Chapter 11: Energy in Thermal Process

Heat and Internal Energy Terminology

• Internal energy U is the energy associated with microscopic components of a system – the atoms and molecules of the system. It includes kinetic and potential energy associated with the random translational, and vibrational motion of the particles that make up the system, and any potential energy bonding the particles together.

• Heat Q is the transfer of energy between a system and its environment due to a temperature difference between them.

Example homework problems: 11,17,27,38,65

Heat and Internal Energy

Units of heat

• Calorie (cal)The energy necessary to raise the temperature of 1 g of waterfrom 14.5o to 15.5oC. “Cal” used for energy content in food is kcal.

• British thermal unit (Btu)

The energy necessary to raise the temperature of 1 lb of waterfrom 63o to 64oF.

• SI unit : joule (J)

1 cal = 4.186 J

Heat and Internal Energy

Example • Example 11.1 : Working off breakfast

Breakfast : two bowls of cereal and milk, 3.20x102 Cal

Exercise : Do curls with a 25.0-kg barbell to consume all the calories from the breakfast

J 1034.1

cal

J 186.4

Cal 00.1

cal 1000.1)Cal 1020.3(

6

32

E

Breakfast calories:

mghmghPEKEW )0(0

Work done by one lifting or lowering barbell:

Number of repetitions n needed:

Emghn )2(

times1084.62

3mgh

En

But see the remark on p.354!

Heat and Internal Energy

Specific heat • Definition of specific heat

Tm

Qc

SI unit : joule per kilo-gram degree Celsius J/(kg oC)

TmcQ T=Tf-Ti

• The energy required to raise the temp. of 0.500 kg of water by 3.00oC.

J 106.28

C)C))(3.00 J/(kg 4186)(kg 500.0(3

Q

Heat and Internal Energy

Specific heat

A steel strut near a ship’s furnace :L0=2.00 m, ms=1.57 kg, A=1.00x10-4 m2, Q=2.50x105 J.

(a) Find the change in temp. of the strut.

C355ss

ss cm

QTTcmQ cs=448 J/(kg oC)

(b) Find the change in length of the strut if it is allowed to expand.

m108.7 60

TLL =11x10-6 oC-1

(c) Find the compressional stress in the strut.

Pa 108.7 8

0

L

LY

A

F Y=2.00x1011 Pa

• Example 11.2 : Stressing a strut

Calorimetry Isolated system • A system whose energy does not leave out of the system is called isolated system.

• The principle of energy conservation for an isolated system requires that the net result of all the energy transfer is zero. If one part of the system loses energy, another part has to gain the energy.

Calorimeter and calorimetry • Imagine a vessel made of good insulating material and containing cold water of known mass and temperature and the temperature of the water can be measure. Such a system of the vessel and water is called calorimeter. If the object is heated to a higher temperature of known value before it is put into the water in the vessel, the specific heat of the object can be measured by measuring the change in temperature of the water when the system (the object, vessel, and water) reaches thermal equilibrium. This measuring process is called calorimetry.

Calorimetry Calorimeter and calorimetry • When a warm object, put into a calorimeter with cooler water described in the previous page, becomes cooler while the water becomes warmer.

Qcold (>0 ) is the heat transferred (energy change) to the cooler object and Qhot (<0) is the heat transferred (energy change) to the warmer object.

hotcold QQ

• In general, in an isolated system consisting of n objects :

])()[(01

ikfkkkk

n

kk TTcmQQ

Tf common to all objects in equilibrium.

Calorimetry Calorimeter and calorimetry • Example 11.3 : Finding a specific heat

A 125-g block of an unknown substance with a temperature of 90.0oCis placed in a Styrofoam cup containing 0.326 kg of water at 20.0oC.The system reaches an equilibrium temperature of 22.4oC. What isthe specific heat, cx , of the unknown substance if the heat capacity ofthe cup is ignored.

hotcold QQ

)()( xxxwww TTcmTTcm

C) J/(kg 388)(

)(

TTm

TTcmc

xx

wwwx

Calorimetry Calorimeter and calorimetry • Example 11.4 : Calculate an equilibrium temperature

Suppose 0.400 kg of water initially at 40.0oC is poured into a 0.300-kgglass beaker having a temperature of 25.0oC. A 0.500-kg block ofaluminum at 37.0oC is placed in the water, and the system is insulated.Calculate the final equilibrium temperature of the system.

0 galw QQQ

0)()()( gggalalalwww TTcmTTcmTTcm

Ccmcmcm

TcmTcmTcmT

ggalalww

gggalalalwww

9.37

Calorimetry Latent heat and phase change • Phase change (change of molecular structure)

- Phases of material : solid, liquid, gas

- A change of phase : phase transition

- For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat and a change of volume and density

Heated added and phase changes of water

Calorimetry Latent heat and phase change • Latent heat

- To change structure of material, change in energy required.

- When a substance goes under phase transition, all the heat transferred is used to change the phase.

Heated added and phase changes of water

mLQ L : latent heat (J/kg) , m : mass (kg)+(-) when energy is absorbed (removed).

Calorimetry Latent heat and phase change • Latent heat

- The latent heat of fusion Lf is used when a phase change occurs during melting or freezing. - The latent heat of vaporization Lv is used when a phase change occurs during boiling or condensing.

Calorimetry Latent heat and phase change • WaterConsider an addition of energy to a 1.00-g cube of ice at -30.0oC ina container held at constant pressure. Suppose this input energyturns ice to steam (water vapor) at 120.0oC.

C) J/(kg 2090 J 7.62

0.30,

ice

ice

c

CTTmcQ

J 333

kg/1033.3, 5

JLmLQ ff

C) J/(kg 4190 J 19.4

100,

water

water

c

CTTmcQ

J 1026.2

kg/1026.2,3

6

JLmLQ vv

C) J/(kg 2010 J 2.40

0.20,

steam

steam

c

CTTmcQ

A:

B:

C:

D: E:

Calorimetry Latent heat and phase change • Example 11.5 : Boiling liquid helium

Liquid helium has a very low boiling point, 4.2 K, as well as a low latentheat of vaporization, 2.09x104 J/kg. If energy is transferred to a containerof liquid helium at the boiling point from an immersed electric heater ata rate of 10.0 W, how long does it take to boil away 2.00 kg of the liquid?

J 104.18J/kg) 1009.2)(kg 00.2( 44 vmLQ

min 69.7s 1018.4 W0.10

J 1018.4 44

P

mL

P

Qt v

Calorimetry Latent heat and phase change • Example 11.6 : Ice water

6.00 kg of ice at -5.00oC is added to a cooler holding 30 liters of waterat 20.0oC. What temperature of the water when it comes to equilibrium?

kg 0.30 Vm waterwater

0 waterwatericemeltice QQQQ

Q m(kg) c(J/(kgoC)) L (J/kg) Tf Ti Exp.

Qice 6.00 2090 0 -5.00 mcT

Qmelt 6.00 3.33x105 0 0 mLf

Qice-water 6.00 4190 T 0 mc

Qwater 30.0 4190 T 20.0 mc

CT 03.3

Calorimetry Latent heat and phase change • Example 11.7 : Ice water

A 5.00-kg block of ice at 0oC is added to an insulated container partiallyfilled with 10.0 kg of water at 15.0oC.(a) Find the final temperature (neglect the heat capacity of the container.)

J 101.67J/kg) 10kg)(3.33 00.5( 65 ficemelt LmQThe amount of energy needed to completely melt the ice:

The maximum energy that can be lost by the initial mass of liquid waterwithout freezing it :

J 1029.6

C)15.0-C))(0J/(kg kg)(4190 0.10(5

TcmQ waterwater

Since this is less than half the energy needed to melt all the ice, so thefinal state of the system is a mixture of water and ice at the freezing point.

(b) Find the mass of the ice melted.

kg 89.1 mmLQ fwater

Energy Transfer Mechanism of heat transfer • Thermal conduction• Convection• Radiation

Thermal conduction • One form of heat transfer when there is temperature difference is thermal conduction.• This process is caused by an exchange of kinetic energy between microscopic particles such as molecules, atoms, and electrons etc. In the process, less energetic particles gain energy as they collide with more energetic particles.

The rate of energy transfer P=Q/t isproportional to cross-sectional area A,the difference in temperature T=Th-Tc>0,and inversely proportional to the thicknessof the slab.

x

TA

t

QP

Energy Transfer Thermal conduction • Consider a substance is in the shape of a long, uniform rod of length L and assume that the rod is insulated so that thermal energy does not escape from the surface.

L

TTkA

t

QP ch

k : proportionality constant thermal conductivity [J/(s m oC)]Th : higher temperature (oC)

Tc : lower temperature (oC)

A : area (m2)

L : length (m)

Q : heat transferred (J)

t : time interval (s)

Energy Transfer Thermal conduction • Thermal conductivities

• Example 11.9 : Energy transfer through a concrete wall

Find the energy transferred in 1.00 h by conductionthrough a concrete wall 2.0 m high, 3.65 m long,and 0.20 m thick if one side of the wall is held at20oC and the other at 5oC.

J 106.2

s) 3600(m 20.0

15)m C))(7.3 m J/(s 3.1(

6

2

C

tL

TTkAtPQ ch

Energy Transfer Thermal conduction • Example : Two rods cases

k1,L1 k2,L2

Tm

k1,L,,A1

k2,L,,A2

22

11 L

TTAk

L

TTAk

t

Q cmmh

tLkA

QTT mh

)/( 11

tLkA

QTT cm

)/( 22

2

2

1

1

k

L

k

L

tA

QTT ch

2

2

1

1

)(

kL

kL

TTA

t

Q ch

L

TTAk

t

Q ch

111

L

TTAk

t

Q ch

222

L

TTkk

t

QQ

t

Q ch

)( 2121

Energy Transfer Thermal conduction • Thermal conduction through multiple layers

k1 k2 k3

L1 L2 L3

Th1 Th2 Th3 Th3

Th Tc1 Tc2 Tc2

11

111

1

111 R

A

P

k

L

A

PTT

L

TTAk

t

QP ch

ch

22

22122 R

A

P

k

L

A

PTTTT ccch

33

33233 R

A

P

k

L

A

PTTTT ccch

n

i i

icnh k

L

A

PTT

11

chcnh TTTT 1

ii

ch

iii

ch

R

TTA

kL

TTA

t

QP

)(

/

)(

Energy Transfer Thermal conduction • Example 11.11 : Staying warm in the arctic

An arctic explore builds a wooden shelter out of wooden planks thatare 1.0 cm thick. To improve the insulation, he covers the shelter witha layer of ice 3.2 cm thick.(a) Compute the R factors for the wooden planks and the ice.

Csm 020.0/ Csm 10.0/ 2ice

2 iceicewoodwoodwood kLRkLR

(b) Find the rate of heat loss.

W830)(

i

i

ch

R

TTA

t

QP

(c) Find the temperature in between the ice and the wood.

TTCTL

TTAkP ch

chwood

,00.5 W,830)(

CT 16

Energy Transfer Convection • The transfer of energy by the movement of a substance is called convection.

Energy Transfer Radiation

• All objects radiate energy because of microscopic movements (accelerations) of charges, which increase with temperature.

Law sStefan' 4TAt

QP

Heat current in radiation

=5.670400(40) x 10-8 W/(m2 K4) :Stefan-Boltzman const.

Heat transfer by radiation

• If an object is at temperature T and its surroundings are at

temperature T0, the net flow of heat radiation between the object and its surroundings is

)( 40

4 TTAt

Q

A :area

: emissivity that depends on nature of surfaceT in K

Energy Transfer Radiation

• Example 11.12 : Polar bear club

A member of the Polar Bear Club, dressed only in bathing trunksof negligible size, prepares to plunge into the Baltic Sea from thein St. Petersburg, Russia. The air is calm, with a temperature of5oC. If the swimmer’s surface body temperature is 25oC, computethe net rate of energy loss from his skin due to radiation. How muchenergy is lost in 10.0 min? Assume his emissivity is 0.900 and hissurface area is 1.50 m2.

K 29827325K 2782735 255 CC TT

W146)( 40

4 TTAPnet

J 1076.8s 600 4 netnet PtPQ