Chapter 11Chapter 11

Energy in Thermal ProcessesEnergy in Thermal Processes

ΔU =Q − PΔV

Vocabulary, 3 Kinds of EnergyVocabulary, 3 Kinds of Energy

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

• WorkW = -FΔx = -PΔV is work done by compression

(next chapter)

• HeatQ = Energy transfer from microscopic contact

next chapter

Q =mcΔT

Temperature and Specific HeatTemperature and Specific Heat

• Add energy -> T rises

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

Property of material

Mass

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

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

Q =mL

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

Property of substance /transition

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-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 11.4Example 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

Three Kinds of Heat TransferThree 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

P =kA⋅ΔT / Δx

ConductionConduction

• Power depends on area A, thickness Δx, temperature difference ΔT and conductivity of material

Conductivity is propertyof material

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

Conductivities and R-valuesConductivities and R-values

• Conductivity (k)

• Property of Material

• SI units are W/(m ºC)

• R-Value• Property of material and thickness Δx.• Measures resistance to heat• Useful for comparing insulation products• Quoted values are in AWFUL units

P =kAΔTΔx

⎛⎝⎜

⎞⎠⎟=A

ΔTR

R=Δx/ k

Conducitivities Conducitivities and R-valuesand R-values

ARGH!

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!

Example 11.6aExample 11.6aAn 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.

a) Reduce the length of the pipe by a factor of 1/2b) Reduce the diameter of the pipe by a factor of 1/2c) Increase the thickness of the pipe by a factor of 2d) All of the abovee) None of the above

In order to reduce the rate of heat loss through the pipe by a factor of 1/2, an engineer could:

Example 11.6bExample 11.6bAn 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 ºCb) Re-design the pipe to double the R-valuec) All of the aboved) None of the above

Example 11.6cExample 11.6cAn 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.

a) Reduce the density of steam in the pipe by a factor of 1/2b) Reduce the temperature of the steam to 124 ºCc) Reduce the velocity of the steam through the pipe by a factor of 1/2d) All of the abovee) None of the above

In order to reduce the rate of heat loss through the pipe by a factor of 1/2, an engineer could:

R-values for layersR-values for layers

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

ΔT = ΔT1 + ΔT2 + ΔT3...

=PR1A+PR2A+PR3A+ ...

=P

A(R1 + R2 + R3 + ...)

R =R1 + R2 + R3 + ...

P =AΔTR

Example 11.7Example 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

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

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

To stop convection!

P =eσAT 4

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

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

Emissivity, 0 < e < 1, usually near 1

Example 11.8Example 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 11.9Example 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 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 K = 3 ºC = 37 ºF

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

Example 11.10aExample 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) (1/4)TA

b) (1/2)TA

c) TA

d) 2TA

e) 4TA

Example 11.10bExample 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) (1/4)TA

b) (1/2)TA

c) (2-1/2)TA

d) (2-1/4)TA

e) TA

Example 11.10cExample 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) (1/4)TA

b) (1/2)TA

c) (2-1/2)TA

d) (2-1/4)TA

e) (2-3/4)TA

Example 11.10dExample 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) 4TA

b) 2TA

c) 21/2TA

d) 21/4TA

e) 23/4TA

Greenhouse GasesGreenhouse 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 Global warmingwarming

•Tearth has risen ~ 1 ºF

• ~ consistent with greenhouse effect•Other gases, e.g. S02, could cool Earth

Mercury and VenusMercury and Venus

Tmercury = 700 K (day) & 90 K (night)

Tvenus = 740 K