Download Physics 231 Topic 14: Laws of Thermodynamics Wade Fisher

Physics 231
Topic 14: Laws of Thermodynamics
Wade Fisher
Nov
- Dec
2012
MSU28
Physics
231 Fall32012
1
Some Housekeeping
The final exam for PHY 231 will be held Dec 11 from 8-10pm.
Sections 002 will take the exam in BPS 1410.
Other sections will go to other rooms, so don’t get
confused by friends in other sections!
MSU Physics 231 Fall 2012
2
Some Housekeeping
RCPD Students: Set up an appointment at Bessey or talk to me.
You MUST do one or the other and let me know.
Make-Up Exams: The make-up exam is scheduled Dec 12 8-10AM
Valid reasons to take the make-up exam:
1) You have 3 exams on Dec 11 and can prove it.
2) You have a direct exam conflict 8-10PM
3) You have evidence of a serious conflict (note from
Doctor, Dean, Coach must be provided).
Invalid reasons to take the make-up exam:
1) I’m leaving for holidays before the exam.
2) I have something else scheduled 8-10PM
If you are sick: Any make-ups after Dec 11 will be at the
discretion of the instructors.
Most likely these will take place Dec 23rd at 7-9AM.
If you do not contact me before the exam, you cannot get a
make-up exam.
MSU Physics 231 Fall 2012
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Question
We will be doing 2-3 days of review before the final exam.
What mixture of course material would you prefer that we
cover in the review?
3 Material Categories:
Topics covered for Exam 1 (Chs 1-5)
Topics covered for Exam 2 (Chs 6-10)
Remaining topics (Chs 10-14)
a)
b)
c)
d)
e)
1/3 Exam 1, 1/3 Exam 2, 1/3 Remaining Topics
50% Exam 1, 30% Exam 2, 20% Remaining Topics
30% Exam 1, 50% Exam 2, 20% Remaining Topics
40% Exam 1, 40% Exam 2, 20% Remaining Topics
50% Exam 1, 50% Exam 2, 0% Remaining Topics
MSU Physics 231 Fall 2012
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Clicker Question!
Ice is heated steadily and becomes liquid and then vapor.
During this process:
a) the temperature rises continuously.
b) when the ice turns into water, the temperature
drops for a brief moment.
c) the temperature is constant during the phase
transformations
d) the temperature cannot exceed 100oC
MSU Physics 231 Fall 2012
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Key Concepts: Laws of
Thermodynamics
Laws of Thermodynamics
1st Law: U = Q + W
2nd Law: Heat flows from hotter  cooler
Thermodynamic Processes
Adiabatic (no heat flow)
Work done in different processes
Heat Engines & Refrigerators
Carnot engine & efficiency
Entropy
Relationship to heat, energy.
Statistical interpretation
Covers chapter 14 in Rex & Wolfson
MSU Physics 231 Fall 2012
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A Piston Engine
Piston is moved downward
slowly so that the gas remains
in thermal equilibrium:
The temperature is the the
same at all times in the gas,
but can change as a whole.
piston
area A
y
P,V,T
T > To
vin
vout
MSU Physics 231 Fall 2012
Vout>Vin
Work is done on the gas
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piston
Isobaric Compression
area A
Let’s assume that the pressure does
not change while lowering the piston
(isobaric compression).
y
P,V,T
P
P
T > To
f
Vf
i
Vi V
W: work done on the gas
W= -Fy = -PAy
(P=F/A)
W= -PV = -P(Vf-Vi) (in Joule)
Sign of the work:
+ if V<0
- if V>0
This corresponds to the area under
the curve in a P-V diagram
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piston
area A
Non-isobaric Compression
In general, the pressure can change
when lowering the piston.
y
P,V,T
P
Pf
T > To
f
The work done by the gas is the
opposite of the work done on the gas.
i
Pi
Vf
The work done on the gas when going
from an initial state (i) to a final
state (f) is the area under the P-V
diagram.
Wgas on piston = -Wpiston on gas
Vi V
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Work Done on Gases:
Getting the Signs Right!
P
Pf
f
i
Pi
Vf
Vi V
If the arrow goes from right to left (volume becomes
smaller), positive work is done on the gas. If the arrow goes
from left to right (volume larger), negative work is done on
the gas (the gas has done positive work on the piston)
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Question
P
P
A
v
B
v
P
C
v
In which case is the work done on the gas largest?
Case B: area under curve is larger and arrow goes from
right to left. Positive work is done on the gas.
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Iso-Volumetric Process
P
v
Work done on/by gas: W = -PV = 0
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Question
Patm
PA
Masslid=50 kg
Alid=100 cm2
Patm=1x105 Pa
a) What is the pressure PA?
b) If the temperature in the volume is raised
the lid moves up by 5 cm. How much work
is done by the gas?
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Example
One mole of an ideal gas initially at 0° C undergoes an
expansion at constant pressure of one atmosphere to
four times its original volume.
a) What is the new temperature?
b) What is the work done on the gas?
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Clicker Quiz!
A gas is enclosed in a cylinder with a moveable piston. The
figures show 4 different PV diagrams. In which case is the
work done by the gas largest?
Work: area under PV diagram
Work done by the gas: volume must become larger, which
leaves a or c. Are is larger for a. Correct answer: a)
MSU Physics 231 Fall 2012
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First Law of
Thermodynamics
By transferring heat to an object
the internal energy can be changed
Think about heat transfer
By performing work on an object
the internal energy can be changed
Think about deformation/pressure
The change in internal energy depends on
the work done on the object and the
amount of heat transferred to the object.
Remember: internal energy is the energy
associated with translational, rotational,
vibrational motion of atoms + PE
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First Law of Thermodynamics
U = Uf - Ui = Q + W
U = change in internal energy
Q = energy transfer through heat
(+ if heat is transferred to the system)
W = energy transfer through work
(+ if work is done on the system)
This law is a general rule for conservation of energy
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First Law: Isobaric Process
A gas in a cylinder is kept at 1.0x105 Pa. The cylinder is
brought in contact with a cold reservoir and 500 J of heat
is extracted. Meanwhile the piston has sunk and the volume
changed by 100cm3. What is the change in internal energy?
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Types of Thermal Processes
A: Isovolumetric V=0
B: Adiabatic
Q=0
C: Isothermal T=0
D: Isobaric
P=0
Ideal Gas:
PV/T = constant
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Iso-volumetric Process: Line A
V=0
W = 0 (area under the curve is zero)
U = Q (Use U=W+Q, with W=0)
In case of ideal gas:
U = 3/2nRT
• if P then T
(PV/T=constant)
so U=negative Q=negative
(Heat is extracted from the gas)
•if P then T (PV/T=constant)
so U=positive Q=positive
(Heat is added to the gas)
Cv=(3/2)R so U=Cvn T
Cv: molar specific heat at const. vol.
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Isobaric Process: Line D ( P = 0 )
Use U=W+Q
In case of ideal gas:
W = -PV & U = 3/2nRT
• if V then T (PV/T=constant)
W: positive (work done on gas)
U: negative Q: negative
(heat extracted)
• if V then T (PV/T=constant)
W: negative (work done by gas)
U: positive
Q: positive
(heat added)
Q = U-W = 3/2nRT+ PV = 3/2nRT+ nRT = 5/2nRT
(used ideal gas law) Q = nCpT with Cp = (5/2)R
Cp = molar heat capacity at constant pressure
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Adiabatic Process: Line B ( Q=0 )
No heat is added or extracted
from the system.
U=W (Use U=W+Q, with Q=0)
In case of ideal gas:
U = 3/2nRT
• if T U=negative W=negative
(The gas has done work)
• if T U=positive W=positive
(Work is done on the gas)
Ideal gas and adiabatic: PV =Constant
Pf Vf − Pi Vi 3
W=
= nR∆T
γ−1
2
MSU Physics 231 Fall 2012
=Cp/Cv=5/3
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Isothermal Processes
T=0 so U=0
The temperature is not changed
Q=-W (Use U=W+Q, with U=0)
• if V
W=positive Q=negative
(Work is done on the gas
and energy extracted)
• if V
W=negative Q=positive
(Work is done by the gas
and energy added)
Using calculus:
𝑉𝑓
W = −nRT ln
𝑉𝑖
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Process
Isobaric
U
nCvT
Q
nCpT
W
-PV
Adiabatic
nCvT
0
U
Isovolumetr
ic
Isothermal
nCvT
U
0
0
-W
-nRTln(Vf/Vi)
General
nCvT
U-W
(PV Area)
negative if V
expands
Ideal (monatomic) gas (only case looked at here…)
Cv=3/2 R
Cp=5/2 R
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P(x105 Pa)
f
6
First Law: General Case
i
3
1
In ideal gas is compressed (see
P-V diagram).
A) What is the change in internal energy
B) What is the work done on the gas?
C) How much heat has been
transferred to the gas?
4 V(m3)
A) Use U = 3/2nRT and PV=nRT so U=3/2PV & U=3/2(PV)
U = 3/2(PfVf - PiVi) = 3/2[(6E+05)x1 - (3E+05)x4] = -9E+5 J
B) Work: area under the P-V graph: (9+4.5)x105 = 13.5x105
(positive since work is done on the gas)
C) U = Q+W so Q = U-W = (-9E+5)-(13.5E+5) = -22.5E+5 J
Heat has been extracted from the gas.
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piston
area A
y
P,V,T
First Law: Adiabatic process
A piston is pushed down rapidly. Because the
transfer of heat through the walls takes a
long time, no heat can escape.
During the moving of the piston, the
temperature has risen 1000C. If the
container contained 10 mol of an ideal gas,
how much work has been done during the
compression?
U=E =3/2nRT
kin
MSU Physics 231 Fall 2012
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Clicker Quiz!
A vertical cylinder with a movable cap
is cooled. The process corresponding
to this is:
a)
b)
c)
d)
e)
CB
AB
AC
CA
Not shown
After the cooling of the gas and the lid
has come to rest, the pressure is … as
before the cooling process.
a) Lower b) higher c) the same
MSU Physics 231 Fall 2012
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Adiabatic gas
An ideal monatomic gas goes from
P1 = 180 atm and V1 = 80 m3 to P2
and V2 via an adiabatic process.
If V2 = 190 m3, what is P2 in atm?
ideal gas and adiabatic: PV =Constant
=Cp/Cv=5/3
(PV)i=(PV)f so (PV)i/Vf =Pf
(180 x 805/3)/1905/3=42.6 atm
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In which of the four paths is…
1) …no work performed on/by
the gas?
2) …heat transfer Q zero
3) …the work done on the gas
non-zero and can the
equation W=-pV can be
used
1) A: no area under the curve, so no work performed
2) B: Adiabatic process: Q=0
3) D: pressure constant, so W=-pV
MSU Physics 231 Fall 2012
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Cyclic processes
P (atm)
A
5.0
1.0
C
10
B
50
V (liter)
In a cyclic process,
The system returns to its original state.
Therefore, the internal energy must be the same after
completion of the cycle (U=0)
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Cyclic Process: Step by Step (1)
Process A-B.
Negative work is done on the gas:
(the gas is doing positive work).
P (atm)
A
5.0
1.0
C
10
W=-Area under P-V diagram
=-([(50-10)10-3][(1.0-0.0)105]
+½[(50-10)10-3][(5.0-1.0)]105)
=-4000-8000 (work done on gas)
B
50
V (liter)
U = 3/2nRT = 3/2(PBVB-PAVA)
= 1.5[(1E+5)(50E-03)-(5E+5)(10E-03)] = 0
The internal energy has not changed
U=Q+W so Q = U-W = 12000 J
Heat that was added to the system was used to do the work!
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Cyclic Process: Step by Step (2)
P (atm)
A
5.0
1.0
Process B-C
W=-Area under P-V diagram
C
10
=-[(10-50)10-3(1.0-0.0)105]
W=4000 J
Work was done on the gas
B
50
V (liter)
U = 3/2nRT = 3/2(PcVc-PbVb)
= 1.5[(1E+5)(10E-3)-(1E+5)(50E-3)] = -6000 J
The internal energy has decreased by 6000 J
U=Q+W so
Q = U-W = -6000-4000 J = -10000 J
10000 J of energy has been transferred out of the system.
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Cyclic Process: Step by Step (3)
P (atm)
A
5.0
1.0
Process C-A
W=-Area under P-V diagram
W=0 J
No work was done on/by the gas.
C
10
B
50
V (liter)
U=3/2nRT=3/2(PcVc-PbVb)=
=1.5[(5E+5)(10E-3)-(1E+5)(10E-3)] = 6000 J
The internal energy has increased by 6000 J
U=Q+W so Q = U-W = 6000-0 J = 6000 J
6000 J of energy has been transferred into the system.
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Summary of the process
Quantity
Process
P (atm)
A
5.0
1.0
B
C
10
50
A-B
Work(W)
Heat(Q)
U
A-B
-12000 J
12000 J
0J
B-C
4000 J
-10000 J
-6000 J
C-A
0J
6000 J
6000 J
-8000 J
8000 J
0
SUM
V (liter)
B-C
MSU Physics 231 Fall 2012
C-A
34
Clicker Quiz!
A vertical cylinder with a movable cap
is cooled. The process corresponding
to this is:
a)
b)
c)
d)
e)
CB
AB
AC
CA
Not shown
MSU Physics 231 Fall 2012
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Cyclic processes
P (atm)
A
5.0
1.0
C
10
B
50
V (liter)
In a cyclic process,
The system returns to its original state.
Therefore, the internal energy must be the same after
completion of the cycle (U=0)
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Summary of the process
Quantity
Process
P (atm)
A
5.0
1.0
B
C
10
50
A-B
Work(W)
Heat(Q)
U
A-B
-12000 J
12000 J
0J
B-C
4000 J
-10000 J
-6000 J
C-A
0J
6000 J
6000 J
-8000 J
8000 J
0
SUM
V (liter)
B-C
MSU Physics 231 Fall 2012
C-A
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P (atm)
A
5.0
1.0
The gas performed net work (8000 J)
while heat was supplied (8000 J):
We have built an engine!
B
C
10
P (atm)
50
A
5.0
1.0
What did we do?
C
10
B
50
V (liter)
The work done on/by the gas is
equal to the area of the loop
What if the process was done in
the reverse way?
Net work was performed on the
gas and heat extracted from the gas.
We have built a heat pump!
(A fridge)
V (liter)
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So…
Clockwise loop:
Engine : heat is supplied to let the gas do work
Counterclockwise loop:
Refrigerator: heat is extracted by doing work on the
gas
Work performed in a cycle: area of loop in PV diagram
Engine: work on gas is negative
Refrigerator: work on gas is positive
MSU Physics 231 Fall 2012
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A Cycle
P
3P0
P0
V0
3V0 V
Consider the cycle in the figure.
A) what is the net work done in
one cycle?
B) What is the net energy added
to the system per cyle?
MSU Physics 231 Fall 2012
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Generalized Heat Engine
Water turned to steam
W = |Qh|-|Qc|
Heat reservoir Th
efficiency: W/|Qh|
e=1-|Qc|/|Qh|
Qh
W
engine
The steam
moves a piston
The steam is
condensed
Qc
Cold reservoir Tc
Work is done
Work
The efficiency is determined
by how much of the heat you
supply to the engine is turned
into work instead of being lost
as waste.
MSU Physics 231 Fall 2012
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Reverse Direction: The Fridge
MSU Physics 231 Fall 2012
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Heat
Pump
heat is expelled to outside
heat reservoir Th
Qh
a piston compresses the coolant work is done
engine
Qc
W
the fridge is cooled
cold reservoir Tc
work
Coefficient of performance
COPheating=|Qh|/W
Qh: amount of energy rejected
into the hot reservoir
W: work performed by the
device
MSU Physics 231 Fall 2012
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Question
Two moles of a perfect gas are
taken along the cycle shown below.
Which of the following is true about
the internal energy U?
a. Not enough information given
b. UD < UA
c. UB > UC
d. UD = UB
e. UC = UA
U=3/2nRT and PV=nRT so U=3/2PV
UA=3/2 x P1V1
UB=3/2 x P1 x 2V1 = 2UA
Uc=3/2 x 2P1 x 2V1 = 4UA
UD=3/2 x 2P1 x V1 = 2UA
MSU Physics 231 Fall 2012
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Clicker Quiz!
P (Pa)
3x105
Consider this clockwise cyclic
process.
Which of the following is true?
1x105
1
a)
b)
c)
d)
e)
3
V (m3)
This is a heat engine and the W done by the gas is +4x105 J
This is a heat engine and the W done by the gas is +6x105 J
This is a heat engine and the W done by the gas is –4x105J
This is a fridge and the W done on the gas is +4x105 J
This is a fridge and the W done on the gas is +6x105 J
Clockwise: work done by the gas, so heat engine
W by gas=area enclosed = (3-1) x (3x105-1x105) = 4x105 J
MSU Physics 231 Fall 2012
45
The 2nd Law of Thermodynamics
1st law: U=Q+W
In a cyclic process (U=0) Q=-W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law: Heat can only flow spontaneously from a hot mass to
a cold mass.
IE, heat cannot move from a cold mass to a hot mass
without work being done “Engines cannot be 100% efficient!”
What is the most
efficient engine we can
make given a heat and
a cold reservoir?
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AB isothermal expansion
BC adiabatic expansion
W-, Q+
W-, T-
Carnot engine
Q=0
T=0
Th
T=0
Q=0
W+, T+
DA adiabatic compression
W+, Q-
Tc
CD isothermal compression
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Carnot cycle
Work done by engine: Weng
Weng = Qhot-Qcold
Efficiency: e = 1-Tcold/Thot
e = 1-|Qc|/|Qh| also holds since
this holds for any engine
inverse Carnot cycle
A heat pump or a fridge!
By doing work we can
transport heat
MSU Physics 231 Fall 2012
48
Carnot engine
General engine:
W=|Qh|-|Qc|
efficiency: W/|Qh|
e=1-|Qc|/|Qh|
e =1-Tcold/Thot carnot only!!
In general: e < ecarnot
The Carnot engine is the most efficient way to operate
an engine based on hot/cold reservoirs because the
process is reversible: it can be reversed without loss
or dissipation of energy
Unfortunately, a perfect Carnot engine cannot be built.
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The 2nd law of thermodynamics
1st law: U=Q+W
In a cyclic process (U=0) Q=-W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law in equivalent forms:
- Heat flows spontaneously ONLY from hot to cold masses
- Heat flow is accompanied by an increase in the entropy
(disorder) of the universe
- Natural processes evolve toward a state of maximum
entropy
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50
Entropy
Lower Entropy
Higher Entropy
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51
Reversing Entropy
We can only reverse the
increase in entropy if we
do work on the system
Do work to compress
the gas back to a
smaller volume
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52
Entropy
Only easy to define the CHANGE in entropy (S):
∆S =
Q
T
(J/K)
Example: Carnot engine
HotCold reservoir: Shot = -Qhot/Thot (entropy is decreased)
ColdHot reservoir: Scold = Qcold/Tcold
We saw: Efficiency for a general engine: e = 1 - Qcold/Qhot
Efficiency for a Carnot engine: e = 1 - Tcold/Thot
So for a Carnot engine: Tcold/Thot = Qcold/Qhot
and thus: Qhot/Thot = Qcold/Tcold
Total change in entropy: Shot + Scold =0
For a Carnot engine, there is no change in entropy
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Entropy and Work
Entropy represents an
inefficiency wherein
energy is “lost” and cannot
be used to do work.
Shot = Qhot/Thot = -24000J / 400K = -60 J/K
Scold = Qcold/Tcold
= +24000J / 300K = +80 J/K
Shot + Scold = -60 J/K + 80 J/K = +20 J/K
Entropy increases!
Cold mass: Gained heat, can do more work.
Hot mass: Lost heat, can do less work.
Cold mass gained less potential to do work than host mass lost.
Net loss in the ability to do work.
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54
Entropy and Work
In the extreme case where thermal contact occurs,
we lose all ability to do work.
Carnot Engine #1
Thot >> Tcold
e =1-Tcold/Thot ~1
Carnot Engine #2
Thot = Tcold
e =1-Tcold/Thot =0
Heat reservoir Thot
Heat reservoir Thot
Engine 1
Cold reservoir Tcold
Work
Engine 2
Work??
Cold reservoir Tcold
MSU Physics 231 Fall 2012
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Examples for this chapter
One mole of an ideal gas initially at 00C undergoes an
expansion at constant pressure of one atmosphere to
four times its original volume.
a) What is the new temperature?
b) What is the work done on the gas?
a) PV/T=constant so if V x4 then T x4 273K*4=1092 K
b) W=-PV
use PV=nRT
before expansion: PV=1*8.31*273=2269 J
after expansion: PV=1*8.31*1092=9075 J
W=-PV=-(PV)=-[(PV)f-(PV)i]=-(9075-2269)=-6806 J
-6806 J of work is done on the gas.
MSU Physics 231 Fall 2012
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Another Example
A gas goes from initial state I to
final state F, given the parameters
in the figure. What is the work done
on the gas and the net energy transfer
by heat to the gas for:
a) path IBF b) path IF c) path IAF
(Ui=91 J Uf=182 J)
The volume expands, positive work is done by the gas.
So negative work is done on the gas. W is defined as
gas: Work on the gas.
a) work done on
-area under graph: note the ‘-’ sign before area
W=-(0.8-0.3)10-3*2.0*105=-100 J U=Uf-Ui=182-91=91
U=W+Q 91=-100+Q so Q=191 J
b) W=-[(0.8-0.3)10-3*1.5*105 + ½(0.8-0.3)10-3*0.5*105]=-87.5 J
U=W+Q 91=-87.5+Q so Q=178.5 J
c) W=-[(0.8-0.3)10-3*1.5*105]=-75 J
U=W+Q 91=-75+QMSUso
Q=166 J
Physics 231 Fall 2012
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Another Example
The efficiency of a Carnot engine is 30%. The engine absorbs
800 J of energy per cycle by heat from a hot reservoir at
500 K. Determine a) the energy expelled per cycle and b)
the temperature of the cold reservoir. c) How much work
does the engine do per cycle?
a) Generally for an engine: efficiency: 1-|Qcold|/|Qhot|
0.3=1-|Qcold|/800, so |Qcold|=-(0.3-1)*800=560 J
b) for a Carnot engine: efficiency: 1-Tcold/Thot
0.3=1-Tcold/500, so Tcold=-(0.3-1)*500=350 K
c) W=|Qhot|-|Qcold|=800-560=240 J
MSU Physics 231 Fall 2012
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A New Powerplant
A new powerplant is designed that makes use of the
temperature difference between sea water at 0m (200) and
at 1-km depth (50). A) what would be the maximum efficiency
of such a plant? B) If the powerplant produces 75 MW, how
much energy is absorbed per hour? C) Is this a good idea?
a) maximum efficiency=carnot efficiency=1-Tcold/Thot=
1-278/293=0.051 efficiency=5.1%
b) P=75*106 J/s W=P*t=75*106*3600=2.7x1011 J
efficiency=1-|Qcold|/|Qhot|=(|Qhot|-|Qcold|)/|Qhot|=
W/|Qhot| so |Qhot|=W/efficiency=5.3x1012 J
c) Yes! Very Cheap!! but… |Qcold|= |Qhot|-W=5.0x1012 J
every hour 5E+12 J of waste heat is produced:
Q=cmT 5E+12=4186*m*1 m=1E+9 kg of water is heated
by 10C.
MSU Physics 231 Fall 2012
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Clicker Quiz!
A person decides to lower his heating cost by replacing
a wall. The new wall is made of a material of which the
thermal conductivity is 4 times smaller than that of
the original wall. However, the material is expensive and
so he makes the wall 3 times thinner than the original wall.
What is the new heat transfer through the wall,
relative to the old wall (Pnew/Pold)?
a)
b)
c)
d)
e)
¼
1/3
¾
3
4
Pold=koldAT/xold
Pnew=(kold/4)AT/(xold/3)=3Pold/4
MSU Physics 231 Fall 2012
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For Next Week
Homework Set 12 Due 12/05
Covers Chapter 14
MSU Physics 231 Fall 2012
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