Physics 207 – Lecture 24
Lecture 24 Goals: • Chapter 17 Employ heat (Q) and energy transfer in materials Recognize adiabatic processes (i.e., Q=0) • Chapter 18 Follow the connection between temperature, thermal energy, and the average translational kinetic energy molecules Understand the molecular basis for pressure and the idealgas law. To predict the molar specific heats of gases and solids.
• Assignment HW11, Due Wednesday 9:00 AM For Thursday, Read through all of Chapter 18 Physics 207: Lecture 24, Pg 1
1st Law of Thermodynamics
Eth =W + Q
(if ∆K & ∆U =0 ) W & Q with respect to the system
Work W and heat Q depend on process by which the
system is changed (path dependent).
The change of energy in the system,
Eth depends only on the total energy exchanged W+Q, not on the process. Physics 207: Lecture 24, Pg 2
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Physics 207 – Lecture 24
1st Law: Work & Heat Work done on system (an ideal
gas….notice minus sign, V is in reference to the system)
Won = −
final
∫ p dV = −(area under curve )
initial
W on system < 0 Moving left to right [where (Vf > Vi)]
If
ideal gas, pV = nRT
W by system
> 0 Moving left to right Physics 207: Lecture 24, Pg 3
1st Law: Work & Heat
Work: Depends on the path taken in the pV-
diagram (It is not just the destination but the path…) Won system > 0 Moving right to left
Physics 207: Lecture 24, Pg 4
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Physics 207 – Lecture 24
1st Law: Work (“Area” under the curve)
Work depends on the path taken in the pV-diagram : 3
W23= -pf (Vf -Vi)
W12= -pi (Vf -Vi)
2 3
1
1
2
(a) W a = W 1 to 2 + W 2 to 3 (1st p then V constant, T changes) W a (on) = - pi (Vf - Vi) + 0 > 0 (b) W b = W 1 to 2 + W 2 to 3 (1st p then V constant, T changes) W b (on) = 0 - pf (Vf - Vi) > W a > 0 (c) Need explicit form of p versus V but W c (on) > 0 Physics 207: Lecture 24, Pg 5
Paths on the pV diagram (with an idea gas) W = - p ∆V ???? W=0 ????
(1) Isobaric (2) Isothermal (3) Isochoric (4) Adiabatic
1 p
3
T1
2 4
T2
Ideal gas
T4 T3
V
Physics 207: Lecture 24, Pg 6
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Physics 207 – Lecture 24
Isothermal processes (in an ideal gas)
Work done when PV = nRT = constant P = nRT / V
W =−
final
∫ p dV
= − (area under curve )
initial
Vf
Vf
Vi
Vi
W = − ∫ nRT dV / V = − nRT ∫ dV / V
W = − nRT ln ( Vf /Vi ) T1
3
p
T2 T4 T3
For this we need access to thermal energy V Physics 207: Lecture 24, Pg 7
Adiabatic Processes (in an ideal gas) An adiabatic process is process in which there is no
thermal energy transfer to or from a system (Q = 0) A reversible adiabatic
process involves a “worked” expansion in which we can return all of the energy transferred. p In this case Cp γ =C PVγ = const. V All real processes are not.
4 2 T1
3 1
T2 T4 T3
V
We need to know Cp & CV Physics 207: Lecture 24, Pg 8
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Physics 207 – Lecture 24
Work and Ideal Gas Processes (on system) Isothermal
W = − nRT ln ( Vf /Vi ) Isobaric
W = − p ( Vf - Vi ) Isochoric
W =0
γ
FYI: Adiabatic (and reversible) PV = const.
W = − ∫VV PdV = − ∫VV 2
1
2
1
const dV Vγ V
(V2−γ − V1−γ ) = const γ Physics 207: Lecture 24, Pg 9
Two process are shown that take an ideal gas from state 1 to state 3. (“by” means “by the system on the world”) Compare the work done by process A to the work done by process B.
A. WA > WB B. WA < WB C. WA = WB = 0 D. WA = WB but neither is zero
p3 p2 p1
ON A 1 3 W12 = 0 (isochoric) B 1 2 W12 = -½ (p1+p2)(V2-V1) < 0 B 2 3 W23 = -½ (p2+p3)(V1-V2) > 0 B 1 3 = ½ (p3 - p1)(V2-V1) > 0
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BY -W12 > 0 -W23 < 0 < 0 Physics 207: Lecture 24, Pg 10
Physics 207 – Lecture 24
Two process are shown that take an ideal gas from state 1 to state 3. Compare the work done by process A to the work done by process B.
A. WA > WB B. WA < WB C. WA = WB = 0 D. WA = WB but neither is zero
A13 B12 B23 B 1 3
ON W12 = 0 (isochoric) W12 = -½ (p1+p2)(V2-V1) < 0 W23 = -½ (p2+p3)(V1-V2) > 0 = ½ (p3 - p1)(V2-V1) > 0
BY -W12 > 0 -W23 < 0 < 0 Physics 207: Lecture 24, Pg 11
Both Q and W can change T We know how work changes the mechanical energy of
a solid “system” Here our system is an ideal gas….only the temperature can change For real materials we can change the temperature or the state We must quantify the response of a system to thermal energy transfer (Q)
Physics 207: Lecture 24, Pg 12
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Physics 207 – Lecture 24
What about Q?
Specific Heat
Latent Heat
What are the relationships between ∆ETh and T.
Latent Heat
Physics 207: Lecture 24, Pg 13
Heat and Latent Heat
Latent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg)
Q = ±ML
Specific heat c of a substance is the energy required to raise the temperature of 1 kg by 1 K. (Units: J / K kg )
Q=Mc T
Molar specific heat C of a gas at constant volume is the energy required to raise the temperature of 1 mol by 1 K.
Q = n CV T If a phase transition involved then the heat transferred is
Q = ±ML+M c T Physics 207: Lecture 24, Pg 14
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Physics 207 – Lecture 24
Q : Latent heat and specific heat The molar specific heat of gasses depends on the
process path CV= molar specific heat at constant volume Cp= molar specific heat at constant pressure Cp= CV+R (R is the universal gas constant)
Physics 207: Lecture 24, Pg 15
Q : Latent heat and specific heat The molar specific heat of gasses depends on the
process path CV= molar specific heat at constant volume Cp= molar specific heat at constant pressure Cp= CV+R (R is the universal gas constant)
γ
Cp = CV Page 8
Physics 207: Lecture 24, Pg 16
Physics 207 – Lecture 24
Latent Heat Most people were at least once burned by hot water
or steam. An equal amount (by mass) of boiling water and
steam contact your skin. Which is more dangerous, the water or the steam?
Physics 207: Lecture 24, Pg 17
Mechanical equivalent of heat Heating
liquid water: Q = amount of heat that must be supplied to raise the temperature by an amount ∆ T . [Q] = Joules or calories. 1 cal = 4.186 J 1 kcal = 1 cal = 4186 J
calorie: energy to raise 1 g of water from 14.5 to 15.5 °C (James Prescott Joule found the mechanical equivalent of heat.) Sign convention: +Q : heat gained - Q : heat lost Physics 207: Lecture 24, Pg 18
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Physics 207 – Lecture 24
Exercise The specific heat (Q = M c T) of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium. Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true?
(a) The iron takes less time than the aluminum to reach 500 K (b) The aluminum takes less time than the iron to reach 500 K (c) The two blocks take the same amount of time to reach 500 K Physics 207: Lecture 24, Pg 19
Exercise The specific heat (Q = M c T) of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium. Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true?
(a) The iron takes less time than the aluminum to reach 500 K (b) The aluminum takes less time than the iron to reach 500 K (c) The two blocks take the same amount of time to reach 500 K Physics 207: Lecture 24, Pg 20
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Physics 207 – Lecture 24
Heat and Ideal Gas Processes (on system)
Isothermal Expansion/Contraction
∆ETh = 0 = W + Q Q = −W = nRT ln ( Vf /Vi )
Isobaric
Q = nC p ∆T = n (CV + R ) ∆T
Isochoric
∆ETh = 0 + Q
Q = nCV ∆T
Adiabatic
Q=0
∆ETh = W + 0 Physics 207: Lecture 24, Pg 21
Combinations of Isothermal & Adiabatic Processes All engines employ a thermodynamic cycle W = ± (area under each pV curve) Wcycle = area shaded in turquoise Watch sign of the work!
Physics 207: Lecture 24, Pg 22
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Physics 207 – Lecture 24
Exercise Latent Heat Most people were at least once burned by hot water or steam. Assume that water and steam, initially at 100°C, are cooled down
to skin temperature, 37°C, when they come in contact w ith your skin. Assume that the steam condenses extremely fast, and that the specific heat c = 4190 J/ kg K is constant for both liquid water and steam. Under these conditions, which of the following statements is true? (a) Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than that of liquid water. (b) Steam burns the skin worse than hot water because the latent heat of vaporization is released as well. (c) Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher than that of steam. (d) Hot water and steam both burn skin about equally badly. Physics 207: Lecture 24, Pg 23
Exercise Latent Heat Most people were at least once burned by hot water or steam.
Assume that water and steam, initially at 100°C, are cooled down to skin temperature, 37°C, when they come in contact with your skin. Assume that the steam condenses extremely fast, and that the specific heat c = 4190 J/ kg K is constant for both liquid water and steam. Under these conditions, which of the following statements is true? (b) Steam burns the skin worse than hot water because the latent heat of vaporization is released as well. How much heat H1 is transferred to the skin by 25.0 g of steam? The latent heat of vaporization for steam is L = 2256 kJ/kg.
H1 = 0.025 kg x 2256 kJ/kg = 63.1 kJ How much heat H2 is transferred to the skin by 25.0 g of water? H2 = 0.025 kg x 63 K x 4190 J/ kg K = 6.7 kJ
Physics 207: Lecture 24, Pg 24
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Physics 207 – Lecture 24
Energy transfer mechanisms Thermal conduction (or conduction) Convection Thermal Radiation
Physics 207: Lecture 24, Pg 25
Energy transfer mechanisms
Thermal conduction (or conduction): Energy transferred by direct contact. e.g.: energy enters the water through the bottom of the pan by thermal conduction. Important: home insulation, etc.
Rate of energy transfer ( J / s or W )
Through a slab of area A and thickness ∆x, with opposite faces at different temperatures, Tc and Th
Q / ∆t = k A (Th - Tc ) / ∆x k :Thermal conductivity (J / s m °C) Physics 207: Lecture 24, Pg 26
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Physics 207 – Lecture 24
Thermal Conductivities J/s m °C
J/s m °C
J/s m °C Aluminum
238
Air
0.0234
Asbestos
0.25
Copper
397
Helium
0.138
Concrete
1.3
Gold
314
Hydrogen
0.172
Glass
0.84
Iron
79.5
Nitrogen
0.0234
Ice
1.6
Lead
34.7
Oxygen
0.0238
Water
0.60
Silver
427
Rubber
0.2
Wood
0.10
Physics 207: Lecture 24, Pg 27
Home Exercise Thermal Conduction Two identically shaped bars (one blue and one green) are placed between two different thermal reservoirs . The thermal conductivity coefficient k is twice as large for the blue as the green. You measure the temperature at the joint between the green and blue bars.
100 C
Tjoint
300 C
Which of the following is true? (A) Ttop > Tbottom (B) Ttop= Tbottom
(C) Ttop< Tbottom
(D) need to know k Physics 207: Lecture 24, Pg 28
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Physics 207 – Lecture 24
Home Exercise Thermal Conduction
Two identically shaped bars (one blue and one green) are placed between two different thermal reservoirs . The thermal conductivity coefficient k is twice as large for the blue as the green.
100 C
Tjoint
300 C
Top: Pgreen = Pblue = Q / ∆t = 2 k A (Thigh - Tj ) / ∆x= k A (Tj - Tlow ) / ∆x 2 (Thigh - Tj ) = (Tj - Tlow ) 3 Tj(top) By analogy for the bottom:
= 2 Thigh – Tlow
3 Tj(bottom) = 2 Tlow – Thigh
3 (Tj(top) - Tj(bottom) = 3 Thigh – 3 Tlow > 0 (A) Ttop > Tbottom Physics 207: Lecture 24, Pg 29
Exercise Thermal Conduction Two thermal conductors are butted together and in contact with two thermal reservoirs held at the temperatures 100 C shown. Which of the temperature vs. position plots below is most physical?
(C)
Temperature
Temperature
Temperature
(B)
(A)
Position
300 C
Position
Position Physics 207: Lecture 24, Pg 30
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Physics 207 – Lecture 24
Exercise Thermal Conduction Two thermal conductors are butted together and in contact with two thermal reservoirs held at the temperatures shown. 100 C Which of the temperature vs. position plots below is most physical?
(C)
Temperature
Temperature
Temperature
(B)
(A)
Position
300 C
Position
Position
Physics 207: Lecture 24, Pg 31
Energy transfer mechanisms Convection:
Energy is transferred by flow of substance 1. Heating a room (air convection) 2. Warming of North Altantic by warm waters from the equatorial regions Natural convection: from differences in density Forced convection: from pump of fan Radiation:
Energy is transferred by photons e.g.: infrared lamps Stefan’s Law
P = σ A e T4 (power radiated) σ = 5.7×10-8 W/m2 K4 , T is in Kelvin, and A is the surface area e is a constant called the emissivity Physics 207: Lecture 24, Pg 32
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Physics 207 – Lecture 24
Minimizing Energy Transfer
The Thermos bottle, also called a Dewar flask is designed to minimize energy transfer by conduction, convection, and radiation. The standard flask is a double-walled Pyrex glass with silvered walls and the space between the walls is evacuated.
Vacuum
Silvered surfaces
Hot or cold liquid
Physics 207: Lecture 24, Pg 33
Anti-global warming or the nuclear winter scenario
Assume P/A = P = 1340 W/m2 from the sun is incident on a thick dust cloud above the Earth and this energy is absorbed, equilibrated and then reradiated towards space where the Earth’s surface is in thermal equilibrium with cloud. Let e (the emissivity) be unity for all wavelengths of light. What is the Earth’s temperature?
P
= σ A T4= σ (4π r2) T4 = P π r2 T = [P / (4 x σ )]¼
σ=
5.7×10-8 W/m2 K4 T = 277 K (A little on the chilly side.) Physics 207: Lecture 24, Pg 34
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Physics 207 – Lecture 24
Lecture 24
• Assignment HW11, Due Wednesday (9:00 AM) Tuesday review Reading assignment through all of Chapter 18
Physics 207: Lecture 24, Pg 35
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