Announcements
1. Schedule –
Chapter 19 – macroscopic view of heat (today)
Chapter 20 – microscopic view of heat (Tuesday – 11/18)
Review Chapters 15-20 – (Thursday – 11/20)
Exam 3 – (Tuesday – 11/25)
2. Physics colloquium today – Professor Gary Pielak from
UNC will speak about “Protein Biophysics in Cells”.
3. Today’s lecture – Chapter 19
Temperature
Heat
The first law of thermodynamics
11/13/2003
PHY 113 -- Lecture 19
1
Dictionary definition: temperature – a measure of the the
warmth or coldness of an object or substance with
reference to some standard value. The temperature of two
systems is the same when the systems are in thermal
equilibrium.
Equilibrium:
Not equilibrium:
T1
T2
T3
“Zeroth” law of thermodynamics:
If objects A and B are separately in thermal equilibrium
with a third object C, then objects A and B are in thermal
equilibrium with each other.
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PHY 113 -- Lecture 19
2
Temperature scales
TF=9/5 TC + 32
500
450
400
350
Kelvin scale:
300
250
T = TC + 273.15o
TF
200
T≥0
150
100
50
0
-5 0
-1 0 0
-1 5 0
-2 0 0
-1 0 0 -5 0
0
50
100 150 200
TC
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PHY 113 -- Lecture 19
3
There is a lowest temperature:
T0 = -273.15o C = 0 K
Kelvin (“absolute temperature”) scale
TC = -273.15 + TK
Example –
Room temperature = 68o F = 20o C = 293.15 K
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PHY 113 -- Lecture 19
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PHY 113 -- Lecture 19
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Effects of temperature on matter
Solids and liquids
Model of a solid composed of
atoms and bonds
Thermal exansion:
∆L = α Li ∆T
∆L
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Li (equilibrium bond length
at Ti)
PHY 113 -- Lecture 19
6
Typical expansion coefficients at TC = 20o C:
Linear expansion: ∆L = α Li ∆T
α = 11 x 10-6/ oC
Steel:
Concrete:
α = 12 x 10-6/ oC
Volume expansion:
V=L3 Î ∆V = 3α Vi ∆T = β Vi ∆T
Alcohol:
Air:
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β = 1.12 x 10-4/ oC
β = 3.41 x 10-3/ oC
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7
Brass
Steel
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Heat – Q -- the energy that is transferred between a “system”
and its “environment” because of a temperature difference that
exists between them.
Sign convention:
Q<0 if T1 < T2
Q>0 if T1 > T2
T1
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T2
Q
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Heat withdrawn
from system
Thermal equilibrium
– no heat transfer
Heat added to system
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PHY 113 -- Lecture 19
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Units of heat: Joule
Other units: calorie = 4.186 J
Kilocalorie = 4186 J = Calorie
Heat capacity: C = amount of heat which must be added
to the “system” to raise its
temperature by 1K (or 1o C).
Q = C ∆T
More generally :
Tf
Qi → f = ∫ C (T )dT
Ti
Heat capacity per mass: C=mc
Heat capacity per mole (for ideal gas): C=nCv
11/13/2003
PHY 113 -- Lecture 19
C=nCp
11
Some typical specific heats
Material
Water (15oC)
Ice (-10oC)
Steam (100oC)
Wood
Aluminum
Iron
Gold
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J/(kg·oC)
4186
2220
2010
1700
900
448
129
PHY 113 -- Lecture 19
cal/(g·oC)
1.00
0.53
0.48
0.41
0.22
0.11
0.03
12
Heat and changes in phase of materials
Example: A plot of temperature versus Q added to
1g = 0.001 kg of ice (initially at T=-30oC)
Q=333J
Q=2260J
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PHY 113 -- Lecture 19
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Some typical latent heats
Material
Ice ⇒Water (0oC)
Water ⇒ Steam (100oC)
Solid N ⇒ Liquid N (63 K)
Liquid N ⇒ Gaseous N2 (77 K)
Solid Al ⇒ Liquid Al (660oC)
Liquid Al ⇒ Gaseous Al (2450oC)
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PHY 113 -- Lecture 19
J/kg
333000
2260000
25500
201000
397000
11400000
14
Peer instruction question
Suppose you have a well-insulated cup of hot coffee
(m=0.25kg, T=100oC). In order to get to class on time you add
0.25 kg of ice (at 0oC). When your cup comes to equilibrium,
what will be the temperature of the coffee.
(A) 0oC
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(B) 10oC
(C) 50oC
PHY 113 -- Lecture 19
(D) 100oC
15
Review question
Suppose you have a well-insulated cup of hot coffee
(m=0.25kg, T=100oC). In order to get to class on time you add
0.25 kg of ice (at 0oC). When your cup comes to equilibrium,
what will be the temperature of the coffee.
Q = m cw (Tf – 100oC) + m Lice + m cw (Tf − 0oC) = 0
2 cw Tf = cw · 100 − Lice
Tf = cw · 100/(2 cw ) − Lice /(2 cw )
Tf = 50 –333000/(2 · 4186) = 10oC
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PHY 113 -- Lecture 19
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Energy in the form of work:
Work done by the “system” due to volume change
xf
xf
Vf
F
W = ∫ Fdx = ∫ Adx = ∫ PdV
xi
xi A
Vi
Sign convention:
W>0 if Vf > Vi (expansion)
W<0 if Vf < Vi (contraction)
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PHY 113 -- Lecture 19
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Work done by a gas:
P (1.013 x 105) Pa
“Isobaric” (constant pressure process)
Po
W= Po(Vf - Vi)
Vi
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Vf
PHY 113 -- Lecture 19
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Work done by ideal gas:
“Isovolumetric” (constant volume process)
P (1.013 x 105) Pa
Pf
W=0
Pi
Vi
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PHY 113 -- Lecture 19
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Work done by a gas:
“Isothermal” (constant temperature process)
P (1.013 x 105) Pa
For ideal gas:
PV = nRT
Wi → f
Vf
Pi
⎛V f
= PiVi ln⎜⎜
⎝ Vi
Vi
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⎛V f
nRT
= ∫ PdV = ∫
dV =nRT ln⎜⎜
Vi
Vi V
⎝ Vi
Vf
⎞
⎟⎟
⎠
Vf
PHY 113 -- Lecture 19
20
⎞
⎟⎟
⎠
Work done by a gas:
“Adiabatic” (no heat flow in the process process) Qi→f = 0
For ideal gas:
P (1.013 x 105) Pa
PVγ = PiViγ
Pi
Wi → f
Vi
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PiVi
PiVi ⎡⎛ V f
= ∫ PdV = ∫ γ dV =
⎢⎜⎜
γ − 1 ⎣⎢⎝ Vi
Vi
Vi V
Vf
Vf
γ
⎞
⎟⎟
⎠
γ −1
⎤
− 1⎥
⎥⎦
Vf
PHY 113 -- Lecture 19
21
Thermodynamic statement of conservation of energy –
First Law of Thermodynamics
∆Eint = Q - W
Work done by system
Heat added to system
“Internal” energy of system
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PHY 113 -- Lecture 19
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“Internal energy” Eint=Eint(T,V,P)
(for an ideal gas, Eint=Eint(T)
W
Tsys
“environment”
can be changed by interaction with the system
Q
“system” Tenv
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∆Eint = Q - W
Note: Thermal equilibrium implies a
uniform temperature. In this example,
the system and the environment are
presumed to be in thermal equilibrium
within themselves but not in thermal
equilibrium with eachother.
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Examples process:
Wnet=W(ABCDA)=WBC+WDA
P (1.013 x 105) Pa
Pf
B
A
Pi
C
D
=(Pf−Pi) (Vf−Vi)
∆Eint(ABCDA)= ∆Eint(AB)
+∆Eint(BC) +∆Eint(CD)
+∆Eint(DA)=0
ÎQnet=Q(ABCDA)=Wnet
Vi
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Vf
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6 m3
1 m3
30 Pa
10 Pa
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Mechanisms of heat transfer:
1. Thermal convection
Heat transmitted by heated fluid flow. (Ex. -- home
heating units, stove top cooking)
2. Radiation
Heat transmitted by electromagnetic radiation. (Ex. –
sun, microwave cooking)
3. Thermal conduction
Heat transfer from one side of a material to another
due to a temperature gradient.
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PHY 113 -- Lecture 19
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Quantitative statement of thermal conduction:
∆Q
∆T
Units: 1 J/s = 1 Watt (W)
= kA
∆t
∆x
cross-sectional area
thermal conductivity coefficient
Material
Copper
Glass
Water
Air
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k (W/(m·oC))
238
0.8
0.6
0.0234
PHY 113 -- Lecture 19
29
Examples of thermal conduction
T1
T −T
dQ
= keff A 2 1
L2 + L1
dt
T2 − T1
dQ
= k2 A
L2
dt
keff
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PHY 113 -- Lecture 19
L2 + L1
=
L2 / k 2 + L1 / k1
30
Example: T1=20oC, T2= 0oC
Single pane glass window:
A=1m2 , L2 = 0.002m, k2=0.8 W/m·oC
T2 − T1
dQ
= k2 A
= 8000 W
dt
L2
Double pane glass window: L1=0.01m, k1=0.0234 W/m·oC (air)
T2 − T1
dQ
= keff A
L2 + L1 + L0
dt
keff
L2 + L1 + L0
=
= 46 W
L2 / k 2 + L1 / k1 + L0 / k0
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