy-driven flow– Newton’s law of cooling appliedNewton’s law of cooling applied

PP = h A(T = h A(T 2 2 – T– T11))h depends on flow conditionsh depends on flow conditions


Radiation:Radiation:

• Stefan-Boltzmann LawStefan-Boltzmann LawPP = = εσεσ AT AT44

•Wien’s Law Wien’s Law

PP ∝∝TT4 4

• σσ = 5.6696 x 10 = 5.6696 x 10-8-8 W/m W/m22‧K‧K44

• Emissivity 0< Emissivity 0< ε ε <1; <1; εε ~ ½ ~ ½• Reflectivity (albedo) R = (1- Reflectivity (albedo) R = (1- εε))• Energy balanceEnergy balance

PP in in - - εσεσ A(T A(Tave ave )) 4 4 = 0= 0


Global Warming?:Global Warming?:

PP in in = ( 1- = ( 1- εελλ) ) PPsunsun

TTave ave = [( 1- = [( 1- εελλ) ) PPsun sun /(/(εεGH GH σσA)]A)]1/41/4

Must understand every parameter Must understand every parameter

to be accurate.to be accurate.


Summary:Summary:

• Heat is transferred by Heat is transferred by o Conduction—energy diffusionConduction—energy diffusiono Convection—mass transportConvection—mass transporto Radiation—electromagnetic wavesRadiation—electromagnetic waves


Guided PracticeGuided PracticeIf the power per unit area (intensity) of sunlight on the If the power per unit area (intensity) of sunlight on the earth is 1.0 kW/mearth is 1.0 kW/m22 and the emissivity is 0.6, what is the and the emissivity is 0.6, what is the expected avearge temperatue of the earth? Comment on expected avearge temperatue of the earth? Comment on the effect of a change in the emissivity or solar radiation.the effect of a change in the emissivity or solar radiation.

PP in in= = ∫∫0 0 22ππ∫∫0 0

π/2π/2 I cos I cos θ θ sinsin θ r θ r 22dθ dφdθ dφ

PP in in= I (2= I (2πr πr 22) (½ ) sin) (½ ) sin22 θ θ ||0 0 π/2π/2 = = I (I (πr πr 22) )

= = 1.0 kW/m1.0 kW/m22 (3.14)(6.4x10 (3.14)(6.4x1066 m) m)22 = 1.3 x 10 = 1.3 x 1017 17 WW

PP in in = = εσεσ A(T A(Tave ave )) 4 4

TTave ave = [= [PP in in / / εσεσ A] A]1/4 1/4

= [1.3 x 10= [1.3 x 1017 17 /(0.6/(0.6‧5.66 x10‧5.66 x10-8-8‧ 5.1x10‧ 5.1x101414)] )] 1/4 1/4 =294 =294 K=21C K=21C


Guided PracticeGuided Practice

If the power per unit area (intensity) of sunlight If the power per unit area (intensity) of sunlight on the earth is 1.0 kW/mon the earth is 1.0 kW/m22 and the emissivity is and the emissivity is 0.6, what is the expected average temperature 0.6, what is the expected average temperature of the earth? Comment on the effect of a of the earth? Comment on the effect of a change in the emissivity or solar radiation.change in the emissivity or solar radiation.


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Physics 1710—Warm-up Quiz In Antarctica the thermometer reads –20 C; what is this temperature on a Fahrenheit thermometer? A.-36 o F B.-11 o F C.-4 o