Temperature, Heat, Thermodynamic Equilibrium, and Heat Transfer, Slides of Physics

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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,

 ,^ 

, …

th^0 law of thermodynamics:2 systems in thermodynamic equilibrium with a 3

2 systems are in thermal contact if heatingone of them changes the other.Otherwise, they are thermally insulated.Two systems have the same temperature ^ they are in thermodynamic equilibriumrd^ system are themselves in equilibrium.

A,B in eqmB,C in eqm

^

A,C in eqm

Temperature Scales

Celsius scale (

C ) :

Melting point of ice

at

P^ = 1 atm

^

T = 0 C^

C.

Boiling point of water at

P^ = 1 atm

^

T = 100 C^

C.

^ Triple point of water = 0.

C

T^ T ^ C

^

^

T^ C

T ^

 

Fahrenheit scale (

F ) :

Melting point of ice

at

P^ = 1 atm

^

T = 32 F^

F.

Boiling point of water at

P^ = 1 atm

^

T = 212 F^

F.

180

32

F^^100

C T^

T ^

^

(^9) F 5

C T^

T ^



^

^  ^



^



Rankine scale (

R ) :

0

(^0) R K ^ ^ R^

F T^

T ^

 

16.2. Heat Capacity & Specific Heat

Q^

C^ T ^ 



Heat capacity

C^ of a body :  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 (

C cal) = heat needed to raise 1 g of water from 14.

C to 15.

C.

1 BTU (

F) = heat needed to raise 1 lb of water from 58.

F to 59.

F.

^

1

1

1055 cal thermochemical

J

BTU

J



^

1

4 kcal

BTU 

The Equilibrium Temperature

Heat flows from hot to cold objects until a common equilibrium temperature is reached.

For 2 objects insulated from their surroundings:

1

(^02) Q^

Q ^

 ^

^

1 1

1

2 2

2

m c

T

m c

T

^

^

^



When the equilibrium temperature

T^ is reached:

^

^

^

1 1

1

2 2

2

0

m c

T^

T^

m c

T^

T

^

^

^



1 1

1

2 2

2

1 1

2 2 m c T

m c T T^

m c

 m c ^





16.3. Heat Transfer

Common heat-transfer mechanisms:

• Conduction• Convection• Radiation

conductor insulator

Specific Heat vs Thermal Conductivity

c ( J/kg

 K )

k (W/m

 K )

Al^

900

237

Cu^

386

401

Fe^

447

Steel

502

46

Concrete

880

1

Glass

753

Water

4184

Wood

1400

Insulating properties of building materials are described by the

-factor (

-value ).

x R A^

 k ^

 R^

= thermal resistance of a slab of unit area

^ ^

2 / m K^

W R

^ 

2

/ ft^

F^ h

BTU   R

U.S.

T  A  H

2

2

1

/^

/

ft^

F^ h

BTU

m^ K

W

 ^ d T H^

k A

d x  ^

T  R  ^

A^ T ^ ^

 R

Example 16.4. Cost of Oil

The walls of a house consist of plaster (

^ = 0.17 ),

-11 fiberglass

insulation, plywood (

^ = 0.65 ), and cedar shingles (

^ = 0.55 ).

The roof is the same except it uses

-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?

0.^

11 0.

 wall

^ 



R^

12.37

0.^

30

0.^

 roof

^

^



R^

31.37

^

^ ^

2 36

28

10

A^ rect

ft^

ft^

ft

^ 

^

^

2 1280

ft 

2 1164

ft 

^

^

14

2 36

cos 30

roof

ft

A^

 ft



^ 

 ^

 ^



^

^ ^

12 28

14

tan 30

2 A^ gable

ft^

ft

^ 

^

^

^

^

2 226

ft ^

2 1506

wall^

rect^

gable

A^

A^

A^

ft

^

^



^

(^) 

2

2

1

/^ /^

/^

1506

70

20

H^ wall

BTU

h^

ft^

F^

ft^

F^

F

^



^

^

^ 



^



^

 ^

(^) 

2

2

1

/^ /^

/^

1164

70

20

H^ roof

BTU

h^

ft^

F^

ft^

F^

F

^



^

^

^ 



^



^



6073

/ BTU

h ^1853

/ BTU

h 

^

^

^

6073

1853

/^

24

/^

30

/

Q^

BTU

h^

h^ d^

d^ month

^

^

5.7^ M

BTU 

^

^

^

5.^

10

/^

$ 2.20 /

Cost

MBTU

gal^

MBTU

gal

^

$ 126

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Radiation

Glow of a stove burner

^ it loses energy by radiation

4 P^

e^ T A

Stefan-Boltzmann law for radiated power:

^ = Stefan-Boltzmann constant = 5.

 10 8 W / m

2 4 K.

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.

Wien‘s displacement law :

max

= b / T

^ P

(^4)  T ^ Radiation dominates at high

T.

Wavelength of peak radiation becomes shorter as

T^ increases.

Sun ~ visible light.Near room T ~

infrared.

4

P^

e^ T A

Stefan-Boltzmann law :

sun^ sun RT

T  TRT

^ 

3

10 b^

 mK ^

 ^

 ^

.

5778 m^ 300

K K

^ 9.

m^ 



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Conceptual Example 15.1.

Energy-Saving Windows

Why do double-pane windows reduce heat loss greatly compared withsingle-paned windows?Why is a window’s

-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

-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 glasswith that of a double-pane window make from the same glass with a5.0-mm air gap between panes.

x  k R^

Glass

k ~ 0.8 W/m

K

Air^

k ~ 0.026 W/m

K

3

single

3.^

10

0.^

^ m  / W m K ^



R^2 double^

0.^

/ m^

K^ W ^



R

2

/ m^

K^ W

^



^

^

^

3

3

3.^

10

5.^

10

2

0.^

/^

/

double

m^

m

R^

W^

m K

A

W^

m K

A

^



^



^ 

 ^

^

^



double^

single (^50)   R^

R

x R^

A^

 k A ^ R 2

0.^

/ m^

K^ W ^ A

