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Latent Heat

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When a solid melts or a liquid boils, energy must be added but the temperature remains constant! (This can be explained by considering that it takes energy to break ... – PowerPoint PPT presentation

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Title: Latent Heat


1
Latent Heat
  • When a solid melts or a liquid boils, energy must
    be added but the temperature remains constant!
    (This can be explained by considering that it
    takes energy to break the bonds holding the
    material together.)
  • The amount of energy it takes to melt or boil a
    certain amount of material is called a latent
    heat.

2
Latent Heat
  • For water, the latent heat of fusion (heat needed
    to melt ice to water) is 79.7 cal/gm.
  • For water, the latent heat of vaporization (heat
    needed to boil water) is 540 cal/gm.
  • For alcohol, the latent heat of vaporization is
    less at 204 cal/gm.

3
Latent Heat - Example
  • Example how much energy does it take to
    vaporize 1 liter of water if the water is
    initially at a temperature of 98oF ?

4
Latent Heat - Example cont.
  • First we need to find the energy to raise the
    temperature of the water up to boiling
  • this involves the heat capacity
  • (which for water is 1 cal/gmoC)
  • (density of water is 1 gm/cc, 1 liter 1000 cc)
  • C Q/(m?T) , with ?T (212-98)5/963oC
  • Q (1 cal/gmoC)(1 gm/1cc)1000 cc63oC
  • 63,333 cal (4.186 J/cal) 265,000 J .

5
Latent Heat - Example cont.
  • Now we add in the latent heat
  • (for water, this is 540 cal/gm)
  • Q Lm (540 cal/gm)(1 gm/cc)(1000 cc)
  • 540,000 cal (4.186 J/cal) 2,260,000 J
  • Total energy required is 265,000 J 2,260,000
    J 2,525,000 J .

6
Latent Heat - Example 2
  • Question how much water would be needed to keep
    cool for 4 hours by evaporation if the outside
    temperature is 100 oF (essentially same as
    bodys) and a power output of 200 Watts (doing
    some work) is desired?

7
Latent Heat - Example cont.
  • Since the body generates 200 Watts, or 200 Joules
    a second, the body must evaporate water to carry
    this energy away.
  • Q (200 J/sec)(4 hs)(3600 sec/hr) 2,880,000 J.
  • From the previous considerations, evaporating 1
    liter of water carries away 2,525,000 J. Thus we
    need 2.88MJ / (2.525MJ/liter)
  • 1.14 liters of water.

8
Latent Heat - Example cont.
  • Would more or less alcohol be needed to keep cool
    for the same energy output?
  • (The heat capacity of alcohol is 2.4 J/gmoC the
    density of ethanol .791 gm/cc the boiling
    point is 78oC latent heat of vaporization is
    854 J/gm).
  • From this you should be able to decide whether
    water or alcohol is better for heat regulation.

9
Heat Transfer
  • There are four ways of moving heat
  • Evaporation (using latent heat - weve
    already looked at this)
  • Convection (moving heat with a material)
  • Conduction (moving heat through a material)
  • Radiation
  • Well develop equations for conduction and
    radiation and talk about convection.

10
Heat Transfer Convection
  • Heat Transfer by Convection is when you heat some
    material and then move that material containing
    the heat.
  • The amount of heat energy moved depends on the
    heat in the material (heat capacity times amount
    of material times the temperature difference) and
    how much material you move per time.
  • The blood and hot air furnaces use this method.

11
Heat Transfer Conduction
  • Heat will flow through a solid material from the
    hot end to the cold end. What is flowing? No
    matter is flowing!
  • We can think of energy as flowing in this case!
    We measure the flow of energy as power 1 Watt
    1 Joule/sec .

12
Heat Transfer Conduction
  • Power Q/t kA?T/L
  • where k is a constant that depends on the
    material, called the thermal conductivity
  • where A is the cross sectional area
  • where L is the distance from the hot end to
    the cold end
  • and ?T is the temperature difference
    between the hot and cold ends.

L
A
k
Thi
Tlow
13
Conduction - R values
  • Units of thermal conductivity from
  • Power Q/t kA?T/L
  • k has units of Wm/(m2K) or J/(secmK), and k
    depends only on the material.
  • Often a material is given an R value, where R
    includes both the material and the thickness of
    the material R L/k, and R has the units of
    m2secK/J (or ft2oFhr/BTU, where 1
    ft2oFhr/BTU 0.176 m2secK/J)

14
Conduction - R values
  • P Q/t kA?T/L A?T / R
  • where we define R L / k .
  • where if we have several different materials and
    thicknesses, we can simply add the individual Rs
    to get the total R
  • Rtotal ?Ri .

15
Conduction - Conductivity
  • approximate k values for some materials
  • metals k 1 cal/(seccmoC)
  • glass k 2 x 10-3 cal/(seccmoC)
  • wood, brick, fiberglass
  • k 1 x 10-4 cal/(seccmoC)
  • air k 5 x 10-5 cal/(seccmoC)

16
Conduction - Example
  • Lets calculate the R value of brick 4 inches in
    thickness L 4 in (.0254 m/in) .10 m
  • k 1.5 x 10-4 cal/(seccmoC)
  • (4.186 J/cal)(100 cm/m) .063 J/(secmoC)
  • R L/k .10 m / .063 J/(secmoC)
  • 1.6 m2K/Watt
  • (1 ft2oFhr/BTU) /( .176 m2K/Watt)
  • 9 (1 ft2oFhr/BTU)

17
Conduction - Example
  • What is the heat loss through the brick walls
    (assume no other insulation) of a house that is
    50 ft x 30 ft (floor area 1500 ft2) x 8 ft when
    the temperature inside is 72oF and the
    temperature outside is 20oF ?

18
Conduction - Example
  • P Q/t kA?T/L A?T / R
  • A 50ft 8ft 30ft 8ft 50ft 8ft
    30ft8ft 1280 ft2 (1 m2/10.76 ft2) 120 m2 .
  • ?T (72-20)oF (5K/9oF) 29 K
  • R 1.6 m2K/Watt
  • P 120 m2 29 K / 1.6 m2K/Watt
  • 2175 Watts 2.175 kW.

19
Blackbody Radiation
  • What is a blackbody?
  • A BLACK object absorbs all the light incident on
    it.
  • A WHITE object reflects all the light incident on
    it, usually in a diffuse way rather than in a
    specular (mirror-like) way.

20
Blackbody Radiation
  • The light from a blackbody then is light that
    comes solely from the object itself rather than
    being reflected from some other source.
  • A good way of making a blackbody is to force
    reflected light to make lots of reflections
    inside a bottle with a small opening.

21
Blackbody Radiation
  • If very hot objects glow (such as the filaments
    of light bulbs and electric burners), do all warm
    objects glow?
  • Do we glow? (Are we warm?
  • Are you HOT?)

22
Blackbody Radiation
  • What are the parameters associated with the
    making of light from warm objects?

23
Blackbody Radiation
  • What are the parameters associated with the
    making of light from warm objects?
  • Temperature of the object.
  • Surface area of the object.
  • Color of the object ? (If black objects absorb
    better than white objects, will black objects
    emit better than white objects?)

24
Blackbody Radiation Color Experiment
  • Consider the following way of making your stove
    hot and your freezer cold

25
Color Experiment
  • Put a white object in an insulated and evacuated
    box with a black object. The black object will
    absorb the radiation from the white object and
    become hot, while the white object will reflect
    the radiation from the black object and become
    cool.
  • Put the white object in the freezer, and the
    black object in the stove.

26
Color Experiment
  • Does this violate Conservation of Energy?

27
Color Experiment
  • Does this violate Conservation of Energy? NO
  • Does this violate the Second Law of
    Thermodynamics (entropy tends to increase) ?

28
Color Experiment
  • Does this violate Conservation of Energy? NO
  • Does this violate the Second Law of
    Thermodynamics (entropy tends to increase) ?
    YES
  • This means that a good absorber is also a good
    emitter, and a poor absorber is a poor emitter.
    Use the symbol ? to indicate the blackness (?1)
    or the whiteness (?0) of an object.

29
Blackbody Radiation
  • What are the parameters associated with the
    making of light from warm objects?
  • Temperature of the object, T.
  • Surface area of the object, A.
  • Color of the object, ??

30
Blackbody Radiation
  • Is the ? for us close to 0 or 1?
  • (i.e., are we white or black?)
  • We emit light in the IR, not the visible.
  • So what is our ? for the IR?

31
Blackbody Radiation
  • So what is our ? for the IR?
  • Have you ever been near a fire on a cold night?
  • Have you noticed that your front can get hot at
    the same time your back can get cold?
  • Can your hand block this heat from the fire?
  • Is this due to convection or radiation?

32
Blackbody radiation
  • For humans in the IR, we are all fairly good
    absorbers (black). An estimated value for ? for
    us then is about .97 .

33
Blackbody Radiation Experimental Results
  • At 310 Kelvin, only get IR

Intensity
IR
blue yellow red
UV
wavelength
34
Blackbody Radiation Experimental Results
  • At much higher temperatures, get visible
  • look at blue/red ratio to get temperature

Intensity
IR
blue yellow red
UV
wavelength
35
Blackbody Radiation Experimental Results
  • Ptotal ??AT4
  • where ? 5.67 x 10-8 W/m2
    K4
  • ??????????peak b/T where b 2.9 x 10-3 mK

Intensity
IR
blue yellow red
UV
wavelength
36
Blackbody Radiation Example
  • Given that you eat 2000 Calories/day,
  • your power output is around 100 Watts.
  • Given that your body temperature is
  • about 90o F , and
  • Given that your surface area is about
  • 1.5 m2,

37
Blackbody Radiation Example
  • Given Ptotal 100 Watts
  • Given that Tbody 90o F
  • Given that A 1.5 m2
  • WHAT IS THE POWER EMITTED VIA
  • RADIATION?

38
Blackbody Radiation Example
  • Pemitted ??AT4
  • ? .97
  • ?? 5.67 x 10-8 W/m2 K4
  • T 273 (90-32)5/9 (in K) 305 K
  • A 1.5 m2
  • Pemitted 714 Watts
  • (compared to 100 Watts generated!)

39
Blackbody Radiation Example
  • need to consider power absorbed at room T
  • Pabsorbed ??AT4
  • ? .97
  • ?? 5.67 x 10-8 W/m2 K4
  • T 273 (72-32)5/9 (in K) 295 K
  • A 1.5 m2
  • Pabsorbed 625 Watts
  • (compared to 714 Watts emitted!)

40
Blackbody Radiation Example
  • Total power lost by radiation
  • 714 W - 625 W 89 Watts
  • (Power generated is 100 Watts.)
  • Power also lost by convection (with air)
  • and by evaporation.

41
Blackbody Radiation Example
  • At colder temperatures, our emitted power stays
    about the same while our absorbed power gets much
    lower. This means that we will get cold unless
  • we generate more power, or
  • our skin gets colder, or
  • we reflect the IR back into our bodies.
  • Use metal foil for insulation!

42
Thermodynamics
  • The First Law of Thermodynamics is a fancy name
    for the Law of Conservation of Energy applied to
    thermal systems. It says
  • DU Q - W
  • where DU indicates the change in the internal
    energy of the system. This internal energy is
    related to the temperature and heat capacity of
    the system Q is the heat energy added to the
    system and W is the work done by the system.

43
Thermodynamics
  • The first law of thermodynamics, like the
    conservation of energy, does not indicate the
    direction. It does not explain why, when cold
    milk is added to hot coffee, the cold milk warms
    up and the hot coffee cools down. The
    conservation of energy (first law of
    thermodynamics) permits the possibility that the
    milk would get even colder while the coffee gets
    hotter after they are mixed.

44
Second Law of Thermodynamics
  • It is the Second Law of Thermodynamics that
    explains why the hot coffee does cool down and
    the cold milk warms up when they are mixed.
  • To understand the second law, however, we need to
    first look a little at probability.

45
Probability
  • Consider flipping four coins. How many heads
    would you expect to get (assuming they were
    honest coins)?
  • Why do you expect this?
  • Lets look at all the possible combinations of
    flipping four coins

46
Flipping Four Coins
  • Four heads (only one way) HHHH
  • Three heads (four ways)
  • THHH HTHH HHTH HHHT
  • Two heads (six ways)
  • TTHH THTH THHT
  • HTTH HTHT HHTT
  • One head (four ways)
  • HTTT THTT TTHT TTTH
  • Zero heads (only one way) TTTT

47
Probability
  • We see that there are more ways of getting two
    heads and two tails than any other combination.
  • The same argument can be made about the
    distribution of energy among many molecules the
    highest probability corresponds to the most ways
    of having that outcome.

48
Probability
  • In the case of distributing the thermal energy
    between the hot coffee and the cold milk, there
    are more ways of distributing the energy equally
    among the many coffee and milk molecules than
    there are ways of giving it all to just the
    coffee or just the milk molecules.

49
Statement of 2nd Law
  • A system will tend to go to its most probable
    state.
  • To measure the ways of having the same state
    (like determining the number of ways of having
    two heads out of four coins), we use the concept
    of entropy.

50
Another Statement
  • Entropy is a measure of the probability of being
    in a state. Since things tend to go to their
    most probable state, we can write the 2nd Law of
    Thermodynamics as systems tend to have their
    entropy increase.

51
About the 2nd Law
  • Note that this is not an absolute law like
    Conservation of Energy. Rather it is a
    probablistic Law. However, when dealing with
    large numbers (recall that Avagadros Number is 6
    x 1023, or almost a trillion trillion), the
    probabilities become essentially certainties.

52
Heat Engines
  • Heat engines are devices that convert some of the
    heat into useful energies such as electrical.
    These engines can only work, however, if there is
    a difference in temperature.
  • We can think by analogy just as gravity tends
    to bring masses together, and we can get work out
    of a gravitational separation so entropy tends
    to increase by bringing temperatures together,
    and we can get work out of a separation of
    temperatures.

53
Efficiency
  • Efficiency is a measure of how much you get out
    versus how much you put in. For heat engines,
    then
  • Efficiency ? Work done / Heat Added
  • By the first law, then, the work done is simply
    the difference in the heat going into the engine
    minus the heat coming out of the engine. The
    total heat added is the heat going into the
    engine. ? (Qhot - Qcold) /Qhot

54
Carnot Efficienty
  • By considering the most efficient way of running
    a heat engine, we come up with the Carnot cycle.
    This is the best we can theoretically do with a
    heat engine. For a Carnot efficiency, we have
    the formula
  • ??Carnot (Thot - Tcold) / Thot .
  • Note that these temperatures must be in absolute
    (Kelvin), not oC or oF .

55
Heat Engines
  • Our heat engines, whether fired by coal, oil or
    nuclear energy, are all limited in their
    efficiencies by this Carnot efficiency. In
    practice we come very close to this theoretical
    maximum.
  • Note that for the best efficiencies, we need Thot
    to be very hot and Tcold to be very cold. Due to
    material limitations on both Thot and Tcold, this
    efficiency is about 30.

56
Heat Engines
  • All heat engines, then, have to get rid of the
    excess heat energy. Most major power stations do
    this via water because of the high heat capacity
    of water relative to air.
  • The biggest structure at nuclear power stations
    is the cooling tower, and this is often depicted
    as a symbol of nuclear power. Yet the same heat
    dump is present at coal and oil facilities!

57
Heat Engines
  • Your car is a heat engine, but it is only about
    15 efficient compared to a major power plants
    efficiency of 30. Your cars engine cannot
    operate at the same high temperature as the power
    plant!
  • Your car employs both a water transfer (via water
    convection inside the engine) as well as an air
    transfer (at the radiator) for the waste heat.
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