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Chapter 11 Energy in Thermal Processes

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Chapter 11 Energy in Thermal Processes. next chapter. Vocabulary, 3 Kinds of Energy. Internal Energy U = Energy of microscopic motion and inter-molucular forces Work W = -F D x = -P D V is work done by expansion (next chapter) Heat Q = Energy transfer from microscopic contact. Mass. - PowerPoint PPT Presentation
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Chapter 11 Chapter 11 Energy in Thermal Processes Energy in Thermal Processes
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Page 1: Chapter 11 Energy in Thermal Processes

Chapter 11Chapter 11

Energy in Thermal ProcessesEnergy in Thermal Processes

Page 2: Chapter 11 Energy in Thermal Processes

Vocabulary, 3 Kinds of EnergyVocabulary, 3 Kinds of Energy

Internal Energy U = Energy of microscopic motion and inter-molucular forces

WorkW = -Fx = -PV is work done by expansion (next chapter)

HeatQ = Energy transfer from microscopic contact

VPQU next chapter

Page 3: Chapter 11 Energy in Thermal Processes

Temperature and Specific HeatTemperature and Specific Heat

Add energy -> T rises

TmcQ

•cH20 = 1.0 cal/(gºC)•1 calorie = 4.186 J

Mass

Property of material

Page 4: Chapter 11 Energy in Thermal Processes

Example 11.1Example 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

Page 5: Chapter 11 Energy in Thermal Processes

Example 11.2Example 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

Page 6: Chapter 11 Energy in Thermal Processes

Phase Changes and Latent HeatPhase 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

mLQ Property of substance /transition

Page 7: Chapter 11 Energy in Thermal Processes

Example 11.3Example 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-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 ¢

Page 8: Chapter 11 Energy in Thermal Processes

Example 11.4Example 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 H20Some fluid drips away

Page 9: Chapter 11 Energy in Thermal Processes

Three Kinds of Heat TranserThree 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

Radiation Light is emitted from hot object Examples: Stars, Incandescent bulbs

Page 10: Chapter 11 Energy in Thermal Processes

ConductionConduction

Power depends on area A, thickness x, temperature difference t and conductivity of material

x

TkAP

Conductivity is propertyof material

Page 11: Chapter 11 Energy in Thermal Processes

Example 11.5Example 11.5A 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

Page 12: Chapter 11 Energy in Thermal Processes

Conductivities and R-valuesConductivities and R-values Conductivity (k)

Property of Material SI units are W/(m ºC)

kxR

R

TA

x

TkAP

/

R-Valueo Property of material and thickness x.o Measures resistance to heato Useful for comparing insulation productso Quoted values are in AWFUL units

Page 13: Chapter 11 Energy in Thermal Processes

Conducitivities Conducitivities and R-valuesand R-values

ARGH!

Page 14: Chapter 11 Energy in Thermal Processes

What makes a good heat conductor?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!

Page 15: Chapter 11 Energy in Thermal Processes

R-values for layersR-values for layers

Consider a layered system, e.g. glass-air-glass

...)(

...

...

321

321

321

RRRA

PA

PR

A

PR

A

PR

TTTT

R

TAP

...321 RRRR

Page 16: Chapter 11 Energy in Thermal Processes

Example 11.6Example 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 ºC and the temperature inside is 20 ºC?DATA: kglass= 0.84 WmºC, kair= 0.0234 Wm ºC

P = 55.7 W

Page 17: Chapter 11 Energy in Thermal Processes

ConvectionConvection

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

Page 18: Chapter 11 Energy in Thermal Processes

Why are windows triple paned?Why are windows triple paned?

To stop convection!

Page 19: Chapter 11 Energy in Thermal Processes

Transfer of heat by radiationTransfer 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

4AeTP = 5.6696x10-8 W/(m2ºK4)is the Stefan-Boltzmann constant

Emissivity, 0 < e < 1, usually near 1

Page 20: Chapter 11 Energy in Thermal Processes

Example 11.7Example 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%

Page 21: Chapter 11 Energy in Thermal Processes

Example 11.8Example 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 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?

a) 1323 W/m2

b) 1.68x1017 W

c) T = 276 ºK = 3 ºC = 37 ºF

Page 22: Chapter 11 Energy in Thermal Processes

What is neglected in estimate?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

NOTE: Venus has a surface T of 900 C

Page 23: Chapter 11 Energy in Thermal Processes

Greenhouse GasesGreenhouse GasesSun is much hotter than Earth so sunlight has much shorter wavelengths than light radiated by Earth (infrared)Emissivity of Earth depends on wavelengthCO2 in Earth’s atmosphere reflects in the infrared

oBarely affects incoming sunlight oReduces emissivity, e, of re-radiated heat

Page 24: Chapter 11 Energy in Thermal Processes

Global Global warmingwarming

Tearth has risen ~ 1 ºF ~ consistent with greenhouse effectOther gases, e.g. S02, could cool Earth


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