213 Misc' Notes

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213 Misc. Notes

• The end is near dont get behind
• All EXcuses must be taken to 231 Loomis before
  noon, Friday, April 24.
• The PHYS 213 (combined) final exam time is 8-10
  am Tues., May 12. The conflict exam is 8-10 am
  Wed., May 13.
• The deadline for changing your final exam time is
  10pm, Tuesday, May 5.
• HW 6 is due Saturday, May 2 at 8 am.
• Course Survey 2 bonus points (accessible at top
  of HW6)

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Lecture 10Heat Pumps,Refrigerators,Available
Work and Free EnergyAgenda for today

• Pumping Heat Refrigerators
• Free Energy and Available Work
• Work from Hot and Cold Bricks

Reference for this Lecture Elements Ch 10
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Ideal Refrigerators and Heat Pumps

• The Carnot cycle is reversible (each step is
  reversible)

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Helpful Hints in Dealing with Engines and Fridges

• Quickly sketch the process. Define Qh and Qc and
  Wby (or Won) as positive and show directions of
  flow.
• Determine which Q is given.
• Write the First Law of Thermodynamics (FLT).
  There are only 3 configurations of Carnot
  engines

Engine (Qh given) Refrigerator (QC given)
Heat Pump (Qh or QC given)
FLT Wby Qh - QC Won Qh - QC
Won Qh - QC
Wby Qh(1- TC/Th) Won QC(Th/TC - 1)
Won Qh(1- TC/Th)
or Fridge eqn.
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Example Refrigerator
There is a 70 W heat leak from a room at
temperature 22 C into an ideal refrigerator.
How much electrical power is needed to keep the
refrigerator at -10 C?
Hint For the fridge, Qc must exactly
compensate the heat leak. So Qc 70 J for each
second of operation. (Watt J/s) Assume Qc/Qh
Tc/Th (Carnot)
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Example Refrigerator
There is a 70 W heat leak from a room at
temperature 22 C into an ideal refrigerator.
How much electrical power is needed to keep the
refrigerator at -10 C?
Kitchen, 295K
Hint For the fridge, Qc must exactly compensate
the heat leak. So Qc 70 J for each second of
operation. (Watt J/s) Assume Qc/Qh Tc/Th
(Carnot) From diagram, Qc Won Qh
QcTh/Tc. Therefore, Won Qc(Th/Tc - 1).
Food, 263K
The work required for each second of operation
is Won (70 J)(295 /263 - 1) 8.5 J Power
required 8.5 W
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Supplement Peltier cooler

• Driving a current (amps) through generates a
  temperature difference. 20-50C typical
• Not so common theyre more costly, take a lot
  of power, and you still have to get rid of the
  heat!
• Hows it work

• Electrons pushed from electron-deficit material
  (p-type) to electron-rich material (n-type) they
  absorb energy, cooling the connector.

Despite the radically different construction,
this heat pump must obey exactly the same limits
on efficiency as the gas-based pumps, because
these limits are based on the 1st and 2nd laws,
not any details.
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Refrigerators and the limits of cooling

• The maximum efficiency is

• The colder you try to go, the less efficient the
  refrigerator gets.The limit as TC goes to zero
  is zero efficiency !
• Since heat leaks will not disappear as the object
  is cooled, you need more cooling power the
  colder it gets.The integral of the power
  required diverges as T? 0.
• Therefore you cannot cool a system to absolute
  zero ?

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Example Heat Pump
Suppose that the heat flow out of your 20-C home
in the winter is 7 kW. If the temperature outside
is -15 C, how much power would an ideal heat
pump require to keep your inside temperature at
20 C?
Hint Now you know Qh. Solve for Won. Assume
Qc/Qh Tc/Th (Carnot)
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Example Heat Pump
Suppose that the heat flow out of your 20C home
in the winter is 7 kW. If the temperature outside
is -15 C, how much power P would an ideal heat
pump require to keep your inside temperature at
20 C?
Hint Now you know Qh. Solve for Won. Assume
Qc/Qh Tc/Th (Carnot)
Qc Won Qh QhTc/Th Won.
Therefore, Won QH(1 - Tc/Th). So, in one
second it takes work Won (7kJ)(1 - 258/293)
0.84 kJ
Power required 840 W
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Quasi-static Heat Flow and Entropy reminder

• We saw for constant V dS dU/T dQ/T
• in every quasi-static case dS dQ/T
• Reason in quasi-static process, total S (of
  system plus environment) doesnt change
• So dS (of system) must cancel dSE of environment
• In quasi-static process, environment is at the
  same T (and p, if volume changes) as the system

therefore
for quasi-static processes of ANY type
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Two blocks with equal masses, each with heat
capacity C 1 J/K are initially at different
temperatures, T1 250K, T2 350K. They are then
placed into contact, and eventually reach a final
temperature of 300K. (Why?)What can you say
about the total change in entropy DStot in this
process?a) DStot lt 0 b) DStot 0 c) DStot gt 0
ACT 1
T1 250K T2 350K Tf 300 K
T1
T2
Tf
Tf
Equal masses each with heat capacity C 1J/K
For a solid, C CV Cp, to good approximation,
since DV ? 0.
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Two blocks with equal masses, each with heat
capacity C 1 J/K are initially at different
temperatures, T1 250K, T2 350K. They are then
placed into contact, and eventually reach a final
temperature of 300K. (Why?)What can you say
about the total change in entropy DStot in this
process?a) DStot lt 0 b) DStot 0 c) DStot gt 0
ACT 1 Solution
T1 250K T2 350K Tf 300 K
T1
T2
Tf
Tf
Equal masses each with heat capacity C 1J/K
For a solid, C CV Cp, to good approximation,
since DV ? 0.
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Home Exercise
We prepare the following system,
Heat capacities C1 100J/K C2 200J/K
a) What is the final equilibrium temperature Tf
when the systems are brought into contact?
Answer 306 K
b) What is the entropy change, DS, when the
systems are brought into contact?
Answer 3.3 J/K
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Home Exercise (a) solution
We prepare the following system,
Heat capacities C1 100J/K C2 200J/K
a) What is the final equilibrium temperature Tf
when the systems are brought into contact?
306.3 K
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Home Exercise (b) solution
We prepare the following system,
Heat capacities C1 100J/K C2 200J/K
b) What is the entropy change, DS, when the
systems are brought into contact?
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ACT 2 Work from a hot brick

• Heat a brick to 400 K. Connect it to a
  Carnot Engine. How much work can we extract if
  the cold reservoir is 300 K? The brick has a
  constant heat capacity of C 1 J/K.
• a. lt 25 J
• b. 25 J
• c. 100 J

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ACT 2 Work from a hot brick

• Heat a brick to 400 K. Connect it to a
  Carnot Engine. How much work can we extract if
  the cold reservoir is 300 K? The brick has a
  constant heat capacity of C 1 J/K.
• a. lt 25 J
• b. 25 J
• c. 100 J

Did you use the relation Wby Qh(1 -Tc/Th)
? If so, you missed that the brick was cooling
during the process. Th became smaller. So the
engine efficiency kept dropping.
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ACT 2 Work from a hot brick

• Heat a brick to 400 K. Connect it to a
  Carnot Engine. How much total work can we
  extract from the brick (as it cools) if the cold
  reservoir is 300 K? The brick has a heat capacity
  C 1 J/K, independent of T.

Did you use the relation Wby Qh(1 -Tc/Th) ?

If so, you missed that the brick was cooling
during the process. You must set Th T (a
variable) and integrate (Remember Qh and Qc
defined positive.) dWby dQh - dQc
dQh - (Tc/T) dQh where dQh ?CdT ?.
You will complete this problem in Discussion.
Here Wby - DU TcDS Is that true
in general?
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Available Work Change in Free Energy

• For a small heat input dQh from the brick, the
  Carnot engine performs work
• For the brick dU -dQh and dS -dQh /Th.
  So
• Its the temperature of the environment that
  enters the work. This equation applies to each
  step of the process. When the brick finally
  reaches Tc, the total work accomplished is
• we define the free energy of the brick (fuel) in
  an environment of temperature T. (In this
  case, T Tc.)

In general
F U - TS Free energy ? Maximum Available Work
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ACT 3 Work from a cold brick?
We obtained positive work Whot from a hot brick,
initially at 400K
If instead the brick were initially at 200K, how
much positive work Wcold could we extract? a)
Wcold gt 0 b) Wcold lt 0 c) Wcold 0
Hint Fill in the diagram
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ACT 3 Work from a cold brick?
We obtained positive work Whot from a hot brick,
initially at 400K
If instead the brick were initially at 200K, how
much positive work Wcold could we extract? a)
Wcold gt 0 b) Wcold lt 0 c) Wcold 0
Before we had the heat flow from the brick to the
environment now we have the heat flow from the
environment to the brick. But theres still a
heat flow ? work! How much? Need to calculate
Wby Fi - Ff
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Exercise Free Energy and Equilibrium
What is the free energy of an object that is
hotter or colder than its environment?
Start with Tbrick temperature of environment (a
thermal reservoir) 300K. Change the brick
temperature and see how its free energy FB
changes.
Heat Capacity of brick C 1kJ / K
DUB C DT C(TB - 300K)
290K 300K 310K
DSB C ln (TB / 300K)
DFB DUB - (300K) DSB
1) Heat the brick to 310 K. DT 10 K.

DFB

2) Cool the brick to 290 K. DT -10 K.
DFB
Plot your results. Conclusion?
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Exercise Free Energy and Equilibrium
What is the free energy of an object that is
hotter or colder than its environment?
Start with Tbrick temperature of environment (a
thermal reservoir) 300K. Change the brick
temperature and see how its free energy FB
changes.
Heat Capacity of brick C 1kJ / K
DUB C DT C(TB - 300K)
290K 300K 310K
DSB C ln (TB / 300K)
DFB DUB - (300K) DSB
1) Heat the brick to 310 K. DT 10 K.

DFB C (10K) - (300K) C ln(310/300) 10kJ -
9.84kJ 0.16 kJ Positive

2) Cool the brick to 290 K. DT -10 K.
DFB C (-10K) - (300K) C ln(290/300) -10kJ
10.17kJ 0.17 kJ Positive!
Plot your results. Conclusion?
Free energy of brick is a minimum at T
Tenvironment
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Available Work Change in Free Energy

• Wby - DU Te DS - DF Fi - Ff
• where F U - TS is known as the Helmholtz free
  energy of the system referenced to the
  temperature Te of the environment (or
  reservoir).
• The free energy of an object is always defined
  with reference to the temperature of a reservoir,
  often the environment.

F U - TS Free energy ? Maximum Available Work
Wby Fi - Ff
Free energy is converted into work!
Which has higher F? hot stone room temp stone
stone on ground raised stone unlit match
burned match
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Converting Chemical Fuels into Work
Fuel Tabulated Free Energy Ethanol 10
kJ/g ? 8000 kJ / liter Methanol 20 KJ/g
? 15,000 kJ / liter Gasoline 40 kJ/g ?
30,000 kJ / liter

• Problem
• If you could convert the free energy of gas
  perfectly into work, how many miles per gallon
  would your car achieve? (Wow, can we really do
  this problem? Sure)
• Solution
• First we need to know how much work it takes
  to drive the car 1 mile. Obviously that depends
  on a number of factors Speed, tire friction,
  wind resistance, etc.
• Actually a simple experiment can give us the
  answer
• Determine the decelerating force! Work force
  x distance.

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Converting Chemical Fuels into Work (contd)
Fuel Tabulated Free Energy Ethanol
10 kJ/g ? 8000 kJ / liter Methanol 20 KJ/g
? 15,000 kJ / liter Gasoline 40 kJ/g ?
30,000 kJ / liter
1 mph 0.45 m/s

• Professor Ks
  Experiment
• I find that when I depress the clutch at 65 mph,
  my car slows to 55 mph in 10 seconds. (Dv ? 5
  m/s)
• Force of wind and friction m DV/Dt (1500 kg)
  x (5 m/s)/(10 sec) 750 N.
• Actual work to drive 1 mile (750 N) x (1600 m)
  1.2 MJ (megajoules).
• If the free energy of gas were converted
  perfectly into work, I would need
• 1.2 MJ / (30 MJ / liter) 0.04 liters 0.01
  gallons of fuel.
• Therefore, if my car were powered by a perfect
  Carnot engine with no friction or extra heat
  loss, I could expect 100 miles per gallon!
• (This calculation is pretty rough. The purpose
  is to demonstrate that Free
  Energy applies to physics, chemistry and
  engineering.)

Actually gas engines are quite
efficient (50) due to large temperature
difference between the hot gases and environment.
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Equilibrium is found at Free Energy Minimum
Equilibrium corresponds to a maximum in total
entropy Stot S SR maximum If
the system does no work (constant V) and draws dU
from the environment dSR - dU /
T dStot dS dSR dS - dU / T 0
in equilibrium. Notice that dStot
- (dU - TdS) / T - dF / T
At fixed V and fixed reservoir T A maximum in
Stot corresponds to a minimum in F U - TS.

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When does maximum Stot mean minimum Fsys?

• A maximum in Stot corresponds to a minimum in F
  U - TS
• (at fixed V and fixed reservoir T)

• But often we want to discuss engines etc. in
  which some fluid V changes.
• So long as dV involves only work on some
  non-entropic piston, theres no entropy change
  other than
• System dS
• Bath (WbydU)/T -(pdVdU)/T

V
bath
So we consider the whole system in some sealed
vacuum jar with a fixed Vbig. We can include the
potential energy changes pdV of the piston
(weight, spring, whatever) as part of dUsys, and
still use F. When p is fixed (open to
environment) we will define a new variable, G
UpV-TS, which is minimized at fixed p, T (rather
than at fixed V, T)
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Why is Free Energy important?1. In many
situations maximizing total S (sometimes hard to
calculate) to reach equilibrium implies
minimizing system free energy (sometimes easier
to calculate). 2. When the system is out of
equilibrium, its excess free energy gives the
amount of work that an ideal engine could extract
from it in a given environment. For the rest
of the course we will consider important
applications of this principle
Paramagnets (revisited)The law of atmospheres
(revisited) Solids defects and
impurities Chemical reactions, especially in
gases Carrier densities in semiconductors Adsorpti
on of particles on surfaces Liquid-gas and
solid-gas phase transitions (using G, not F)