Transcript Chapter23 The first law of thermodynamics.ppt

Chapter 23 The first law of
thermodynamics
23-1 Heat: energy in transit(传输)
1. “Heat is energy that flows between a system
and
its environment because of a temperature
difference between them”
We choose our sign convention so that Q is
positive in the case that the internal energy of the
system tends to be increased.
Like other forms of energy, heat can be
expressed in the SI unit of Joules (J).
2. Misconceptions(误解) about heat
Neither heat nor work is an intrinsic property of a system.
We cannot say that a system “contains” a certain amount
of heat or work. They are not state functions.
Both heat and work associated with a “thermodynamic
process”
3. The understanding of heat in history
See动画库\力学夹\5-03作功与传热
23-2 The transfer of heat
1.Thermal conduction
Fig 23-2
Consider a thin slab of a
homogeneous material of
A
thicknessx and area A (Fig
23-2). One face is held at T  T T
T and the other at a
somewhat higher constant
temperature
T  T
x
.
Q
Q
Experiment shows that
is
H ( )
t
ΔT
H  kA
Δx
(23-1)
H is the “rate of heat transfer”
k is the “thermal conductivity” of the material
A is the area of the slab
x is the thickness of the slab
Considering the direction of H and infinitesimal thickness
of the slab, we have:
dT
(23-4)
H   kA
dx
Sample Problem 23-2
A thin, cylindrical metal pipe is carrying steam at a
temperature of Ts=1000c. The pipe has a diameter of
5.4 cm and is wrapped with a thickness of 5.2 cm of
fiberglass(玻璃丝) insulation. A length D=6.2 m of
the pipe passes through a room in which the
temperature is TR=110c. At what rate does heat
energy pass through the insulation?
2. Convection(对流)
3. Radiation
Problem
Two identical rectangular rods of metal are
welded end to end as shown in Fig. a), and 10 J
of heat flows through the rods in 2.0 min. How
long would it take for 30 J to flow throught the
rods if they are welded as shown in Fig. b?
a)
b)
00C
1000C
0 0C
1000C
23-3 The first law of thermodynamics
1.For a thermodynamic system, internal energy is
the only type of energy the system may have.
The law of conservation of energy of the system
can be expressed as (First law of
thermodynamics)
Q  W  Eint
(23-6)
(i) Q is the energy transferred (as heat)
between
the system and its environment because of a
temperature different.
(ii) W is the work done on ( or by) the system
by
forces that act through
the system boundary.
Fig 23-9
(a)
(b)
(c)
W
initial
state
Boundary
Environment
process
Eint,i
final
state
Q
Eint,f
Eint  Eint, f  Eint,i  Q  W
(iii) In any thermodynamic process between
equilibrium state i and f, the quantity Q+W has
the
same value. ThisEquantity
is equal to the change
int
in
the internal energy
.
(iv) The first law of the thermodynamics is a
general result that is thought to apply to every
process in nature that proceeds between
equilibrium states.
23-4 Heat capacity (热容)and specific heat（比热）
1. Heat capacity C：
Q
C
T
(23-7)
2. Specific heat：
The heat capacity per unit mass of a body
C
Q
c 
m mT
(23-8)
The heat capacity (C) is characteristic of a particular object,
but the specific heat (c) characterizes a kind of substance.
Usually both C and c depend on the temperature and
condition under which the heat Q is added to the material.
Q   m  c(T )  dT  m  c(T )dT
(23-10)
3. Molar heat capacity
If we multiply the specific heat by the molar mass M, we
obtain the “molar heat capacity”.
cmol
Q
Q
M

mT nT
n--- the molar number
Larger
3R
*Measured at room temperature and atmospheric pressure.
*4. Heats of Transformation (Latent heat(潜热))
When heat enters a sample, the sample may change from one
phase or state to another. In this case, the temperature of the
sample does not rise. Vice versa.
Sample Problem 23-3
A cube of copper of mass mc=75 g is placed in an oven
at a temperature of T0=3120C until it comes to thermal
equilibrium. The cube is then dropped quickly into an
insulated beaker(烧杯) containing a quantity of water
of mass mw=220 g. The heat capacity of the beaker
alone is Cb=190 J/K. Initially the water and the beaker
are at a temperature of Ti=12.00c. What is the final
equilibrium temperature Tf of the system consisting of
the copper +water+beaker?
23-5 Work done on or by an ideal gas
1. Work done on an ideal gas
Fig 23-13
p
ideal gas
Fx
W
v1
x
(a)
W  P  V

v2
v
(b)
dW   P(V)  dV
vf
W    P(V)  dV
vi
Work done on gas in a more general form:
(23-15)
W    PdV
(a) If PV relationship is known, the work done on the
gas is equal to the area under the curve
representing the process.
(b) The pressure force is not a conservative force.
P
Fig 23-14
B
D
A
C
vi
vf V
Two paths:
A B D
A C D
W    PdV
2. Several typical thermal processes
(a). Work done at constant volume (V is const.)
p
p vf
W 0
(23-16)
W 0
W=0
vi  v f
v
vi v
(b). Work done at constant pressure (P is const.)
W    P(V)dV   P  dV
  P(v f  vi )   P  ΔV
p
(23-17)
vi
vfv
(c). Work done at constant temperature (T is const.)
namely “isothermal (等温) process”,
PV  constant
P(V) curve is hyperbolic(双曲线). P
Fig 23-15
The curve of PV=const. is
called an “isotherm (等温线)”.
W
vi
vf
V
vf
v f dv
nRT
W    PdV   
dv  nRT 
 nRT ln
vi v
v
vi
nRT
P
v
(23-18)
(d). Work done in thermal isolation
Thermal isolation is also called “adiabatic”
process. Q=0. T can be changed.
PV   const. (23-19)
P
pv  const.
(   1.1 ~ 1.8）
If we know  , and the initial
Pi ,Vi , we have

PV   PiVi  constant
pv   const.
Pi
Pf
vi
vf
Fig 23-16
V
PiVi
P 
V

(23-20)
We can now find the adiabatic work:
W  -  PdV
γ
PiVi
γ dV
dV


P
V
i i 
Vi
Vγ
Vγ
γ
PiVi
PiVi Vi γ 1
1 γ
1 γ

(Vi  V f ) 
[( )  1]
γ 1
γ  1 Vf
Vf
 


(23-21)
By further using PiVi  Pf V f , Eq(23-21) becomes
1
W
( Pf V f  PiVi )
(23-22)
 1
Sample problem 23-4
A sample of gas consisting
of 0.11 mol is compressed
P (Pa) v
( f , Pf )
3
3
1.0m 40
from a volume of4.0m to
3
while its pressure increase 30
from 10 to 40 Pa. Compare 20
2
the work done along the
10
1
( v i , Pi )
three different paths shown
1 2 3 4 V (m 3 )
in Fig 23-17.
Fig 23-17
Solution:
Path 1
W1   Pi (Vf  Vi )  (10 Pa)(1.0m 3  4.0m 3 )  30 J
Path 2 represents an isothermal process
W2  nRT ln
vf
  P iVi ln
vi
vf
vi
1.0
 (10 Pa)( 4.0m ) ln
 55 J
4.0
3
Path 3 W3  0  Pf (Vf  Vi )  120 J
23-6 The internal energy of an ideal gas
1. Internal energy of ideal monatomic gas
Translational kinetic energy is
(23-23)
3
K trans 
2
KT
No potential energy. No rotational kinetic energy.
So 3 KT is the entire store of internal energy.
2
The total internal energy of n moles of an ideal
monatomic gas is
Eint  (nN A ) K trans
3
3
 nN A KT  nRT
2
2
(23-24)
y
2. Internal energy of
molecule consisting
of two particles
such asO2 , N 2 , CO
Fig 23-18
d
X
z
The total kinetic energy of a diatomic molecule is
1
1
1
1
1
2
2
2
2
2
K  mv x  mv y  mv z  I x  x  I y  y (23-26)
2
2
2
2
2
U=0
Etot=K
3. Degrees of freedom
The five terms in Eq23-26 represent
independent
ways in which a molecule can absorb energy
and
are called ‘degree of freedom”.
4. Equipartition of energy theorem (能量均分定
理)
1
KT
2
Maxwell derived a theorem:
“When the number of molecule is large, the
average
energy per molecule is
for each
See动画库\力学夹\4-13能量均分定理
independent degree of freedom”
5. The internal energy for different kinds of ideal
gases:
3
3
(a) Monatomic
Eint  N ( ideal
KT ) gas
nRT
2
2
(23-27)
5
5
(b) Diatomic
Eint  Ngas
( KT )  nRT
2
2
(23-28)
(c) A polyatomic gas 6generally has six degrees of
Eint  N ( KT )  3nRT
2
freedom,
(23-29)
(d) If the molecule is not a rigid one, there are
also oscillating degrees of freedom.
Notes: 1* Eint here dependents only on T and degree of freedom.
2* Equipartition of energy theorem only apply in classical physics.
23-7 Heat capacities of an ideal gas
1. Molar heat capacity at constant volume
Constant volume
W 0
Q  Eint
(23-30)
If C v represents the molar heat capacity at constant
volume,
Eint
Q
(23-31)
Cv 

nT
nT
Using Eqs(23-27),(23-28),(23-29), we have
3
C v  R  12.5 J
Mol  K
2
5
C v  R  20.8 J
Mol  K
2
Cv  3R  24.9 J
Mol  K
( monatomic gas)
( diatomic gas )

( polyatomic gas)
Rigid
model
2. Molar heat capacity at constant pressure
Fig 23-19 shows two isotherms P
B
differing in temperature by
T .
n is fixed in the process.
Fig 23-19
C
A
D
V
Path AB is the constant-volume process; P
Path AC is a constant pressure process.
B
A
C
The change in internal energy is the same for path AB V
and AC.
Eint,AB  Eint,AC
Along AB: Eint,AB  nCV T
Along AC: E int,AC  Q  W
W  PV
Q  nCP T
(23-35)
(23-36)
(23-27)
Using the ideal gas law
W  PV  nRT
(23-38)
Eint,AC  nC P T  nRT  nCV T
or
C P  CV  R
5
C P  R  20.8 J
Mol  K
2
7
(23-40)
C P  R  29.1 J
Mol  K
2
C P  4 R  33.3 J
41)
Mol  K
(23-39)
( monatomic gas)
( diatomic gas ) (23( polyatomic gas)
(23-42)
Sample problem 23-7
An ideal gas with 0.11 mole begins at the initial
Pi  10 Pa
Vi  4.0m3
point with volume
and pressure
. Let us compress the gas adiabatically until
volume
V f  1.0m3
is
. Find the change in internal energy
of
  1.66
the gas, assuming it to be helium ( a monatomic
gas with
).
Solution:
Using Eq(23-19)

PiVi
(10 Pa)( 4.0m 3 )1.66
Pf 

 100 Pa

3 1.66
(1.0m )
Vf
The initial and final temperature are
PiVi
Ti 
 44K
nR
Tf 
Pf V f
nR
 109 K
The change in internal energy is
3
Eint  nRT  89 J
2
23-8 Applications of the first law of thermodynamics
(for ideal gases)
1. Adiabatic process ( Q  0 )
Let’s derive:
PV   const .
(23-19)
Q  0
Eint  W
(23-47)
dEint  dW   PdV
 dEintcan always be expressed as:
dEint  nCv dT , (W  0)
(23-48)
PdV  nCv dT
 PV  nRT
d ( PV )  d (nRT )
PdV  VdP  nRdT
(23-49)
Take the ratio between Eqs(23-49) and (23-48)
PdV  VdP
nRdT
R


PdV
 nC v dT
Cv
dP
dV
CV  R
CP
VdP
 


 
P
V
PdV
Cv
Cv
Pf
Vf
Pf dP
V f dV
ln
  ln



Pi P
Vi V
Pi
Vi
Pf
V f 
ln
 ln( )
Pi
Vi


CP
(23-51)
PiVi  Pf Vf

Cv
Since i and f are arbitrary points, then we obtain
(23-52)
PV   const.
We can rewrite these results in terms of
temperature, using the ideal gas equation of state
(PV=nRT) TV  1  cons.
Vi  1
(23-53)
T f  Ti ( )
Vf
or
(23-54)
2. Isothermal
Eint  0 process
or
Q W  0
(23-55)
3. Constant-volume processes
W 0
Eint  Q
(23-56)
In this case all the heat that enters the gas (Q>0) is
stored as internal energy.
A
P
4. Cyclical processes
In a cyclical process, the system
1
restores to its initial state (A-B-C-A).B
Eint  0
or
Q W  0
3
W>0
2
C
V
(23-57)
Fig 23-21
5. Free expansion
Stopcock
The gas in Fig(23-22) is initially insulating (隔离阀)
in one side of the container, and
when the stopcock is opened,
gas vacu.
the gas expands into the
previously evacuated half.
In this process, no work is
done.
The container is insulated, so the
process is adiabatic.
Fig 23-22
Hence W  0 Q  0
Eint  0 (T  0)
(23-58)
The initial state ( all gas on one side ) is an
equilibrium state, as is the final state.
But the intermediate process during intial and final
state is nonequilibrium. T and P do not have
unique value, and we can not plot this process on a
PV diagram.
Summary of typical thermal processes:
Sample problem 23-9
The cycle shown in Fig 23-21 consists of three
processes, starting at point A:
A->B: constant-volume;
P
A
B->C: constant-pressure;
PA
C->A: isothermal compression.
1 3
n=0.75mol diatomic gas with
C
PB  PC B 2
3
3
PA  3.2  10 Pa V, A  0.21m
,
Eint
PB  1.2  10 3 Pa . Find Q,W and
. V A  VB Vc V
Fig 23-21
Solution: T  PAVA  108 K  T
A
C
nR
PBVB
TB 
 48K
nR
at point C:
nRTC
VC 
 0.56m 3
PC
process 1 (A->B)
Q1  nCV (TB  TA )  1060 J
W1  0
Eint,1  1060 J
process 2 (B->C)
Q2  nC P (TC  TB )  1480 J
W2   P(VC  VB )  420 J
Eint,2  Q2  W2  1060 J
process 3 C->A
VA
W3  nRTC ln
 660 J
VC
Eint,3  0
Q3  Eint,3  W3  660 J
For the cycle, we
have:

Q  Q1  Q2  Q3  240 J
W  W1  W2  W3  240 J
Eint  Q  W  0