Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710...

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Main points of today’s lecture: Heat and heat capacity: Phase transitions and latent heat: Heat flow Examples of heat conductivity, R values for insulators Convection Radiation Physic 231 Lecture 33 T cm Q Δ = m L Q Δ = ( ) L T T kA t Q H 1 2 = Δ Δ = P = σ AeT 4 where σ = 5.67x10 -8 W /( A K 4 ) i i i k L R / =

Transcript of Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710...

Page 1: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

•  Main points of today’s lecture: •  Heat and heat capacity:

•  Phase transitions and latent heat: •  Heat flow

•  Examples of heat conductivity, R values for insulators

•  Convection •  Radiation

Physic 231 Lecture 33

TcmQ Δ=

mLQ Δ=

( )LTTkA

tQH 12 −=ΔΔ=

P =σ AeT 4

where σ = 5.67x10-8W / ( A ⋅K 4 )

iii kLR /=

Page 2: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Heat and heat capacity •  We saw that for atoms or molecules in an ideal gas is <KE>=f/2kBT. In

general, the atoms or molecules in matter increase in energy if the object is heated. This is heat energy. f is the number of degrees of freedom (3 for atom).

•  Heat Q is the energy that flows from a hot object to a cold object solely because of the difference in T.

•  When the heat flow is sufficient, the two objects reach the same temperature and we say they are in thermal equilibrium at the same T.

•  For a monatomic ideal gas at constant volume

•  For many other materials, the relationship between heat transferred and temperature change is given by: (usually transferred at constant V or P)

object. theofcapacity heat total the and mass theis massunit per capacity heat specific theis

cmmc

TcmQ Δ=

Q = ΔEth = ΔU = Uf − Ui =f2

NkbTf −f2

NkbTi =f2

Nkb Tf − Ti( )Q = f

2NkbΔT = f

2nRΔT = ncmolar,VΔT cmolar,V=f/2R is the heat capacity per mole at

constant volume for an ideal gas

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Example of equilibration and energy conservation

•  At a fabrication plant, a hot metal forging has a mass of 75 kg and a specific heat capacity of 430 J/(kg �°C). To harden it, the forging is quenched by immersion in 710 kg of oil that has a temperature of 32°C and a specific heat capacity of 645 cal/(kg �°C). The final temperature of the oil and forging at thermal equilibrium is 47°C. Assuming that heat flows only between the forging and the oil, determine the initial temperature of the metal forging.

metal oil

Energy conservation:Q Q 0 Δ + Δ =

( )metal metal metal f metal Q c m T T Δ = − ( )oil oil oil f oil Q c m T TΔ = −

o ooil

4.19Jc 645cal / (kg C) 2700kJ / (kg C)cal

= ⋅ ⋅ = ⋅

( ) ( )metal metal f metal oil oil f oilc m T T c m T T− = − −

( )oil oil f oilf metal

metal metal

c m T TT T

c m− −

− =

( )oil oil f oilmetal f

metal metal

c m T TT T

c m−

⇒ = + ( ) ( )( )

o oo o

o

2700J / (kg C) 710kg 47 32 C47 C 939 C

430J / (kg C) 75kg⋅ −

= + =⋅

mmetal 75 kg moil 710 kg cmetal 430J/kg/ oC coil 645cal/kg/ oC Tf 47 oC Toil,0 32 oC Tmetal,0 ?

metal oilQ Q ⇒Δ = −Δ

Page 4: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Quiz •  A 100 kg mass of water at a temperature of 30°C is dropped into a

thermally isolated vessel containing 100 kg of water which is a temperature of 10°C. After system comes to thermal equilibrium the final temperature is (useful information: cwater=4186 J/(kg• °C)) –  a) 10°C –  b) 15°C –  c) 20°C –  d) 25°C

cwater100kg Tf − 30oC( ) + cwater100kg Tf −10oC( ) = 0

Tf − 30oC +Tf −10oC = 0

Tf = 30oC +10oC( ) / 2 = 20oC

/ / / /

Page 5: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Two objects are made of the same material, but have different masses and temperatures. If the objects are brought into thermal contact, which one will have the temperature change of greater magnitude?

a) the one with the higher initial temperature

b) the one with the lower initial temperature

c) the one with the greater mass

d) the one with the smaller mass

e) the one with the higher specific heat

Question 13.1 Thermal Contact I

Qheavy +Qlight = 0 cmheavy Tf −Theavy( ) + cmlight Tf −Tlight( ) = 0

cmheavy Tf −Theavy( ) = −cmlight Tf −Tlight( )

Tf −Tlight( ) = −

mheavy

mlight

Tf −Theavy( )

Page 6: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Q

T

Phase transitions

•  The caloric curve that describes T vs. Q. Regions with a linear increase correspond to a constant heat capacity where

)/(cmQTTcmQ Δ=Δ⇒Δ=Δ•  The flat regions occur when the system has a phase transition. For water,

there is one where ice changes to water and other where water changes to steam. If the pressure remains constant during the phase change, the temperatures will remain constant. The heat required to change a mass Δm of the matter is given by the latent heat L for the phase change.

kgJxLkgJxLmLQ onvaporizatifusion /106.22 /105.33 54 ==Δ=Δ

Page 7: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Example

•  When it rains, water vapor in the air condenses into liquid water, and the energy is released. (a) How much energy is released when 0.0254 m (one inch) of rain falls over an area of 2.59x106 m2 (one square mile)? (b) If the average energy needed to heat one home for a year is 1.5x1011 J, how many homes can be heated with the energy determined in part (a)?

a) mwater = ρwaterVwater = 1000kg / m3 2.59x106 m2( ) .0254m( ) = 6.58x107 kg

ΔQreleased = Lvaporizationmwater = 22.6x105 J / kg( ) 6.58x107 kg( ) = 1.5x1014 J

b) nhouses ⋅1.5x1011 J = 1.5x1014 J ⇒ nhouses = 1000

Page 8: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Heat flow

•  There are three major processes that transfer heat from one point to another. –  Conduction –  Convection –  Radiation

•  Convection results from the fact that hot objects generally expand. This decreases their density. If this occurs in a fluid, the less dense hot fluid rises and the colder denser fluid falls.

Page 9: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Conduction

•  Conduction concerns the transfer of heat through materials without convection.

•  Consider an concrete wall of a heated garage. The outside of the garage is at temperature T1 and the interior of the garage is a temperature T2. The conductive heat flow through a portion of the wall with area A is given by:

( )LTkA

LTTkA

tQH Δ=−=ΔΔ= 12

•  H is the heat flow through the wall, k is the thermal conductivity of concrete, and L is the thickness of the concrete wall.

•  Example: Calculate the heat flow through a 2 m2 section of a 20 cm thick concrete wall when the outside temperature is 0 oC and the inside temperature is 20 oC. Assume the thermal conductivity of the concrete is 1.3 J/(s� m � oC)

H = ΔQ

Δt= kAΔT

L=

1.3J / (s ⋅m ⋅ oC) 2m2( ) 20 oC( )0.2m

= 260W

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Quiz

•  Two materials have the same insulating value if the same amount of heat per second per square meter flows through each due to the same temperature difference. Ignoring air convection, what thickness of body fat is required to give the same insulating value as a 0.010 inch thickness of air? (kfat=0.2 J/(s�m �oC), kair=0.023 J/(s�m �oC)) –  a) 0.09 inches –  b) 0.7 inches –  c). 2.3 inches –  d) 4.2 inches

want H fat = Hair

k fat AΔTLfat

=kair AΔT

Lair

⇒ Lfat = Lair

k fat

kair

= 0.01inch( ) 0.20.023

= .09inch

⇒k fat

Lfat

=kair

Lair

/ / / /

multiply both sides of eq. by

Lfat ⋅ Lair

kair

⇒k fat

Lfat

⋅Lfat ⋅ Lair

kair

=kair

Lair

⋅Lfat ⋅ Lair

kair/ / /

/ / /

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Layered materials – R values

•  Consider the layered insulating structure at the right. The area of each layer is the same. Each layer can have a different thickness Li and heat conductivity ki.

•  Here the important thing to remember is that we are concerned with a steady state solution. There is no build up of heat in any of these layers. For each layer one has:

H

A

Li

ii

ii

i

ii RAH

kL

AHT

LTAk

tQH ⋅=⋅=Δ⇒Δ=ΔΔ=

•  Here Ri it the “R value” of the ith insulating layer. Large R value means better insulation and less heat loss in winter. The total temperature difference for a layered structure is given by the sum.

∑∑∑Δ=

ΔΔ=⇒

⋅=Δ⋅=Δ=Δ

ii

ii

ii

ii

RTA

tQH

RAHTR

AHTT i.e.

Page 12: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Example of a layered structure •  Two rods, one of aluminum and the other of copper are joined end to

end. The cross-sectional area of each is 4.0x10-4 m2, and the length of each is 0.04 m. The free end of the aluminum rod is kept at 302 oC, while the free end of the copper rod is kept at 25 oC. The loss of heat through the sides of the rods may be ignored. (a) What is the temperature at the aluminum-copper interface? (b) How much heat is conducted through the unit in 2.0 s? (kalum = 238 J/(s � m �oC), kCu = 397 J/(s � m �oC))

a) ΔTalum = H

A⋅Ralum

= H

A⋅

Lalum

kalum

ΔTCu =

HA⋅

LCu

kCu

⇒ΔTalum

ΔTCu

=

HA⋅

Lalum

kalum

HA⋅

LCu

kCu

=kCu

kalum

= 397238

= 1.67

ΔT = ΔTalum + ΔTCu = ΔTCu (

ΔTalum

ΔTCu

+1) = 2.67ΔTCu

ΔTCu =

ΔT2.67

= 302 oC − 25oC2.67

= 103.7 oC

b) H =

kCu AΔTCu

LCu

= 397 ⋅4x10−4 ⋅103.7.04

W = 412W

Tinterface = 25oC +103.7 oC = 128.7 oC

Page 13: Physic 231 Lecture 33 - Michigan State Universitykg C) 75kg ⋅− =+ = ⋅ m metal 75 kg m oil 710 kg c metal 430J/kg/ oC c oil 645cal/kg/ oC T f o47 C T oil,0 32 ...

Example of a layered structure •  Two rods, one of aluminum and the other of copper are joined end to

end. The cross-sectional area of each is 4.0x10-4 m2, and the length of each is 0.04 m. The free end of the aluminum rod is kept at 302 oC, while the free end of the copper rod is kept at 25 oC. The loss of heat through the sides of the rods may be ignored. What is the temperature at the aluminum-copper interface between Copper and Aluminum? Useful information: kalum=238J/(s�m�oC) and : kCu=397J/(s�m�oC).

a) ΔTalum = H

A⋅Ralum

Ralum =Lalum

kalum

=.04m

238J / (s ⋅m ⋅ oC) =1.681x10−4 s ⋅m2 ⋅ oC / J

ΔTCu =

HA⋅RCu

RCu =

LCu

kCu

=.04m

397J / (s ⋅m ⋅ oC) =1.008x10−4 s ⋅m2 ⋅ oC / J

ΔTtot =

HA⋅Rtot Rtot = RCu + RAlum = 2.69x10−4 s ⋅m2 ⋅ oC / J

H

A =

ΔTtot

Rtot

Tint erface = 25oC + ΔTCu = 25oC + H

A⋅RCu

Tint erface = 25oC +

ΔT tot

Rtot

⋅RCu = 25oC + (302− 25)1.0082.69

= 129 oC