# Chapter 11 Energy in Thermal Processes ```Chapter 11
Energy in Thermal Processes
Vocabulary, 3 Kinds of Energy
 Internal Energy


U = Energy of microscopic motion and intermolucular forces
Work
W = -FDx = -PDV is work done by expansion (next
chapter)
Heat
Q = Energy transfer from microscopic contact
DU  Q  PDV
next chapter
Temperature and Specific Heat
 Add energy -&gt; T rises
Q  mcDT
Mass
Property of material
•cH20 = 1.0 cal/(g&ordm;C)
•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/g&ordm;C. If 110
g of hot water at 90 &ordm;C is added to an aluminum cup of
mass 50 g which is originally at a temperature of 23
&ordm;C, what is the final temperature of the equilibrated
water/cup combo?
T = 87.3 &ordm;C
Phase Changes and Latent Heat
 T does not rise when phases change (at constant P)
 Examples: solid -&gt; liquid (fusion), liquid -&gt; 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 &ordm;C to 100 &ordm;C,
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-hrs, what was
the cost of boiling the water?
a) Q = 8.8x104 cal = 3.68x105 J
b) Q = 5.4x105 cal = 2.26x106 J
c) 5.5 &cent;
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 &ordm;C)
= 193 g
A can of soda has ~ 325 g of H20
Some fluid drips away
Three Kinds of Heat Transer
 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
 Light is emitted from hot object
 Examples: Stars, Incandescent bulbs
Conduction
 Power depends on area A, thickness Dx, temperature
difference Dt and conductivity of material
 DT 
P  kA

 Dx 
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 &ordm;C while the temperature of the inside
of the pot is 100 &ordm;C. What mass of water is boiled
away every minute?
DATA: kCu = 397 W/m&ordm;C
m=1.43 kg
Conductivities and R-values
 Conductivity (k)
 Property of Material
 SI units are W/(m &ordm;C)
DT
 DT 
P  kA
 A
R
 Dx 
R  Dx / k
 R-Value
o Property of material and thickness Dx.
o Measures resistance to heat
o Useful for comparing insulation products
o 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!
R-values for layers
Consider a layered system, e.g. glass-air-glass
DT
PA
R
DT  DT1  DT2  DT3 ...
PR1 PR2 PR3



 ...
A
A
A
P
 ( R1  R2  R3  ...)
A
R  R1  R2  R3  ...
Example 11.6
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 &ordm;C
and the temperature inside is 20 &ordm;C?
DATA: kglass= 0.84 Wm&ordm;C, kair= 0.0234 Wm &ordm;C
P = 55.7 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!
 All objects emit light if T &gt; 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  AeT
4
Emissivity,
 = 5.6696x10-8 W/(m2&ordm;K4)0 &lt; e &lt; 1, usually near 1
is the Stefan-Boltzmann constant
Example 11.7
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 11.8
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 &ordm;K = 3 &ordm;C = 37 &ordm;F
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
•Hot underground (less so in Canada)
NOTE: Venus has a surface T of 900 C
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
oBarely affects
incoming sunlight
oReduces emissivity, e,