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Chapter 11 Energy in Thermal Processes

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c) Find average T of earth. T = 276 K = 3 C = 37 F. Why is the Earth warmer? ... Emissivity lower at Earth's thermal wavelengths than at Sun's wavelengths ... – PowerPoint PPT presentation

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Title: Chapter 11 Energy in Thermal Processes


1
Chapter 11Energy in Thermal Processes
2
Vocabulary, 3 Kinds of Energy
  • Internal Energy U Energy of a system due to
    microscopic motion and inter-molucular forces
  • WorkW -FDx -PDV is work done by expansion
    (next chapter)
  • HeatQ Energy transfer from microscopic contact

3
Temperature and Specific Heat
  • Add energy -gt T rises

4
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!
  • 10 of Q gets converted to PE

Nfloors 25
5
Example (Calorimetry)
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?
Solution
  • Equate heat loss of water with heat gain of cup
  • Solve for T

T 87.3 ºC
6
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

7
Example Boiling water
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 1000
kW-hrs, what was the cost of boiling the water?
8
Solution
a) Given m1000 g, c1.0 cal/g, DT88 Find Q
Q 8.8x104 cal 3.68x105 J
9
Solution continued
b) Given L540 cal/g, m1000gFind Q
Q 5.4x105 cal 2.26x106 J
c) Given Q 2.26x1063.68x105 J Rate
75/(1000 kW-hr)Find cost
  • First, find rate in dollars/J
  • Then find net cost Q multiplied by rate

5.5
10
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

11
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 ºC)
Solution
Given Q 0.9x1.3x105 cal, L 540 cal/g Find
mevap
mevapQ/L
217 g
A can of soda has 350 g of H20 Some fluid drips
away
12
Three Kinds of Heat Transer
  • Conduction
  • Shake your neighbor - pass it down
  • Examples Heating a skillet
    Losing heat through the walls of a house
  • 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

13
Conduction
  • Power depends on area, length, temperature
    difference and conductivity of material

14
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 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
Solution
Given Dx 0.005 m, A pr2 (r.12 m), Th 115
ºC, Tc 100 ºC time 60 sec, kCu
397 W/mºC, L 540 cal/g
  • First, find the power, P 5.39x104 Watts
  • Next, find Q Ptime 3.23x106 J
  • Finally, find m of vaporized water

Remember (L4.186540 J/g)
m1.43 kg
15
Conductivities and R-values
  • Conductivity
  • Property of Material
  • SI units are W/(m ºC)
  • R-Value
  • Property of material and thickness Dx.
  • Measures resistance to heat
  • Useful for comparing insulation products
  • Quoted values are in AWFUL units

16
Conducitivities and R-values
17
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!
18
R-values for layers
Consider a layered system, e.g. glass-air-glass
19
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 ºC and the temperature inside is 20 ºC? DATA
kglass 0.84 WmºC, kair 0.0234 Wm ºC
20
Solution
Known kglass0.84, Dxglass0.005, kair0.0234,
Dxair0.025, A2.0, DT60 Find P
  • First, find Rglass and Rair for one layer of each

Rglass0.00595, Rair1.068
  • Next, find R for all the layers
  • Finally, find the power

P 55.7 W
21
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

22
Why are windows triple paned?
To stop convection!
23
Transfer of heat by radiation
  • 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)
  • Stefans Law give power of emitted radiation

24
Example
If the temperature of the Sun fell 5, and the
radius shrank 10, what would be the percentage
change of the Suns power output?
Solution
- 34
25
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
  • What is the intensity (power/m2) of sunlight when
    it reaches Earth?
  • How much power is absorbed by Earth in sunlight?
    (assume that none of the sunlight is reflected)
  • What average temperature would allow Earth to
    radiate an amount of power equal to the amount of
    sun power absorbed?

26
Solution
Given Psun, Rearth-sun, Rearth
a) Find IP/A of sunlight at the Earths orbit
1320 W/m2
b) Find power absorbed by earth
1.67x1017 W
c) Find average T of earth
T 276 ºK 3 ºC 37 ºF
27
Why is the Earth warmer?
  • Earth is not at one single temperature
  • Emissivity lower at Earths thermal wavelengths
    than at Suns wavelengths
  • Radioactive decays inside Earth
  • Hot underground (less so in Canada)
  • Most of Jupiters radiation

28
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 Earths atmosphere reflects in the
    infrared
  • Barely affects incoming sunlight
  • Reduces emissivity, e, of re-radiated heat

29
Global warming
  • Tearth has risen 1 ºF
  • consistent with greenhouse effect
  • Other gases, e.g. S02, could cool Earth
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