Chapter 11
Energy in Thermal Processes
Vocabulary, 3 Kinds of Energy
•
•
•
Internal Energy
U = Energy of microscopic motion and intermolucular forces
Work
W = -Fx = -PV is work done by compression
(next chapter)
Heat
Q = Energy transfer from microscopic contact
U  Q  PV
next chapter
Temperature and Specific Heat
•
Add energy -> T rises
Q  mcT
Mass
Property of material
•cH20 = 1.0 cal/(gºC)
•1 calorie = 4.186 J
Example 11.1
Bobby Joe drinks a 130 “calorie” can of soda. If the
efficiency for turning energy into work is 20%, how
many 4 meter floors must Bobby Joe ascend in order
to work off the soda and maintain her 55 kg mass?
Nfloors = 50.4
Example 11.2
Aluminum has a specific heat of .0924 cal/gºC. If 110
g of hot water at 90 ºC is added to an aluminum cup of
mass 50 g which is originally at a temperature of 23
ºC, what is the final temperature of the equilibrated
water/cup combo?
T = 87.3 ºC
Phase Changes and Latent Heat
•
•
•
T does not rise when phases change (at constant P)
Examples: solid -> liquid (fusion), liquid -> vapor
(vaporization)
Latent heat = energy required to change phases
Q  mL
Property of substance /transition
Example 11.3
1.0 liters of water is heated from 12 ºC to 100 ºC,
then boiled away.
a) How much energy is required to bring the water
to boiling?
b) How much extra energy is required to vaporize
the water?
c) If electricity costs $75 per MW-hr, what was
the cost of heating and boiling the water?
a) Q = 8.8x104 cal = 3.68x105 J
b) Q = 5.4x105 cal = 2.26x106 J
c) 5.5 ¢
Example 11.4
Consider Bobby Joe from the previous example. If
the 80% of the 130 kcals from her soda went into
heat which was taken from her body from
radiation, how much water was perspired to
maintain her normal body temperature? (Assume a
latent heat of vaporization of 540 cal/g even
though T = 37 ºC)
= 193 g
A can of soda has ~ 325 g of H20
Some fluid drips away
Three Kinds of Heat Transfer
•
•
•
Conduction
• Shake your neighbor - pass it down
• Examples: Heating a skillet, losing heat
through the walls
Convection
• Move hot region to a different location
• Examples: Hot-water heating for buildings
Circulating air
Unstable atmospheres
Radiation
• Light is emitted from hot object
• Examples: Stars, Incandescent bulbs
Conduction
•
Power depends on area A, thickness x, temperature
difference T and conductivity of material
P  kA  T / x
Conductivity is property
of material
Example 11.5
A copper pot of radius 12 cm and thickness 5 mm sits
on a burner and boils water. The temperature of the
burner is 115 ºC while the temperature of the inside
of the pot is 100 ºC. What mass of water is boiled
away every minute?
DATA: kCu = 397 W/mºC
m=1.43 kg
Conductivities and R-values
•
Conductivity (k)
• Property of Material
• SI units are W/(m ºC)
T
 T 
P  kA 
A

 x 
R
R  x / k
• R-Value
• Property of material and thickness x.
• Measures resistance to heat
• Useful for comparing insulation products
• Quoted values are in AWFUL units
Conducitivities
and R-values
ARGH!
What makes a good heat conductor?
•“Free” electrons (metals)
•Easy transport of sound (lattice vibrations)
•Stiff is good
•Low Density is good
•Pure crystal structure
Diamond is perfect!
Example 11.6a
An large pipe carries steam at 224 ºC across a large
industrial plant. The outside of the pipe is at room
temperature, 24 ºC. The pipe is 120 m long and has
a diameter of 70 cm. The pipe is constructed of an
insulating material of conductivity k= 2.62 W/mºC.
In order to reduce the rate of heat loss through
the pipe by a factor of 1/2, an engineer could:
a)
b)
c)
d)
e)
Reduce the length of the pipe by a factor of 1/2
Reduce the diameter of the pipe by a factor of 1/2
Increase the thickness of the pipe by a factor of 2
All of the above
None of the above
Example 11.6b
An large pipe carries steam at 224 ºC across a large
industrial plant. The outside of the pipe is at room
temperature, 24 ºC. The pipe is 120 m long and has
a diameter of 70 cm. The pipe is constructed of an
insulating material of conductivity k= 2.62 W/mºC.
In order to reduce the rate of heat loss through
the pipe by a factor of 1/2, an engineer could:
a) Make the pipe using a new material with twice the
conductivity, 5.24 W/m ºC
b) Re-design the pipe to double the R-value
c) All of the above
d) None of the above
Example 11.6c
An large pipe carries steam at 224 ºC across a large
industrial plant. The outside of the pipe is at room
temperature, 24 ºC. The pipe is 120 m long and has
a diameter of 70 cm. The pipe is constructed of an
insulating material of conductivity k= 2.62 W/mºC.
In order to reduce the rate of heat loss through
the pipe by a factor of 1/2, an engineer could:
a) Reduce the density of steam in the pipe by a
factor of 1/2
b) Reduce the temperature of the steam to 124 ºC
c) Reduce the velocity of the steam through the pipe
by a factor of 1/2
d) All of the above
e) None of the above
R-values for layers
Consider a layered system, e.g. glass-air-glass
T
PA
R
T  T1  T2  T3 ...
PR1 PR2 PR3



 ...
A
A
A
P
 (R1  R2  R3  ...)
A
R  R1  R2  R3  ...
Example 11.7
Consider three panes of glass, each of thickness 5
mm. The panes trap two 2.5 cm layers of air in a
large glass door. How much power leaks through a 2.0
m2 glass door if the temperature outside is -40 ºC
and the temperature inside is 20 ºC?
DATA: kglass= 0.84 W/mºC, kair= 0.0234 W/m ºC
P = 55.8 W
Convection
•
•
•
If warm air blows across the room, it is convection
If there is no wind, it is conduction
Can be instigated by turbulence or instabilities
Why are windows triple paned?
To stop convection!
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
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
•Emissivity lower at Earth’s thermal wavelengths
than at Sun’s wavelengths
•Radioactive decays inside Earth
•Hot underground (less so in Canada)
•Most of Jupiter’s radiation
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
• ~ consistent with greenhouse effect
•Other gases, e.g. S02, could cool Earth
Mercury and Venus
Tmercury = 700 K (day) & 90 K (night)
Tvenus = 740 K