# Chapter 11 Energy in Thermal Processes

```Chapter 11
Energy in Thermal Processes
Vocabulary, 3 Kinds of Energy
 Internal Energy
U = Energy of a system due to microscopic motion and
inter-molucular 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 of material
Property of the material
•cH20 = 1.0 cal/(g&ordm;C)
•1 calorie = 4.186 J
Example: Converting calories
Bobby Joe drinks a 130 “calorie” can of soda. If the
efficiency for turning energy into work is 10%, how
many 4 meter floors must Bobby Joe ascend in order to
work off the soda and maintain her 55 kg mass?
Solution
•First, note that the “calories” listed in food are actually
kilocalories!
J
Q  130 kcal  1.3 10 cal  4.186
 5.44 105 J
cal
5
•10% of Q gets converted to PE
Q / 10  mgh, h  101 m
Nfloors = 25
Example (Calorimetry)
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?
Solution
•Equate heat loss of water with heat gain of cup
DU  Q  mcDT
mcupc Al (T  23 C )  mH 20cH 20 (90 C  T )
•Solve for T
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 and
type of transition
Example: Boiling water
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 1000 kW-hrs, what was
the cost of boiling the water?
Solution
a) Given m=1000 g, c=1.0 cal/g, DT=88
Find Q
Q  mcDT
Q = 8.8x104 cal = 3.68x105 J
Solution continued
b) Given L=540 cal/g, m=1000g
Find Q
Q  mL f
Q = 5.4x105 cal = 2.26x106 J
c) Given Q = 2.26x106+3.68x105 J
Rate = \$75/(1000 kW-hr)
Find cost
•First, find rate in dollars/J
\$75
Rate 
dollars/J  2.08 10 8 dollars/J
1000 1000  3600
•Then find net cost = Q multiplied by rate
= 5.5 &cent;
Announcements
•Midterms graded on scale of 11
•Who wants extra review/recitation sessions?
•You can pick up your exams (not bubble sheet)
in Friday helproom
Example: Body cooling
Consider Bobby Joe from the previous example. If the
90% 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)
Solution
Given: Q = 0.9x1.3x105 cal, L = 540 cal/g
Find: mevap
Q  mL
mevap=Q/L
= 217 g
A can of soda has ~ 350 g of H20
Some fluid drips away
Three Kinds of Heat Transer
 Conduction
o Shake your neighbor - pass it down
o Examples: Heating a skillet
Losing heat through the walls of a house
 Convection
o Move hot region to a different location
o Examples: Hot-water heating for buildings
Circulating air
Unstable atmospheres
o Light is emitted from hot object
o Examples: Stars
Incandescent bulbs
Conduction
 Power depends on area, length, temperature
difference and conductivity of material
 DT 
P  kA

 Dx 
Conductivity is property
of material
Example
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
Solution
Given: Dx = 0.005 m, A = pr2 (r=.12 m), Th = 115 &ordm;C, Tc = 100 &ordm;C
time = 60 sec, kCu = 397 W/m&ordm;C, L = 540 cal/g
 DT 
P  kA

 Dx 
Q  mL f
•First, find the power, P = 5.39x104 Watts
•Next, find Q = P&middot;time = 3.23x106 J
•Finally, find m of vaporized water
m  Q / Lf
Remember
(L=4.186&middot;540 J/g)
m=1.43 kg
Conductivities and R-values
 Conductivity
o Property of Material
o 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: Glass Door
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
Solution
Known: kglass=0.84, Dxglass=0.005, kair=0.0234, Dxair=0.025,
A=2.0, DT=60
Find: P
•First, find Rglass and Rair for one layer of each
R  Dx / k
Rglass=0.00595, Rair=1.068
•Next, find R for all the layers
R  R1  R2  R3  ...
R  3Rglass  2 Rair  2.155
•Finally, find the power
DT
PA
R
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,
0 &lt; e &lt; 1, usually near 1
 = 5.6696x10-8 W/(m2&ordm;K4)
is the Stefan-Boltzmann constant
Example
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?
P  AeT
4
Solution
4
2
4
P AeT
 4pR eT


4
2
4
P0 A0 eT0  4pR 0 eT0
 0.9  0.95  0.660
2
4
- 34%
Example: Power of the Sun
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 when
it reaches Earth?
b) How much power is absorbed by Earth in sunlight?
(assume that none of the sunlight is reflected)
c) What average temperature would allow Earth to
radiate an amount of power equal to the amount of
sun power absorbed?
Solution
Given: Psun, Rearth-sun, Rearth
a) Find I=P/A of sunlight at the Earth’s orbit
I sun
Psun

2
4pRearth
-sun3
= 1320 W/m2
b) Find power absorbed by earth
P  Ia, a  pR
2
earth
= 1.67x1017 W
c) Find average T of earth
P  AeT
4
Aearth  4pR
2
earth
P
T 
Ae
4
, e 1
T = 276 &ordm;K = 3 &ordm;C = 37 &ordm;F
Why is the Earth warmer?
•Earth is not at one single temperature
•Emissivity lower at Earth’s thermal wavelengths
than at Sun’s wavelengths
•Hot underground (less so in Canada)
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,