Unit-11--POP-Heat Temp Heat Transfer Thermal Revised

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.htm


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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

Unit-11--POP-Heat Temp Heat Transfer Thermal Revised

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