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8/3/2019 Unit-11--POP-Heat Temp Heat Transfer Thermal Revised
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Heat, Temperature,
Heat Transfer, ThermalExpansion &Thermodynamics
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Heat vs. TemperatureHeat A form of energy
Measured incalories or Joules
There is nocoldness energy
Any object withtemperature
above zero Kelvinhas heat energy
Temperature Avg. Kinetic Energy of
the particles
Measured in C, F, K
hot & cold arerelative terms
Absolute zero is zeroKelvin
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Heat Transfer
1.Conduction - requires direct contact orparticle to particle transfer of energy;
usually occurs in solids
2.Convection - heat moves in currents; hotair rises and cold air falls; only occurs in
fluids
3.Radiation - heat waves travel through
empty space, no matter needed; sun
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CONDUCTIONCONVECTION
RADIATION
METHODS OF HEAT TRANSFER
www.mech.northwestern.edu/. ../ME377/ME377.htm
http://www.mech.northwestern.edu/dept/courses/info/ME377/ME377.htmhttp://www.mech.northwestern.edu/dept/courses/info/ME377/ME377.htm8/3/2019 Unit-11--POP-Heat Temp Heat Transfer Thermal Revised
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Thermal EquilibriumA system is in thermalequilibrium when all of its partsare at the same temperature.
Heat transfers only from high tolow temperatures and only untilthermal equilibrium is reached.
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Temperature Scales There are 3 temperature scales Celsius
(Centigrade), Kelvin, Fahrenheit 1. Celsius, C metric temp. scale
2. Kelvin, K
metric absolute zero temp.scale 3. Fahrenheit, F customary (English)
temp. scale
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Handout s:1.NEW YELLOW FORMULA CHART2. Temp. Scale/Questions on back
3.Phase Change Graph/Energyconversions on back
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Absolute Zero:theoreticaltemp. at whichall particlemotion stops!!
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Converting C to F and F to CYELLOW FORMULA CHART
C F C x 1.8 + 32Celsius x 1.8 + 32 = Fahrenheit
EX: 20 degrees CelsiusConversion: 20 C x 1.8 + 32 = 68 F
F C F -32 / 1.8Fahrenhiet
32 / 1.8 = Celsius
EX: 95 degrees Fahrenheit
Conversion: 95 F 32 / 1.8 = 35 C
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Converting K to C and C to KYELLOW FORMULA CHART
C K K = C + 273EX: 20 C
K = 20 + 273 = 293 K
K C C = K 273EX: 5 K
C = 5
273 = -268 C
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Change of State
Increasing Heat Energy (Joules)
-20
100
0
melting
vaporization
condensation
freezing
*Asheat is added to a substance it will either be absorbed to
raise the temperature ORto change the state of matter.
*It can NEVERdo both at the same time!
*Temperature will NOTchange during a phase change!
Heat of
vaporization
Heat of
fusion
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Specific Heat
The amount of heat energy needed to raisethe temperature of 1 gram (or kg) of a
substance by 1C (or 1 K).
FYI
Substances with higher specific heats, suchas water, change temperature more slowly.
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Specific Heat ( add this formulaand the spec. heat data (nextslide) to your new Yellow formulachart )Formula Q = Cm TUnitsQ = JC = J/kg*C or Km = kgT = C or KReminder () = ( Tf Ti )
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Substance J/kg*K Substance J/kg*K
Water (l) 4186 Copper 385
Steam 1870 Gold 129Ammonia (g) 2060 Iron 449
Ethanol (l) 2440 Mercury 140
Aluminum 897 Lead 129
Carbon 709 Silver 234
SPECIFIC HEATSAT 25C
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Solve the Problem
A 0.59 kg brass candlestick has aninitial temperature of 98 C. If 2.11 E4Joules of energy is removed for thecandlestick to lower its temperature to6.8 C, what is the specific heat of thebrass?Think! Order of operation:1. What variable are you solving for? ____2. Write out the original formula and thegivens.3. Is there a need to re-write the formula?4. Solve and put the proper unit w/ answer
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Solving the Problem1. Solve for variable C = specific heat.2. Formula Q = C m TGivens : m = .59 kgQ = 2.11 E4 JT = ( 6.8C 98C )
C = J / Kg * C3. Re-write formula to solve for CC = Q / m T4. SolveC = 2.11 E4 J / .59 kg * - 91.2CC = 2.11 E4 J / - 53.8 kg * CC = -392.2 J / kg * C
* Why is the specific heat a negative #?
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Latent Heat
(Latent) Heat of fusionthe heat energyneeded to melt (solidliquid) or freeze
(liquid solid) one gram (or kg) of a
substance.EX:For water: Hf=334,000 J/kg or 80 cal/g
Hf
= heat of fusion, J/k
( solidliquid ) OR ( liquidsolid )
Melt Freeze
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Latent Heat cont.
(Latent) Heat of vaporizationthe heat energyneeded to vaporize (liquidgas) or condense
(gasliquid) one gram (or kg) of a substance.
EX:For water: Hv = 2.26 x 105 J/kg or 540 cal/g
Hv = heat of vaporization, J/kg
( liquidgas ) OR ( gasliquid )Vaporize Condense
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Q = mHf
Q = mHv
Q = heat absorbed or released, J
m = mass of substance changing phase, kg
Hf = Heat of Fusion, Given in J/kg
Hv = Heat of Vaporization, Given in J/kg
Latent Heat Phase Changesadd these formulas and
info to yourYELLOW formula chart
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Thermodynamics
The study of changes in thermalproperties of matter
Follows Law of Conservation ofEnergy
1st Law of Thermodynamicsthe total increase in the thermal
energy of a system is the sum ofthe work done on it and the heatadded to it (also calledLaw of
conservation of energy)
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2nd Law of Thermodynamics
All natural processes go in a directionthat increases the total entropy of theuniverse.
Entropy - a measure of the disorder of asystem.
If heat is added, entropy is increased.
If heat is removed, entropy is decreased.
Work with NO T, entropy is unchanged.
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PRACTICE PROBLEM WORKSHT.
Complete the practice problem worksheet
for Temperature conversions
Specific Heat
Heat of Fusion/Vaporization
Due Date: Mon. March 28th