Chapter 20 Heat Heat Transfer Phase Changes Specific Heat Calorimetry First Law of Thermo Work

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1 Chapter 20 Heat Heat Transfer Phase Changes Specific Heat Calorimetry First Law of Thermo Work

2 Heat Energy is a flow of energy from hotter to colder because of a difference in temperature. Objects do not have heat. [Heat] = Joule Heat Energy entering or leaving a system will cause either a Temperature Change: Q = mcδt or a Phase Change: Q = ml

3 Any two systems placed in thermal contact will have an exchange of heat energy until they reach the same temperature. If the systems are in thermal equilibrium then no net changes take place.

4 Heat Energy is a flow of energy from hotter to colder because of a difference in temperature. Objects do not have heat. [Heat] = Joule Internal Energy of a system is a measure of the total Energy due to ALL random molecular motions INTERNAL of the system (Translations KE, Rotational KE, Vibrational KE) and internal POTENTIAL energies due to interactive forces (electromagnetic, strong, weak, gravitational) Objects have energy. Mechanical Energy is due to the kinetic and potential energies of the system itself in an external reference frame. Mechancial Equivalent of Heat: mechanical energy converted to heat energy by doing work on the system: kcal = 4186J

5 The change of internal energy of a system due to a temperature or phase change is given by: Temperature Change: Q = mcδt Phase Change: Q = ml Q is positive when the system GAINS heat and negative when it LOSES heat.

6 Specific Heat: Thermal Inertia The Specific Heat of a substance is the amount of Energy it requires to raise the temperature of 1 kg, 1 degree Celsius. Q Q J = mcδt c = = 0 mδt kg C The higher the specific heat, the more energy it takes and the longer it takes to heat up and to cool off. The lower the specific heat, the less energy it takes and the quicker it takes to heat up and cool off. Substances with HIGH specific heat STORE heat energy and make good thermal moderators. (Ex: Water, Oceans)

7 Some Specific Heat Values

8 More Specific Heat Values

9 Specific Heat c c c water glycerin iron J = 4186 kg Why does water have such a high specific heat? Heat goes into other modes of energy so that temperature changes slowly. 0 0 J = 2410 kg J = 452 kg C C 0 C

10 A 10,000 kg truck applies the brakes and descends 75.0 m at a constant speed, causing the brakes to smoke as shown. If the brakes have a mass of kg and a specific heat of 800 J/kg C, calculate the temperature increase of the brakes. m gh= m cδt truck m m brakes Δ truck T = = brakes gh c 92.0 o C

11 Q = mcδt How much heat is required to raise the temperature of a 0.750kg aluminum pot containing 2.50kg of water at 30ºC to the boiling point? Q= malcalδ T + mwcwδt = m c + m c ΔT ( ) Al Al w w o o o =.75 kg(900 J / kg C) kg(4186 J / kg C) (70 C) Q= 7.798x10 5 J

12 Phase Change Q = ml A change from one phase to another A phase change always occurs with an exchange of energy! A phase change always occurs at constant temperature!

13 Sample Latent Heat Values

14 Phase Change Energy goes into the system and breaks molecular bonds.. Energy is given up by the system by forming molecular bonds

15 Phase Change: Melting & Freezing Melting: Energy goes into the system and breaks molecular bonds.. Freezing: Energy is given up by the system by forming molecular bonds

16 Phase Change: Melting & Freezing

17 Phase Change: Melting & Freezing Melting: Solid to the melting temperature Melting is a cooling process Freezing: Liquid to the melting temperature Freezing is a warming process.

18 Why do farmers spray peaches with water to save them from frost? Freezing is a warming process!

19 If you were in an igloo on a freezing night. You would be warmed more by a) a bucket of ice melting. b) a bucket of water freezing c) the same either way. d) neither - are you nuts?

20 Phase Change: Evaporation Takes place at the surface of a liquid due to escaping molecules. Occurs at all temperatures Evaporation occurs when water vapor pressure in the liquid exceeds the pressure of water vapor in the surrounding air. Evaporation is a cooling process.

21 Evaporation is a Cooling Process

22 Phase Change: Boiling Boiling is evaporation under the surface of the liquid. Liquid boils at the temperature for which its vapor pressure exceeds the external pressure (mostly atmospheric pressure.) Boiling point depends on temperature AND 1 atm, bp of water is 5atm, bp of water is 374 ºC Boiling is a cooling process. At low pressures, liquids are boiled ( freeze-dried ) into solids.

23 Increasing the Pressure Does increasing the external pressure increase or decrease the boiling temperature of water? Increases! Boiling happens when vapor pressure in the liquid exceeds the external vapor pressure - now greater due to the increased pressure so the boiling temperature increases!

24 Phase Change: Condensation Gas molecules condense to form a liquid. Condensation is a warming process Why is a rainy day warmer than a cloudy or clear day in winter? Why do we feel uncomfortable on a muggy day?

25 Condensation is a Warming Process

26 Phase Change: Humidity Vapor is the gas phase of a substance below its boiling temperature. Air can hold only so much water vapor before it becomes saturated and condensation occurs. Humidity is a measure of vapor density. Warm air can hold more water vapor. More condensation occurs at cooler temperatures because the molecules are moving slower. Slow moving water molecules coalesce upon collision.

27 Windward: Wet Leeward: Dry Cools and condenses at Top Warm Humid Air Pushed Up Warm Dry Air Falls Down

28 Stormy Weather When warm air rises, it expands and cools. The water vapor in the air soon condenses into water droplets, which form clouds and eventually these droplets fall from the sky as rain.

29 Phase Change:Sublimation The conversion of a solid directly to a gas & visa versa Examples: snowflakes, Moth Balls, dry ice

30 Phase Change: Triple Point A temperature and pressure at which all three phases exist in equilibrium. Lines of equilibrium Freezing-Melting Evaporation -Condensation Sublimation

31 Phase Change Q = ml Phase change occurs at a Constant Temperature! Latent Heats of: Fusion & Evaporation L f, L v Water: L L f v = 334 kj / kg solid-liquid = ( ) 2256 kj / kg liquid-gas ( )

32 Phase Change: Water Q = ml How much 100 C does it take to melt 1kg of ice at -30 C? How much energy is needed to raise the ices to 0 C How much energy is needed to melt 1kg of ice? How much energy is given up by the steam? What happens to the steam that is melting the ice? L L f v = 334 kj / kg = 2256 kj / kg = 0 cice 2090 J / kg C = 0 cwater 4186 J / kg C

33 Phase Change: Water Q = ml How much 100 C does it take to melt 1kg of ice at -30 C? How much energy is needed to raise the ices to 0 C L L f v = 62700J = 334 kj / kg = 2256 kj / kg = 0 cice 2090 J / kg C Q kg J kg C C = 0 cwater 4186 J / kg C = 1 (2090 / )(30 )

34 Phase Change: Water Q = ml How much 100 C does it take to melt 1kg of ice at -30 C? How much energy is needed to melt 1kg of ice? Q2 = ml= 1 kg(334 kj / kg) Q2 = 334kJ L L f v = 334 kj / kg = 2256 kj / kg = 0 cice 2090 J / kg C = 0 cwater 4186 J / kg C Q1 = 62700J Q2 = 334kJ

35 Phase Change: Water Q = ml How much 100 C does it take to melt 1kg of ice at -30 C? How much energy is given up by the steam? What happens to the steam that is melting the ice? m= Qtotal / L m= (397 kj) /(2256 kj / kg) =.18kg L L f v = 334 kj / kg = 2256 kj / kg = 0 cice 2090 J / kg C = 0 cwater 4186 J / kg C Q1 = 62700J Q2 = 334kJ Qtotal = 397kJ

36 Heat flows from HOT to COLD Conduction (solids) Convection (liquids & gases) Radiation (solids, gases, plasma)

37

38 Energy transferred via molecular collisions

39 Heat energy is transferred in solids by collisions between free electrons and Good Conductors: Most Metals (free electrons!) Bad Conductors: Organic & Inert Materials Good Insulators: Air, Water, Wood Good Conductors are BAD Insulators & Visa Versa vibrating atoms.

40 The heat Q conducted during a time t through a material with a thermal conductivity k. dt/dx is the Temperature Gradient. P dt = ka dx

41 Some Thermal Conductivities

42 Temperature Gradient The quantity dt / dx is called the temperature gradient Q = = Δt ka dt dx dt T T = h dx L c

43 Compound Slab: R values For a compound slab containing several materials of various thicknesses (L 1, L 2, ) and various thermal conductivities (k 1, k 2, ) the rate of energy transfer depends on the materials and the temperatures at the outer edges: A T = i ( h Tc) ( L k ) i i Substances are rated by their R values R = L / k and the rate becomes = ( T ) A T i h For multiple layers, the total R value is the sum of the R values of each layer Wind increases the energy loss by conduction in a home R i c

44 Conduction Problem T L T h c = ka A bar of gold is in thermal contact with a bar of silver of the same length and area as shown. One end of the compound bar is maintained at 80.0 C while the opposite end is at 30.0 C. When the energy transfer reaches steady state, what is the temperature at the junction? Ignore thermal expansion of the metals.

45 In the same room, at the same temperature, the tile floor feels cooler than wood floor. How can they be the same temperature?

46 HW13.45 A fire walker runs across a bed of hot coals without sustaining burns. Calculate the energy transferred by conduction into the sole of one foot of a a fire walker given that the bottom of the foot is a 3.00mm thick callous with a conductivity of 0.08 J/smC and its density is 300kg/m 3. The area of contact is 25.0 cm 2, the temperature of the coals is 700C, and the time in contact is 1.00s. Assume his foot has an initial temperature of 37C (98.6F) E dt E P= ka = dx t E = kaδtt L kaδtt J smc x m C C s = = L.003m 4 2 (0.08 / )( )( )(1 ) E = 44.2J What temperature increase is produced on his foot?

47 dt E P= ka = dx t A fire walker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a a fire walker given that the bottom of the foot is a 3.00mm thick callous with a conductivity of 0.08 J/smC and its density is 300kg/m3. The area of contact is 25.0 cm 2, the temperature of the coals is 700C, and the time in contact is 1.00s. Assume his foot has an initial temperature of 37C (98.6F) E = 44.2J kaδtt L What temperature increase is produced on his foot? Q = Q = mcδt = ρvcδt Q o => Δ T = = 1.68 C ρvc (~35F)

48 Electromagnetic Radiation is emitted and absorbed via atomic excitations. All objects absorb and emit EM waves.

49 Electromagnetic Radiation is emitted and absorbed via atomic excitations. All objects absorb and emit EM waves.

50 Frequency ~ Temperature When an object it heated it will glow first in the infrared, then the visible. Most solid materials break down before they emit UV and higher frequency EM waves. Long Short

51 Radiation Radiation does not require physical contact All objects radiate energy continuously in the form of electromagnetic waves due to thermal vibrations of their molecules Rate of radiation is given by Stefan s law

52 Stefan s Law P = σaet 4 P is the rate of energy transfer, in Watts σ = x 10-8 W/m 2. K 4 A is the surface area of the object e is a constant called the emissivity e varies from 0 to 1 The emissivity is also equal to the absorptivity T is the temperature in Kelvins

53 A good absorber reflects little and appears Black A good absorber is also a good emitter.

54 Ideal Absorbers An ideal absorber is defined as an object that absorbs all of the energy incident on it e = 1 This type of object is called a black body An ideal absorber is also an ideal radiator of energy

55 Ideal Reflector An ideal reflector absorbs none of the energy incident on it: e = 0

56 P = 4 eσ T A Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m 2 of 1200C fresh lava into 30.0C surroundings, assuming lava s emissivity is 1. The net heat transfer by radiation is: P = eσ A( T T ) = P = eσ A( T T ) (5.67x10 J / smk )1 m (( K) ( K) ) P = 266kW

57

58 The heating effect of a medium such as glass or the Earth s atmosphere that is transparent to short wavelengths but opaque to longer wavelengths: Short get in, longer are trapped!

59 CO 2 : Greenhouse Gas Unless we change our direction, we are likely to end up where we are headed.

60

61 Hot Air rises, expands and cools, and then sinks back down causing convection currents that transport heat energy. Hot air rises because fast moving molecules tend to migrate toward regions of least obstruction - UP - into regions of lesser density! Rising air cools because a decrease in density reduces number of collisions & speeds decrease. As the air cools, it becomes denser, sinking down, producing a convection current.

62 Uneven heating on the earth and over water cause convection currents in the atmosphere, resulting in WINDS. Global wind patterns (Trade Winds, Jet Streams) are due to convection current from warmer regions (equator) to cooler regions (poles) plus rotation of Earth. Convection Currents in the Ocean (Gulf Stream) transport energy throughout the oceans. Air & Ocean Convection causes the WEATHER.

63 How do fur coats keep you warm? Fur is filled with air. Convection currents are slow because the convection loops are so small.

64 Convection between water and land causes the Winds.

65 Sea Breeze

66 High Pressure Dry Warm Weather Low Pressure Stormy Weather

67

68

69 Happy Earth Day! The organic and inorganic components of Planet Earth have evolved together as a single living, self-regulating system Life maintains conditions suitable for its own survival. - James Lovelock

70 It is much too late for sustainable development; what we need is a sustainable retreat. -James Lovelock, The Revenge of Gaia

71 Our Spaceship Earth One island in one ocean...from space...we re all astronauts aboard a little spaceship called Earth - Bucky Fuller

72 67,000 miles/hr 500,000 miles/hr "We are on a spaceship; a beautiful one. It took billions of years to develop. We're not going to get another. - Bucky Fuller, Operating Manual for Spaceship Earth

73 Space Ecology

74

75 Art by Ray Troll

76 Work in Thermodynamics Work can be done on a deformable system, such as a gas Consider a cylinder with a moveable piston A force is applied to slowly compress the gas The compression is slow enough for all the system to remain essentially in thermal equilibrium This is said to occur quasi-statically dw = F dr = Fˆj dyˆ j= Fdy = PA dy = PdV dw = PdV

77 Work Interpreting dw = - P dv If the gas is compressed, dv is negative and the work done on the gas is positive If the gas expands, dv is positive and the work done on the gas is negative If the volume remains constant, the work done is zero The total work done is: W = V V f i PdV

78 PV Diagrams Used when the pressure and volume are known at each step of the process W = V V f i PdV The state of the gas at each step can be plotted on a graph called a PV diagram This allows us to visualize the process through which the gas is progressing The curve is called the path

79 PV Diagrams The work done on a gas in a quasi-static process that takes the gas from an initial state to a final state is the negative of the area under the curve on the PV diagram, evaluated between the initial and final states This is true whether or not the pressure stays constant The work done does depend on the path taken

80 The First Law of Thermodynamics Δ E = Q+ W int The First Law of Thermodynamics is a special case of the Law of Conservation of Energy It takes into account changes in internal energy and energy transfers by heat and work Although Q and W each are dependent on the path, Q + W is independent of the path

81 Work Done By Various Paths W = V V f i PdV Not necessarily an isotherm! W = P( V V) i f i W = P ( V V) f f i W = P( V) dv The work done depends on the path taken!

82 HW Problem 21 W = V V f i PdV

83 Δ Eint = Q+ W Adiabatic Process Δ E = W int W = V V f i PdV An adiabatic process is one during which no energy enters or leaves the system by heat: Q = 0 This is achieved by: Thermally insulating the walls of the system Having the process proceed so quickly that no heat can be exchanged Since Q = 0, ΔE int = W If the gas is compressed adiabatically, W is positive so ΔE int is positive and the temperature of the gas increases If the gas expands adiabatically, the temperature of the gas decreases Examples of adiabatic processes related to engineering are: The expansion of hot gases in an internal combustion engine The liquefaction of gases in a cooling system The compression stroke in a diesel engine Adiabatic free expansion of a gas The gas expands into a vacuum, no piston: W = 0 Since Q = 0 and W = 0, ΔE int = 0 : initial and final states are the same, no change in temperature is expected.

84 Special Case: Adiabatic Free Expansion This is an example of adiabatic free expansion The process is adiabatic because it takes place in an insulated container Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0 Since Q = 0 and W = 0, ΔE int = 0 and the initial and final states are the same and no change in temperature is expected. No change in temperature is expected

85 Isothermal Process At right is a PV diagram of an isothermal expansion The curve is a hyperbola The curve is called an isotherm The curve of the PV diagram indicates PV = constant The equation of a hyperbola Because it is an ideal gas and the process is quasi-static, PV = nrt and nrt W = P dv = dv = nrt V V V V f f f V V V i i i W V i = nrtln V f dv V

86 Δ Eint = Q+ W PV = nrt W = V V f i PdV Isobaric Processes An isobaric process is one that occurs at a constant pressure The values of the heat and the work are generally both nonzero The work done is W = P (V f V i ) where P is the constant pressure

87 Δ Eint = Q+ W PV = nrt W = V V f i PdV Isovolumetric Processes An isovolumetric process is one in which there is no change in the volume Since the volume does not change, W = 0 From the first law, ΔE int = Q If energy is added by heat to a system kept at constant volume, all of the transferred energy remains in the system as an increase in its internal energy

88 Δ Eint = Q+ W PV = nrt W = V V f i PdV Isothermal Process An isothermal process is one that occurs at a constant temperature Since there is no change in temperature, ΔE int = 0 Therefore, Q = - W Any energy that enters the system by heat must leave the system by work

89 Thermo Processes Δ Eint = Q+ W Adiabatic No heat exchanged Q = 0 and ΔE int = W Isobaric Constant pressure W = P (V f V i ) and ΔE int = Q + W Isovolumetric Constant Volume W = 0 and ΔE int = Q Isothermal Constant temperature ΔE int = 0 and Q = -W W V i = nrtln V f

90 Was ist das? Fig. 20-9, p. 569

91 HW Problem 22 Fig. P20-22, p. 582

92 Cyclic Processes Δ Eint = Q+ W A cyclic process is one that starts and ends in the same state On a PV diagram, a cyclic process appears as a closed curve If ΔE int = 0, Q = -W In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram

93 Δ Eint = Q+ W W = V V f i PdV A gas is taken through the cyclic process as shown. (a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed that is, the process follows the path ACBA what is the net energy input per cycle by heat?

94 Δ Eint = Q+ W PV = nrt W = V V f i PdV HW Problem 28 A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kj of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kj of energy leaving the system by heat. Determine the difference in internal energy E(B) E(A).

95 HW Problem 34 Fig. P20-34, p. 582

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