entropy, energy, and temperature · 15/12/2010 · focus on the possible arrangements of energy in...
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LIVE INTERACTIVE LEARNING @ YOUR DESKTOP
December 15, 20106:30 p.m. - 8:00 p.m. Eastern time
ENTROPY, ENERGY, and TEMPERATURE
Presented by: Jerry Bell, Pat Deibert and Bonnie Bloom
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20102
ENTROPY, ENERGY, and TEMPERATURE
Energy Exchange
Jerry Bell, ACS (retired)Bonnie Bloom, Hilliard Davidson HS, OHPat Deibert, Sheboygan Falls HS, WI
Pat Bonnie Jerry
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20103
Quiz: What is likely to be the result, if two identical blocks of copper metal, one hot and the other cold, are brought together and allowed to exchange energy before separating again.
?→
A
B
C
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20104
Quiz: What is likely to be the result, if two identical blocks of copper metal, one hot and the other cold, are brought together and allowed to exchange energy before separating again.
How general is this result?
→ B
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20105
Activity: Gather your cup of warm water and second cup of an equal volume of cool water. Take and record the temperature of the warm water and then of the cool water. While the thermometer is still in the cool water, pour in all the warm water and stir until the temperature is constant. Pat does a similar experiment in this video.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20106
Activity: Gather your cup of warm water and second cup of an equal volume of cool water. Take and record the temperature of the warm water and then of the cool water. While the thermometer is still in the cool water, pour in all the warm water and stir until the temperature is constant.
Result: The final temperature isA. lower than that of the cool water.B. higher than that of the warm water.C. intermediate between the warm and cool water temperatures.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20107
Activity: Gather your cup of warm water and second cup of an equal volume of cool water. Take and record the temperature of the warm water and then of the cool water. While the thermometer is still in the cool water, pour in all the warm water and stir until the temperature is constant.
Result: The final temperature isA. lower than that of the cool water.B. higher than that of the warm water.C. intermediate between the warm and cool water temperatures.
Our experience is that the spontaneous direction of energy exchange is from the warmer to the cooler body. Whatever model we propose to explain energy exchange must be consistent with this experience.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20108
The Direction of Change
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 20109
Model for change from previous Web Seminar--Entropy: Mixing and Oil Spills--restated in terms of energy:
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201010
Model for change from previous Web Seminar--Entropy: Mixing and Oil Spills--restated in terms of energy:
If a system can exist in more than one observable state (hot-cold or warm-warm, for example), spontaneous changes will be in the direction toward the state that is most probable.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201011
Model for change from previous Web Seminar--Entropy: Mixing and Oil Spills--restated in terms of energy:
If a system can exist in more than one observable state (hot-cold or warm-warm, for example), spontaneous changes will be in the direction toward the state that is most probable.The number of distinguishable arrangements, W, of the energy that give a particular state of a system is a measure of the probability that this state will be observed.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201012
Model for change from previous Web Seminar--Entropy: Mixing and Oil Spills--restated in terms of energy:
If a system can exist in more than one observable state (hot-cold or warm-warm, for example), spontaneous changes will be in the direction toward the state that is most probable.The number of distinguishable arrangements, W, of the energy that give a particular state of a system is a measure of the probability that this state will be observed.Fundamental assumption: Each distinguishably different energy arrangement of a system is equally probable.Two arrangements are distinguishable if you can tell them apart. Exchanging identical objects does not produce a new arrangement.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201013
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201014
Energy at the Molecular Scale
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201015
Thermal energy in a solid sample:
The atomic cores in a piece of copper metal are fixed in space in a sea of delocalized electrons, but vibrate according to how much thermal energy there is in the system, as in this animation.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201016
Thermal energy in a solid sample:
The atomic cores in a piece of copper metal are fixed in space in a sea of delocalized electrons, but vibrate according to how much thermal energy there is in the system, as in this animation.
All energy is quantized, that is, comes only in discrete packets.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201017
Thermal energy in a solid sample:
The atomic cores in a piece of copper metal are fixed in space in a sea of delocalized electrons, but vibrate according to how much thermal energy there is in the system, as in this animation.
All energy is quantized, that is, comes only in discrete packets.
Thermal energy can be transferred to and from a sample of metal like this in quantized amounts that are characteristic of the particular metal and atomic structure.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201018
Focus on the possible arrangements of energy in a tiny sample, four atoms, of copper metal.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201019
Focus on the possible arrangements of energy in a tiny sample, four atoms, of copper metal.
Each atom can vibrate in place and its vibrational energy is quantized. It can have any number of quanta of energy(shown like this to represent vibrational energy).
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201020
Focus on the possible arrangements of energy in a tiny sample, four atoms, of copper metal.
Each atom can vibrate in place and its vibrational energy is quantized. It can have any number of quanta of energy(shown like this to represent vibrational energy).
For example, the third atom could be vibrating with one quantum of energy, while the others have none.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201021
Focus on the possible arrangements of energy in a tiny sample, four atoms, of copper metal.
Each atom can vibrate in place and its vibrational energy is quantized. It can have any number of quanta of energy(shown like this to represent vibrational energy).
For example, the third atom could be vibrating with one quantum of energy, while the others have none.
Quiz: How many distinguishable arrangements, W1,4, are there for one quantum among the four atoms? A = 1 B = 2 C = 3 D = 4
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201022
The four distinguishable arrangements, W1,4, for one quantumamong the four atoms are:
(1,0,0,0)
(0,1,0,0)
(0,0,1,0)
(0,0,0,1)
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201023
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201024
Adding More Energy
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201025
How many distinguishable arrangements, W2,4, are there for twoquanta among the four atoms? Two arrangements are:
(0,2,0,0)
(0,1,0,1)
Quiz: What is another distinguishable arrangement of two quanta among four atoms, where each atom can have any number of quanta? Place a stamp twice on one atom or once each on two atoms where you want to place quanta.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201026
Number of distinguishable arrangements for two quanta among thefour atoms: W2,4 = 10
(2,0,0,0) (1,0,1,0)
(0,2,0,0) (1,0,0,1)
(0,0,2,0) (0,1,1,0)
(0,0,0,2) (0,1,0,1)
(1,1,0,0) (0,0,1,1)
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201027
For a system of n identical objects (quanta) allowed to occupy anyof N boxes (atoms), any number of objects per box, the number of distinguishable arrangements, Wn,N is given by
Wn,N = (N +n−1)!(N −1)!n!
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201028
For a system of n identical objects (quanta) allowed to occupy anyof N boxes (atoms), any number of objects per box, the number of distinguishable arrangements, Wn,N is given by
Wn,N =
W2,4 =
(N +n−1)!(N −1)!n!
5!(3!)2!=
5⋅4 ⋅(3!)2⋅1⋅(3!) =
5⋅42⋅1 =
5⋅21 = 5⋅2=10
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201029
For a system of n identical objects (quanta) allowed to occupy anyof N boxes (atoms), any number of objects per box, the number of distinguishable arrangements, Wn,N is given by
Wn,N =
W2,4 =
Quiz: What is the number of distinguishable arrangements, W6,4, for 6 quanta distributed among the 4 atoms in the solid?
W6,4 =
(N +n−1)!(N −1)!n!
5!(3!)2!=
5⋅4 ⋅(3!)2⋅1⋅(3!) =
5⋅42⋅1 =
5⋅21 = 5⋅2=10
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201030
For a system of n identical objects (quanta) allowed to occupy anyof N boxes (atoms), any number of objects per box, the number of distinguishable arrangements, Wn,N is given by
Wn,N =
W2,4 =
Quiz: What is the number of distinguishable arrangements, W6,4, for 6 quanta distributed among the 4 atoms in the solid?
W6,4 =
Take a moment to copy this table for later use in the seminar.n 0 1 2 3 4 5 6Wn,4 1 4 10 20 35 56 84
(N +n−1)!(N −1)!n!
5!(3!)2!=
5⋅4 ⋅(3!)2⋅1⋅(3!) =
5⋅42⋅1 =
5⋅21 = 5⋅2=10
9!(3!)6!=
9⋅8⋅7⋅(6!)3⋅2⋅1⋅(6!) =
3⋅4 ⋅71 =84
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201031
Quiz: Consider these two identical pieces of copper, one with 2quanta of energy and the other with 6. What observable property is different for the two pieces?
A. EnergyB. EntropyC. TemperatureD. Atomic motion
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201032
Quiz: Consider these two identical pieces of copper, one with 2quanta of energy and the other with 6. What observable property is different for the two pieces?
A. EnergyB. EntropyC. TemperatureD. Atomic motion
If one object has more energy than an another identical object, the one with more energy will have a higher temperature. Also, the atoms or molecules will be moving with higher average speed and the entropy will be higher, but these are not observable properties.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201033
Quiz: Consider these two identical pieces of copper, one with 2quanta of energy and the other with 6. What observable property is different for the two pieces?
A. EnergyB. EntropyC. TemperatureD. Atomic motion
If one object has more energy than an another identical object, the one with more energy will have a higher temperature. Also, the atoms or molecules will be moving with higher average speed and the entropy will be higher, but these are not observable properties.
The total number, Wtotal, of distinguishable arrangementsfor this system is
Wtotal = W2,4·W6,4 = 10·84 = 840
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201034
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201035
Energy Exchange
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201036
Quiz:What would happen if the metals collided, two quanta were exchanged, and the metals separated?
A. Four quanta would be distributed over each group of four atoms in such a way that the total number of arrangements would be <840.
B. Four quanta would be distributed over each group of four atoms in such a way that the total number of arrangements would be =840.
C. Four quanta would be distributed over each group of four atoms in such a way that the total number of arrangements would be >840.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201037
This is one of the 1225 distinguishable arrangements of quanta inthe system after two quanta have been transferred from the hotter to the colder piece of copper.
Wtotal = W4,4·W4,4 = 35·35 = 1225
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201038
This is one of the 1225 distinguishable arrangements of quanta inthe system after two quanta have been transferred from the hotter to the colder piece of copper.
Wtotal = W4,4·W4,4 = 35·35 = 1225
Although the number of distinguishable arrangements for the hotter piece decreased in the energy transfer, this was more than offset by the increase for the colder piece, so Wtotal after the transfer (1225) is larger than before the transfer (840). The transfer of energy from the hotter to colder body is favored and is observed for actual processes in the macroscopic world.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201039
This is one of the 1225 distinguishable arrangements of quanta inthe system after two quanta have been transferred from the hotter to the colder piece of copper.
Wtotal = W4,4·W4,4 = 35·35 = 1225
Although the number of distinguishable arrangements for the hotter piece decreased in the energy transfer, this was more than offset by the increase for the colder piece, so Wtotal after the transfer (1225) is larger than before the transfer (840). The transfer of energy from the hotter to colder body is favored and is observed for actual processes in the macroscopic world.
Both pieces of copper now have the same energy (same number of energy quanta), so their temperatures are the same and intermediate between the original temperatures as is also observed macroscopically.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201040
This is one of the 1225 distinguishable arrangements of quanta inthe system after two quanta have been transferred from the hotter to the colder piece of copper.
Wtotal = W4,4·W4,4 = 35·35 = 1225
Quiz: What would Wtotal be if only one quantum had been transferred? Does this make sense?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201041
This is one of the 1225 distinguishable arrangements of quanta inthe system after two quanta have been transferred from the hotter to the colder piece of copper.
Wtotal = W4,4·W4,4 = 35·35 = 1225
Quiz: What would Wtotal be if only one quantum had been transferred? Does this make sense?.
Wtotal = W3,4·W5,4 = 20·56 = 1120
W2,4·W6,4 [840] < W3,4·W5,4 [1120] < W4,4·W4,4 [1225]
As energy is transferred from the hotter to the colder body, the number of arrangements (probability) increases until the temperatures are the same. Reaching a common temperature is observed experimentally for such a system. The model makes sense.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201042
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201043
Entropy Change in Energy Exchange
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201044
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201045
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial = k·lnWfinal – k·lnWinitial = k·ln(Wfinal/Winitial)
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201046
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial = k·lnWfinal – k·lnWinitial = k·ln(Wfinal/Winitial)
If Wfinal > Winitial, then k·ln(Wfinal/Winitial) > 0 and ΔS > 0
That is, ΔS > 0 for changes that are favored to occur.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201047
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial = k·lnWfinal – k·lnWinitial = k·ln(Wfinal/Winitial)
If Wfinal > Winitial, then k·ln(Wfinal/Winitial) > 0 and ΔS > 0
That is, ΔS > 0 for changes that are favored to occur.Apply these ideas to the energy transfer process just analyzed.
ΔS(cold→warm) = k·ln(Wfinal/Winitial) = k·ln(35/10) > 0ΔS(hot→warm) = k·ln(Wfinal/Winitial) = k·ln(35/84) < 0
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201048
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial = k·lnWfinal – k·lnWinitial = k·ln(Wfinal/Winitial)
If Wfinal > Winitial, then k·ln(Wfinal/Winitial) > 0 and ΔS > 0
That is, ΔS > 0 for changes that are favored to occur.Apply these ideas to the energy transfer process just analyzed.
ΔS(cold→warm) = k·ln(Wfinal/Winitial) = k·ln(35/10) > 0ΔS(hot→warm) = k·ln(Wfinal/Winitial) = k·ln(35/84) < 0
For the overall process, the net entropy change, ΔSnet, is the sum of all the entropy changes involved
ΔSnet = ΔS(cold→warm) + ΔS(hot→warm) = k·ln(35/10) + k·ln(35/84)ΔSnet = k·ln[(35/10)·(35/84)] = k·ln(1225/840) > 0
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201049
Consider the entropy change, ΔS, in a process, where S ≡ k·lnWΔS ≡ Sfinal – Sinitial = k·lnWfinal – k·lnWinitial = k·ln(Wfinal/Winitial)
If Wfinal > Winitial, then k·ln(Wfinal/Winitial) > 0 and ΔS > 0
That is, ΔS > 0 for changes that are favored to occur.Apply these ideas to the energy transfer process just analyzed.
ΔS(cold→warm) = k·ln(Wfinal/Winitial) = k·ln(35/10) > 0ΔS(hot→warm) = k·ln(Wfinal/Winitial) = k·ln(35/84) < 0
For the overall process, the net entropy change, ΔSnet, is the sum of all the entropy changes involved
ΔSnet = ΔS(cold→warm) + ΔS(hot→warm) = k·ln(35/10) + k·ln(35/84)ΔSnet = k·ln[(35/10)·(35/84)] = k·ln(1225/840) > 0
Note that the same final result would have been obtained simply by using the values of Wtotal for the initial and final states.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201050
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201051
Entropy, Energy, and Temperature
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201052
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201053
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
The change in entropy for the cooler system is larger than that for the warmer.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201054
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
The change in entropy for the cooler system is larger than that for the warmer.The change in enthalpy, ΔH, in each case is the same, +1 quantum.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201055
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
The change in entropy for the cooler system is larger than that for the warmer.The change in enthalpy, ΔH, in each case is the same, +1 quantum.A relationship that is consistent with these results is
ΔS = ΔH/T
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201056
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
The change in entropy for the cooler system is larger than that for the warmer.The change in enthalpy, ΔH, in each case is the same, +1 quantum.A relationship that is consistent with these results is
ΔS = ΔH/TFor a given energy (enthalpy) change, the entropy change is larger for a colder body than for a hotter body.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201057
The relationship among entropy, energy (= enthalpy), and temperature.Consider the entropy changes when 1 quantum of energy is added to our 4-atom model system containing 2 quanta and 1 quantum to another containing 5 quanta.
ΔS(2,4→3,4) = k·ln(W3,4/W2,4) = k·ln(20/10) = k·ln(2.0)ΔS(5,4→6,4) = k·ln(W6,4/W5,4) = k·ln(84/56) = k·ln(1.5)
The change in entropy for the cooler system is larger than that for the warmer.The change in enthalpy, ΔH, in each case is the same, +1 quantum.A relationship that is consistent with these results is
ΔS = ΔH/TFor a given energy (enthalpy) change, the entropy change is larger for a colder body than for a hotter body.The units here are consistent with the definition of S, since the units of the Boltzmann constant, k, are energy per kelvin.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201058
Questions?
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201059
Application: Solution Freezing Point
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201060
Activity: Gather your two cups of (crushed) ice, 50 mL of water, 50 mL of salt water, a thermometer (or two) and a stout stirrer.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201061
Activity: Gather your two cups of (crushed) ice, 50 mL of water, 50 mL of salt water, a thermometer (or two) and a stout stirrer.
Add water to one cup of ice and salt water to the other. Stir the ice liquid mixtures vigorously while measuring their temperatures and watching Pat do a similar activity.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201062
Activity: Gather your two cups of (crushed) ice, 50 mL of water, 50 mL of salt water, a thermometer (or two) and a stout stirrer.
Add water to one cup of ice and salt water to the other. Stir the ice liquid mixtures vigorously while measuring their temperatures.
What was your result?A. T(ice-water) = T(ice-salt water)B. T(ice-water) > T(ice-salt water)C. T(ice-water) < T(ice-salt water)
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201063
Activity: Gather your two cups of (crushed) ice, 50 mL of water, 50 mL of salt water, a thermometer (or two) and a stout stirrer.
Add water to one cup of ice and salt water to the other. Stir the ice liquid mixtures vigorously while measuring their temperatures.
What was your result?A. T(ice-water) = T(ice-salt water)B. T(ice-water) > T(ice-salt water)C. T(ice-water) < T(ice-salt water)
The observations are that the mixture of ice and salt water is at a lower temperature than the ice and pure water mixture.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201064
Consider these changes: ice ⇔ waterice ⇔ solution (nonvolatile solute)
Quiz: Which of these diagrams represents the relative values of the entropies of ice, water, and solution?
A B C
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201065
ΔSsoln > ΔSwater
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201066
ΔSsoln > ΔSwater
ΔHsoln = ΔHwater because the change in both cases is ice going to liquid water
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201067
ΔSsoln > ΔSwater
ΔHsoln = ΔHwater because the change in both cases is ice going to liquid water
Rearrange ΔS = ΔH/T to give T = ΔH/ΔS
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201068
ΔSsoln > ΔSwater
ΔHsoln = ΔHwater because the change in both cases is ice going to liquid water
Rearrange ΔS = ΔH/T to give T = ΔH/ΔS
For these changesTsoln = (ΔHsoln)/(ΔSsoln) < (ΔHwater)/(ΔSwater) = Twater
That is, the temperature at which ice and solution are in equilibrium is lower than the freezing point of water where ice and water are in equilibrium. The freezing point is lowered by nonvolatile solutes.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201069
ΔSsoln > ΔSwater
ΔHsoln = ΔHwater because the change in both cases is ice going to liquid water
Rearrange ΔS = ΔH/T to give T = ΔH/ΔS
For these changesTsoln = (ΔHsoln)/(ΔSsoln) < (ΔHwater)/(ΔSwater) = Twater
That is, the temperature at which ice and solution are in equilibrium is lower than the freezing point of water where ice and water are in equilibrium. The freezing point is lowered by nonvolatile solutes.
Freezing point depression is a colligative property that depends on how much solute is mixed into the solvent, but not on the solute identity.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201070
This result is the basis for using solutes to melt ice on sidewalks incold climates and for the old fashioned ice-cream makers cooled with ice-salt mixtures.
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201071
To check your understanding of the topics in this web seminar, use an entropy-energy-temperature analysis to determine whether the boiling point of an aqueous solution of a nonvolatile solute is higher, lower, or the same as the boiling point of pure water. Model your approach on the one just done for the freezing point of water and an aqueous solution.
See the post-seminar notes that will be posted with the archive for this seminar for the solution.
We welcome your feedback on whether you would like to attend further web seminars exploring more applications of entropy and this model to phase changes, osmosis, chemical reactions, etc. You can do this in the chat here or off-line to me: [email protected].
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Entropy, Energy, and Temperature
ACS-NSTA Entropy Web Seminar 2, Fall 201072
Questions?
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Thank you to the sponsor of tonight's Web Seminar:
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74http://learningcenter.nsta.org
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http://www.elluminate.com
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National Science Teachers AssociationDr. Francis Q. Eberle, Executive Director
Zipporah Miller, Associate Executive Director Conferences and Programs
Al Byers, Assistant Executive Director e-Learning
LIVE INTERACTIVE LEARNING @ YOUR DESKTOP
NSTA Web SeminarsPaul Tingler, Director
Jeff Layman, Technical Coordinator