PHYSICS 231
INTRODUCTORY PHYSICS I
Lecture 17
Recap - Heat Transfer
•
Heat: Q = Energy transferred due to microscopic
contact
•
Heat can:
•
•
Change temperature
Q  mcT
•
c = specific heat
•
For water: c= 1.0 cal/(g°C)
Change state of matter
Q  mL
•
L = Latent heat of fusion or vaporization
•
For water: LF=79.7 cal/g, LV=540 cal/g
Example 11.11
A 50 g ice cube at 0ºC is put into a styrofoam cup
containing 300 g of coffee at 90ºC. Assuming no
heat is lost to the cup or the air, what is the final
temperature of the coffee after the ice melts?
T = 65.8°C
Recap - Kinds of Heat Transfer
• Conduction
• Hot and cold objects in physical contact
• Examples: Heating a skillet, losing heat
through the walls
• Convection
• Hot objects move (gas or liquid)
• Examples: Hot-water heating for buildings
Circulating air
Unstable atmospheres
• Radiation
• Energy transferred by light (UV, IR,…)
• Examples: Stars, Incandescent bulbs
Conduction
• Rate of heat transfer
Q
P   kA T /x
t
 A  T / R
• Conductivity k is property of
material
• R-value R  x /k also depends
on thickness. It adds for
layered objects (R=R1+R2)

Example 11.12
What is the ratio of heat transfer for a single pane
of glass (1 cm thick) to that of a double pane of
glass (each 0.5 cm thick with 1 mm air between)?
DATA: kglass= 0.84 W/mºC, kair= 0.0234 W/m ºC
P1/P2 = 4.59
Convection
• Due to movement of hot gas or liquid
• Hot air rises from radiator causing air currents
• Air trapped between glass panes cannot transfer
heat by convection, only conduction.
Transfer of heat by radiation
• All objects emit light if T > 0
• Colder objects emit longer wavelengths
(red or infra-red)
• Hotter objects emit shorter wavelengths
(blue or ultraviolet)
• Stefan’s Law give power of emitted radiation
P  e AT
Emissivity,
0 < e < 1, usually near 1
4
 = 5.6696x10-8 W/(m2ºK4)
is the Stefan-Boltzmann
constant
T must be in Kelvin !!!
Example 11.8
If the temperature of the Sun fell 5%, and the radius
shrank 10%, what would be the percentage change of
the Sun’s power output?
- 34%
Example 11.9
DATA: The sun radiates 3.74x1026 W
Distance from Sun to Earth = 1.5x1011 m
Radius of Earth = 6.36x106 m
a) What is the intensity (power/m2) of sunlight
a) 1323 W/m2
when it reaches Earth?
b) How much power is absorbed by Earth in
sunlight? (assume that none of the sunlight is
b) 1.68x1017 W
reflected)
c) What average temperature would allow Earth to
radiate an amount of power equal to the amount
of sun power absorbed?
c) T = 276 K = 3 ºC = 37 ºF
What is neglected in estimate?
•Earth is not at one single temperature
•Some of Sun’s energy is reflected
•Reduces TE ~ 20°K
•Emissivity lower at Earth’s thermal wavelengths
than at Sun’s wavelengths (due to atmosphere)
•Increases TE ~ 40°K
•Natural greenhouse effect - necessary for
life on Earth
•Radioactive decays inside Earth are additional
source of energy
•Small effect for Earth
•Jupiter radiates much more energy than it
receives from the sun
Example 11.10a
Two Asteroids A and B orbit the Sun at the same radius
R. Asteroid B has twice the surface area of A. (Assume
both asteroids absorb 100% of the sunlight and have
emissivities of 1.0)
The average temperature of B, TB = _____
a)
b)
c)
d)
e)
(1/4)TA
(1/2)TA
TA
2TA
4TA
Example 11.10b
Two identical asteroids A and B orbit the sun. Asteroid
B is located twice as far from sun as Asteroid A.
RB=2RA
(Assume both asteroids absorb 100% of the sunlight and
have emissivities of 1.0)
The average temperature of B, TB = _____
a)
b)
c)
d)
e)
(1/4)TA
(1/2)TA
(2-1/2)TA
(2-1/4)TA
TA
Example 11.10c
Two Asteroids A and B orbit the Sun at the same
radius R. Asteroid B is painted with reflective paint
which reflects 3/4 of the sunlight, while asteroid A
absorbs 100% of the sunlight. Both asteroids have
emissivities of 1.0.
The average temperature of B, TB = _____
a)
b)
c)
d)
e)
(1/4)TA
(1/2)TA
(2-1/2)TA
(2-1/4)TA
(2-3/4)TA
Example 11.10d
Two Asteroids A and B orbit the Sun at the same radius
R. Asteroid B has an emissivity of 0.25, while the
emissivity of asteroid A is 1.0. Both asteroids absorb
100% of the sunlight.
The average temperature of B, TB = _____
a)
b)
c)
d)
e)
4TA
2TA
21/2TA
21/4TA
23/4TA
Greenhouse Gases
• Sun is much hotter than Earth so sunlight has much
shorter wavelengths than light radiated by Earth
(infrared)
• Emissivity of Earth depends on wavelength
• CO2 in Earth’s atmosphere reflects in the infrared
• Barely affects
incoming sunlight
• Reduces emissivity, e,
of re-radiated heat
Global
warming
•Tearth has risen ~ 1 ºF in past 100 years
• “most of observed increase […] is very likely due to
the observed increase in anthropogenic greenhouse gas
concentrations” Intergovernmental Panel on Climate Change (IPCC)
Mercury and Venus
Tmercury = 700 K (day)
& 90 K (night)
Tvenus = 740 K