• Main points of last lecture:• Heat and heat capacity:

• Phase transitions and latent heat:

• Main points of today’s lecture:• Heat flow

• Examples of heat conductivity, R values for insulators

• Convection• Radiation

Physic 231 Lecture 27

TcmQ ∆=

mLQ ∆=

( )L

TTkAtQH 12 −=∆∆

=

)/(5.67x10 where

48-

4

KAWAeTP

⋅=

=

σ

σ

iii kLR /=

Conduction

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

• Consider an concrete wall of a heated garage. The outside of thegarage 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:

( )L

TkAL

TTkAtQH ∆

=−

=∆∆

= 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 conductive of the concrete is 1.3 J/(s⋅ m ⋅ oC)

( )( ) Wm

CmCmsJL

TkAtQ oo

2602.0

202)/(3.1 2

=⋅⋅

=∆

=∆∆

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

( ) inchinchkk

LL

Lk

Lk

LTAk

LTAk

HH

air

fatairfat

air

air

fat

fat

air

air

fat

fat

airfat

09.023.02.001.0

want

===⇒

=⇒∆

=∆

=

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 ⋅=⋅=∆⇒

∆=

∆∆

=

• Here Ri it the “R value” of the ith insulating layer. The total temperature difference is given by the sum.

∑

∑∑∑∆

=∆∆

=⇒

⋅=∆⋅=∆=∆

ii

ii

ii

ii

RTA

tQH

RAHTR

AHTT i.e.

Example

• 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))

WWxL

TAkH

CCCCCCTT

TTTTTTT

kk

kL

AH

kL

AH

TT

kL

AHT

kL

AHR

AHT

Cu

CuCu

oooooo

Cu

CuCu

alumCuCualum

alum

Cu

Cu

Cu

alum

alum

Cu

alum

Cu

CuCu

alum

alumalumalum

41204.

7.103104397 b)

7.1287.10325T 7.10367.2

2530267.2

67.2)1(

67.1238397 a)

4

interface

=⋅⋅

=∆

=

=+==−

=∆

=∆

∆=+∆∆

∆=∆+∆=∆

===⋅

⋅=

∆∆

⇒⋅=∆⋅=⋅=∆

−

Conceptual quiz

• Two large metal blocks made of the same material and at the sametemperature are brought together and placed into thermal contact. Block A contains twice as much heat as Block B. Which of the following statements is true– a) There will be net flow of heat from A to B– b) Net flow of heat from B to A– c) No net heat flow will occur– d) Not enough information to tell

Radiation• The Earth is warmed by the

radiation from the sun.• The rate at which an object

radiates energy is given by Stefan’s Law which provides the power radiated from a object with surface area A at temperature T.

)/(5.67x10 where 4-84 KAWAeTP ⋅== σσ

• Here e is emissivity (0≤e≤1) which describes how the radiation or absorption can be reduced below that for a “black” surface that absorbs light completely.

• Example: The filament of a light bulb has a temperature of 3.0x 103 oCand radiates sixty watts of power. The emissivity of the filament is 0.36. Find the surface area of the filament.– a) 3.4 m2

– b).063 m2

– c) 2.6x10-5 m2

– d) 9.8x10-6 m2

( )25

44284

106.23273)36)(./(1067.5

60

mxAKKmWx

WeTPA

−

−

=⇒

⋅==

σ

Example• A solid aluminum sphere is coated with lampblack (emissivity=0.97)

and hung inside an evacuated container. The sphere has a radius of 0.020 m and is initially at 20.0 oC. The container is maintained at a temperature of 70.0 oC. (a) Assuming that the temperature of the sphere does not change very much, what is the net energy gained by the sphere in 10.0 s? (b) Estimate the change in temperature of the sphere.

( )( ) ( ) ( )

( )

( ) ( )C

mmkgKkgJJT

mc∆ETTVcTmc∆E

JstPPAeE

tPPEKAeP

KAeP

alumalumalumalumalumalumalum

033

3

4428-2070

2070

420

470

22.03/02.4/2700))/(900(

183/02.4

change atureget temper ocapacity theat need work),(no QE Haveb)1810)293343)[97(.)02.0(45.67x10

sphereby gainedenergy is emittedpower )293(

absorbedpower )343(absorbed.power emittedpower mequilibriu In :balance Detaileda)

=⋅

=∆

=∆⇒∆=∆=

∆=∆=−=∆−⋅=∆

∆−=∆=⋅=

=⋅=

=

π

πρρ

πσ

σ

σ

Temperature of the Planets

• The surface temperature of the Sun is about 5800 K. a) Taking the Sun’s radius to be 6.96 x 108 m, calculate the total energy radiated by the Sun each second. (Assume e = 0.965.) b) Calculate the equilibrium temperature of the Earth, Mercury, Venus and Mars. Assume for simplicity that the emissivity of the Earth, Mercury, Venus and Mars are all also 0.965. The orbital radii of these planets are given in the table.

~5.8x1010 mMercury

2.3x1011 mMars

1.5x1011 mEarth 1.1x1011 mVenus

Orbital radiusPlanet

( ) ( )( )WxP

KmxKmWxAeTP

sun

sunsun26

4284284

1076.3

5800965.1096.64)/(1067.5 a)

=⇒

⋅== − πσ

• The earth is at temperature Tearth and radiates in all directions with a power:

Temperature of planets continued• Let’s calculate for the earth first. At a distance equal to the earth’s

radius, rearth, Psun is spread over a spherical area of 4πr2earth.. The earth

covers a circular area of πR2earth where Rearth is the radius of the earth.

• The power absorbed on earth therefore is:

2

2

__ 4 b)

earth

earthearthsunearthonsun r

RePPππ

=

424 4 earthearthearthearthearthearthearth TeRTeAP πσσ ==

CKTCKTCKT

rPT

rPT

rPT

CKT

Kxr

PeRr

RePT

TeRPrRePP

marsvenusmerc

mars

sunmars

venus

sunvenus

merc

sunmerc

earth

earth

sun

earthearthearth

earthearthsunearth

earthearthearthearthearth

earthearthsunearthonsun

000.

24

24

2.

4.

0

49222

24

422

2

__

50223 50323 172445

16

16

16:Note

0277

1088.51644

44

statesteady In

−======⇒

===

==⇒

===⇒

===

πσπσπσ

πσπσππ

πσππ re

2Re

Sometimes

• Even though Mercury is much closer to the Sun than Venus is, Venus is the hottest planet in our Solar System. Temperatures on the plains of Venus often reach 850 degrees Fahrenheit (450 degrees Celsius). This is because of the thick layer of carbon dioxide clouds that surround Venus. Much of the heat from the Sun that hits Mercury escapes back into space because Mercury does not have a large atmosphere withclouds to trap in the warmth. But the carbon dioxide clouds on Venus are so thick they create what scientists call a Greenhouse Effect.

• We cannot calculate the Greenhouse effect using Stefan’s equation. It occurs because the emissivity of the Venusian atmosphere is much larger for long wavelength infrared light and smaller for shorter wavelength light.

443K445 K5.8x1010 m Mercury

2.3x1011 m

1.5x1011 m1.1x1011 m

Orbital radius

223 K

277 K323 K

Calculated T

230 K

287 K723 K

Measured <T>

Mars

Earth Venus

Planet

Wavelength dependence of emissivity of earth

e

If e were constant, it would not matter what value it is.

Mars Details• The Temperature of Mars• Mars is about one and a half times farther from the Sun as the Earth. This

means it receives 44% of the sunlight that we do on Earth, making the planet cold like an icebox. The Mars Pathfinder spacecraft, sitting on Mars' equator in the summer, recorded a high of 14º Fahrenheit.

• Mars undergoes strong seasonal temperature changes. The equator has a summer high temperature of 10-40º Fahrenheit while the Martian poles have a winter temperature of -195º Fahrenheit.

• However, there are also strong changes from day to day as well: Mars Pathfinder recorded the 14º Fahrenheit in the afternoon and went on to record a nighttime low of -105º Fahrenheit! This is the same as the difference between Chicago's summer and winter temperatures, occurring every single day in the same place!

• Lastly, the air on Mars is warmed from the ground up so at early morning your feet can be 70º Fahrenheit warmer than your nose!

• The ground temperatures affect the winds at ground level. Since cold air is denser (heavier), pressure increases at night in the cold and decreases in the day when it is warmer. The pressure difference between night and day drives thermal tide winds. These winds move east to west from the coldest, densest part of the planet's night side towards the hottest part of the planet's dayside (usually early afternoon). So, at ground level, winds blow at 4-16 miles per hour and change direction due to the large daily temperature change following the path of the Sun around Mars every day.

Venus details

• Even though Mercury is much closer to the Sun than Venus is, Venus is the hottest planet in our Solar System. Temperatures on the plains of Venus often reach 850 degrees Fahrenheit (450 degrees Celsius). This is because of the thick layer of carbon dioxide clouds that surround Venus. Much of the heat from the Sun that hits Mercury escapes back into space because Mercury does not have a large atmosphere withclouds to trap in the warmth. But the carbon dioxide clouds on Venus are so thick they create what scientists call a Greenhouse Effect. The Sun's heat enters the atmosphere and then the clouds trap much of it, not allowing it to escape out into space. So even though Venus is 31 million miles (or almost twice as far) from the Sun as Mercury, it is the hotter of the two planets