Transcript Short Version : 16. Temperature & Heat 短版: 16.温度&熱量

Short Version :
16. Temperature & Heat
16.1. Heat , Temperature & Thermodynamic Equilibrium
Thermodynamic equilibrium:
State at which macroscopic properties of system remains unchanged over time.
Examples of macroscopic properties:
L, V, P, , , …
2 systems are in thermal contact if heating
one of them changes the other.
Otherwise, they are thermally insulated.
A,B in eqm
B,C in eqm

Two systems have the same temperature
A,C in eqm
 they are in thermodynamic equilibrium
0th law of thermodynamics:
2 systems in thermodynamic equilibrium with a 3rd system are themselves in equilibrium.
Gas Thermometers & the Kelvin Scale
Constant volume gas thermometer T  P
Kelvin scale:
P = 0  0 K = absolute zero
Triple point of water  273.16 K
Triple point: T at which solid, liquid &
gas phases co-exist in equilibrium
Mercury fixed at this level
by adjusting h  P  T.
All gases behave similarly as P  0.
Temperature Scales
Celsius scale ( C ) :
Melting point of ice
at P = 1 atm  TC = 0 C.
Boiling point of water at P = 1 atm  TC = 100 C.
 Triple point of water = 0.01C
TC  T  273.15
 TC  T 
Fahrenheit scale ( F ) :
Melting point of ice
at P = 1 atm  TF = 32 F.
Boiling point of water at P = 1 atm  TF = 212 F.
TF 
180
TC  32
100
Rankine scale ( R ) :
0R  0 K
TR  TF
9



T


T
F
C 

5


16.2. Heat Capacity & Specific Heat
Heat capacity C of a body :
Q  C T
Q = heat transferred to body.
C  J / K
Specific heat c = heat capacity per unit mass
Q  m c T
c  J /  kg K 
1 calorie (15C cal) = heat needed to raise 1 g of water from 14.5C to 15.5C.
1 BTU (59F) = heat needed to raise 1 lb of water from 58.5F to 59.5F.
1 cal  thermochemical   4.184 J
1 BTU  1055 J
1 kcal  4 BTU
c = c(P,V) for gases  cP , cV .
The Equilibrium Temperature
Heat flows from hot to cold objects until a common equilibrium temperature is reached.
For 2 objects insulated from their surroundings:
Q1  Q2  0  m1 c1 T1  m2 c2 T2
When the equilibrium temperature T is reached:
m1 c1 T  T1   m2 c2 T  T2   0

T
m1 c1 T1  m2 c2 T2
m1 c1  m2 c2
16.3. Heat Transfer
Common heat-transfer mechanisms:
• Conduction
• Convection
• Radiation
Conduction
Conduction: heat transfer through direct physical contact.
Mechanism: molecular collision.
Heat flow H , [ H ] = watt :
H
dQ
dt
 k A
Thermal conductivity k ,
[ k ] = W / mK
T
x
conductor
insulator
Specific Heat vs Thermal Conductivity
c ( J/kgK )
k (W/mK )
Al
900
237
Cu
386
401
Fe
447
80.4
Steel
502
46
Concrete
880
1
Glass
753
0.8
Water
4184
0.61
Wood
1400
0.11
H  k A
T
x
applies only when T = const over each (planar) surface
For complicated surface, use
H  k A
dT
dx
Prob. 72 & 78.
Composite slab:
H must be the same in both slabs to prevent
accumulated heat at interface
H  k1 A
T T
T2  T1
 k2 A 3 2
x1
x2
R
Thermal resistance :

T1  T2  H R1
T2  T3  H R2

H 
H
T
R
T1  T3
R1  R2

x
kA
[R]=K/W
T2  T1
T T
 3 2
R1
R2
Resistance in series
Insulating properties of building materials are described by the R-factor ( R-value ) .
R  R A
x T

A
H
k
R   m2 K / W
U.S.
R   ft 2  F  h / BTU
1 ft 2  F  h / BTU  0.176 m2 K / W
H  k A
T
A
dT

  T
R
R
dx
= thermal resistance of a slab of unit area
Example 16.4. Cost of Oil
The walls of a house consist of plaster ( R = 0.17 ), R-11 fiberglass
insulation, plywood (R = 0.65 ), and cedar shingles (R = 0.55 ).
The roof is the same except it uses R-30 fiberglass insulation.
In winter, average T outdoor is 20 F, while the house is at 70 F.
The house’s furnace produces 100,000 BTU for every gallon of oil,
which costs $2.20 per gallon.
How much is the monthly cost?
R wall  0.17  11  0.65  0.55  12.37
Arect  2  36 ft  28 ft   10 ft   1280 ft 2
1
Agable  2    28 ft   14 ft  tan 30   226 ft 2
2
R roof  0.17  30  0.65  0.55  31.37
 14 ft 
2
Aroof  2   36 ft   
  1164 ft
 cos 30 
Awall  Arect  Agable  1506 ft 2
 1

H wall  
BTU / h / ft 2 / F  1506 ft 2   70F  20F   6073 BTU / h
 12.37

 1

H roof  
BTU / h / ft 2 / F  1164 ft 2   70F  20F   1853 BTU / h
 31.37

Q   6073 1853 BTU / h 24 h / d 30 d / month  5.7 MBTU
Cost  5.7 MBTU 10 gal / MBTU
$ 2.20 / gal 
 $126
Convection
Convection = heat transfer by fluid motion
T      rises
Convection cells in liquid film between glass plates
(Rayleigh-Bénard convection, Benard cells)
Radiation
Glow of a stove burner  it loses energy by radiation
Stefan-Boltzmann law for radiated power:
P
 e T4
A
 = Stefan-Boltzmann constant = 5.67108 W / m2 K4.
A = area of emitting surface.
0 < e < 1 is the emissivity ( effectiveness in emitting radiation ).
e = 1  perfect emitter & absorber ( black body ).
Black objects are good emitters & absorbers.
Shiny objects are poor emitters & absorbers.
Stefan-Boltzmann law :
P
 e T4
A
Wien‘s displacement law : max = b / T
b  2.898  103 mK
 P  T4  Radiation dominates at high T.
Wavelength of peak radiation becomes shorter as T increases.
Sun ~ visible light.
Near room T ~ infrared.
RT 

sunTsun
TRT
.502 m  5778K 
300 K
 9.66 m
Example 16.5. Sun’s Temperature
The sun radiates energy at the rate P = 3.91026 W, & its radius is 7.0 108 m.
Treating it as a blackbody ( e = 1 ), find its surface temperature.
P  e  AT 4
 = 5.67108 W / m2 K4
1/4


P

T 
 e   4  R2  



3.9 1026 W


2
8
2
4
8
5.67

10
W
/
m
K
4


7.0

10
m






 5.8 103 K
1/4






Conceptual Example 15.1. Energy-Saving Windows
Why do double-pane windows reduce heat loss greatly compared with
single-paned windows?
Why is a window’s R-factor higher if the spacing between panes is small?
And why do the best windows have “low-E” coatings?
Thermal conductivity (see Table 16.2):
Glass
k ~ 0.8 W/mK
Air
k ~ 0.026 W/mK
Layer of air reduces heat loss greatly & increases the R-factor .
This is so unless air layer is so thick that convection current develops.
“low-E” means low emissivity, which reduces energy loss by radiation.
Making the Connection
Compare the for a single pane window made from 3.0-mm-thick glass
with that of a double-pane window make from the same glass with a
5.0-mm air gap between panes.
x
R 
k
R single
Rdouble
Glass
Air
3.0  103 m

0.8 W / m  K
k ~ 0.8 W/mK
k ~ 0.026 W/mK
R
R x

A kA
 0.004 m2  K / W
3.0  103 m
5.0  103 m
 2

 0.8 W / m  K   A  0.026 W / m  K   A
R double  0.2 m2  K / W
R double  50  Rsingle

0.2 2
m  K /W
A
16.4. Thermal Energy Balance
A house in thermal-energy balance.
System with fixed rate of energy input
tends toward an energy- balanced state
due to negative feedback.
Heat from furnace balances
losses thru roofs & walls
Example 16.7. Solar Greenhouse
A solar greenhouse has 300 ft2 of opaque R-30 walls,
& 250 ft2 of R-1.8 double-pane glass that admits solar energy at the rate of 40 BTU / h / ft2.
Find the greenhouse temperature on a day when outdoor temperature is 15 F.
H
T A T

R
R
 300 ft  T
2
H wall 
30 ft  F h / BTU
2
 250 ft  T
2
H glass 
1.8 ft  F h / BTU
2
 10 BTU / h / F  T
 139 BTU / h / F  T
H sun   40 BTU / h / ft 2  250 ft 2 
104 BTU / h
 67 F
T 
149 BTU / h / F
 104 BTU / h
 Hwall  H glass
T  15 F  67 F  82 F
Application: Greenhouse Effect & Global Warming
Average power from sun :
Total power from sun :
S  960 W / m2
HS  S  R2E
Power radiated (peak at IR) from Earth :
H E  e   4  RE2  T 4
HS  HE
e 1
1/4

C.f.  T   15 C


960 W / m 2

T 
  5.67 108 W / m 2 K 4   4 


 255 K  18 C
natural greenhouse effect
Mars: none
Greenhouse gases: H2O, CO2 , CH4 , …
passes incoming sunlight, absorbs outgoing IR.
Venus: huge
CO2 increased by 36%
0.6 C increase during 20th century.
1.5 C – 6 C increase by 2100.