2 all about heat (students_ copy)

8/10/2019 2 All About Heat (Students_ Copy)

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HEAT

THERMODYNAMICSBy: Engr. Lyza Marie G. Luchico


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THERMODYMANICS

Thermodynamics is thestudy of the motion of heatenergy as it is transferredfrom the system to thesurrounding or from thesurrounding to the system.

The transfer of heat could be due toa physical change or a chemical

change.

There are three laws of chemicalthermodynamics.


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CHEMICAL THERMODYMANICS

The first law of thermodynamics:

Energy and matter can be neither created nor destroyed;only transformed from one form to another. The energy

and matter of the universe is constant.

The second law of thermodynamics:

In any spontaneous process there is always an increase inthe entropy of the universe. The entropy is increasing.

The third law of thermodynamics:The entropy of a perfect crystal at 0 K is zero.

There is no molecular motion at absolute 0 K.


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HEAT (AFORMOFENERGY)The energy that flows into or out of a system

because of a difference in temperaturebetween the thermodynamic system and its

surrounding.

Symbolized by Q".When heat is evolved by a system, energy

is lost and Q is negative (-).

When heat is absorbed by the system, theenergy is added and Q" is positive (+).


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MOLECULESANDMOTION

The motion of moleculesproduces heat

The more motion, the moreheat is generated


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INTERNALENERGY

It is the grand total of all energies

inside a substance.

1. Molecular KE

Random motion of the

molecules in a substance

2. Rotational KE

3. KE due to internal movements of atoms

within molecules4. PE due to forces between

molecules


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There will be a transfer of

energyfrom one substance

to another.

Transferof energy

Heat Flow

Heat Transfer


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Does it mean that the flow of energyis from a

substance with more internal energyto a

substance with less internal energy?

Which has more internal energy?

A bowl of warm water

or

A red-hot thumb tack

Heat Flow:

Tto

T


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HOWMUCHENERGYREQUIREDTOHEAT

OBJECT?

Heat (energy) flows because of

temperature difference

Bigger temperature difference bigger

heat flow

Less insulation give more heat flow

for the same temperature difference

Heat will not flow between two bodies

of the same temperature


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HEAT FLOW

Heat can flow in one of two directions:

ExothermicTo give off heat; energy is lost from the system: (-

Q)

Endothermic

To absorb heat; energy is added to the system:(+Q)


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IFTHEHEATTRANSFERINVOLVESACHEMICALREACTIONTHENQISCALLED:

HEAT OF REACTION

The heat energy (

H; enthalpy) required toreturn a system to the given temperature at thecompletion of the reaction.

Q = H at constant pressure

The heat of reaction can be specific to a reaction like:

HEAT OF COMBUSTIONThe quantity of heat energy given off when aspecified amount of substance burns inoxygen.

UNITS: kJ/mol (kilojoules per mole) or kcal/mol (kilocalories per mole)


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HEAT CAPACITY & SPECIFIC HEAT

HEAT CAPACITY: The quantity of heat neededto raise the temperature of a substance one

degree Celsius (or one Kelvin).

Q = Cp T

SPECIFIC HEAT: The quantity of heat required

to raise the temperature of one gram of a

substance by one degree Celsius (or one

Kelvin).

Q = C m T

Both Cp & c are chemical specific constants

found in the textbook or Handbooks.


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SOMETHINGSAREEASIERTOHEAT(SPECIFIC

HEATCAPACITY)

More water in the kettle needs longer time to boil

Alcohol needs less energy to heat it than water

Energy required (Q) proportional desired change intemperature (DT) x mass (m) of material Q = mc DT

c called the specific heat cwater= 4190 J/(kg K) - very difficult to heat

cice= 2000 J/(kg K)

cmercury= 138 J/(kg K) - very easy to heat cethanol= 2428 J/(kg K) - very easy to heat


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UNITS FORHEAT ENERGY

Heat energy is usually measured in either Joules,given by the unit (J), and kilojoules (kJ) or in

calories, written shorthand as (cal), and kilocalories

(kcal).

1 cal = 4.186 J

NOTE: This conversion correlates to thespecific heat of water which is 1 cal/g oC or

4.186 J/g oC.


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SIRJAMESJOULE

James Joule 1818-1889

Stirring water made itwarm Change in temperature

proportional to work done

Showing equivalence ofheat and energy

Also that electricalcurrent flow through aresistor gives heating


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For a weight watcher, a peanut

contains 10 Calories;But to a physicist, a peanut can

release 10,000 calorieswhen

burned or consumed.


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ctivity Calorie burned per pound per

minute

Badminton 0.044Stationary Bike 0.0227

Calisthenics 0.0265

Stair Machine 0.0529

Aerobics, general 0.0492

Yoga 0.0303

Jump Rope 0.074

Swimming, slow 0.058Walking, normal,

asphalt

0.036

Sitting Still 0.009


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The quantity of heat, Q, required to increase

the temperature of a mass, m, of a certainmaterial is proportional to the change in

temperature.


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The quantity of heat, Q, is also proportional

to the mass, m of the material.


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where:

Q is heat, J

m is mass, kgDT is the change in temperature, Co

c is the specific heat capacity of the

substance, J / (kgCo)


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SPECIFIC HEAT PROBLEMS

Determine the energy (inkJ) required to raise the

temperature of 100.0 g ofwater from 20.0 oC to 85.0oC?


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EQUILIBRIUM

+ =Hot black

coffee

Cold

milk

Warm white

coffee

Two objects of different temperature when

placed in contact will reach the same

temperature


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HEATTRANSFER= ENERGYTRANSFER

Energy measured in Joules but heat often

measured in Calories

One cal raises one gram of water from 14.5C to

15.5C 1 cal - 4.186J

Doing work on something usually makes it

hot

Splash in the bath and the water will get warm!

1st law of thermodynamics heat and work

are both forms of energy

LAW OF CONSERVATION OF


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LAW OF CONSERVATION OF

ENERGYThe law of conservation of energy (the

first law of thermodynamics), whenrelated to heat transfer between twoobjects, can be stated as:

The heat lost by the hot object = the heat

gained by the cold object

-Qhot= Qcold

-mh ch Th= mc cc Tcwhere T = Tfinal- Tinitial


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LAW OF CONSERVATION OF ENERGY

Assuming no heat is lost, what

mass of cold water at 0.00oC is

needed to cool 100.0 g of waterat 97.6oC to 12.0oC?


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SPECIFIC HEAT PROBLEMS

LAW OF CONSERVATION OF ENERGY

1. Determine the specific heat of anunknown metal that required 2.56 kcalof heat to raise the temperature of150.00 g of the metal from 15.0oC to

200.0

o

C?2. Calculate the specific heat of anunknown metal if a 92.00 g piece at100.0oC is dropped into 175.0 mL ofwater at 17.8 oC. The final temperatureof the mixture was 39.4oC.


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ORIGINAL

GROUP

1 yellow paper

[accessed on July28, 2014]


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PRACTICE PROBLEMS 21. Iron metal has a specific heat of 0.449 J/goC. How much heat

is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when

it is placed in a beaker of boiling water at 1 atm?2. How many calories of energy are given off to lower the

temperature of 100.0 g of iron from 150.0 oC to 35.0 oC?

3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0oC, what would

be the final temperature of the water?

4. A 100. g sample of water at 25.3 oC was placed in a

calorimeter. 45.0 g of lead shots (at 100 oC) was added to the

calorimeter and the final temperature of the mixture was 34.4oC. What is the specific heat of lead?

5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It

was then added to 28.05 g of water in an insulted cup. The

water temperature rose from 21.04 oC to 23.98oC. What is the

specific heat of the metal in J/goC?


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EXPLANATION

OFANSWERS

Write the solution on the

yellow paper as you explain.


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PRACTICE PROBLEMS 21. Iron metal has a specific heat of 0.449 J/goC. How much heat is transferred

to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker ofboiling water at 1 atm?

2. How many calories of energy are given off to lower the temperature of100.0 g of iron from 150.0 oC to 35.0 oC?

3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0oC, what would be the final

temperature of the water?

4. A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g oflead shots (at 100 oC) was added to the calorimeter and the finaltemperature of the mixture was 34.4 oC. What is the specific heat of lead?

5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It was thenadded to 28.05 g of water in an insulted cup. The water temperature rosefrom 21.04 oC to 23.98oC. What is the specific heat of the metal in J/goC?

0.792 J/goC

180. J

1.28 J/go

C

1.23 x 103cal

31.1 oC


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REFERENCES:

Padgett, Miles. 2013. Thermal Physics. Physics 9.

Available at [email protected]

[accessed on July 28, 2014]

No Author. 2013. 11 Heat. Available at

http://faculty.lacitycollege.edu[accessed on July 28,2014]
mailto:[email protected]://faculty.lacitycollege.edu/http://faculty.lacitycollege.edu/mailto:[email protected]


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PHASE

CHANGEEngr. Lyza Marie G. Luchico


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PHASECHANGES(E.G. SOLIDTOLIQUID)

time

Tempe

rature

ice

water

steam

MP

BP


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ENERGYREQUIREDFORPHASECHANGE

Heat of fusion (Q), solid TOliquid

Q = mLf

(Lf is latent heat of fusion)Lf (water) = 334 x10

3J/kg

Lf (mercury) = 11.8 x103

J/kg


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Heat of vaporization (Q), liquid TO gas

Q = mLv(Lv is latent heat of vaporization)Lv (water) = 2256 x10

3J/kg

Lv (mercury) = 272 x103J/kg

Heat of sublimation (Q), solid TO gas

Q = mLs(Ls is latent heat of sublimation)


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EXAMPLE

A 20-g ice cube at 0 0C is

heated until 20.0 g has

become water at 1000

C and5.0 g has been converted to

steam. How much heat is

added to do this system?


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TRY ME (PROBLEM1)

What mass of steam [c = 2010

J/(kgCo)] initially at 130oC is

needed to warm 200-g of water [c= 4186 J/(kgCo)] in a 100-g glass

container [c = 837 J/(kgCo)] from

20o

C to 50o

C? The latent heat ofvaporization of steam is 2.26 x 106

J/kg


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PAIR SHARE

What are the similarities

between heat and God?

How about differences?


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Heat is being used for

some other purpose thanraising the temperature of

the ice-water mixture.

Heat is being used to melt

the ice by separatingthe

ice molecules from each

other.

There is a change in phasebecause the potential

energyof the ice molecules changes but its

average kinetic energy remains the same.


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Phaseit describes a state of matter ---

solid, liquid, gas or plasma

Phase Changea physical change fromone phase to another.

A phase change always require a transfer

of energy.

Energy-in-Transit Heat


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Fusion/Melting

Solidification/

Freezing

Evaporation

Condensation

Sublimation

Deposition


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The amount of heat that must be added or

removed depends on: Type of material

Nature of the phase change


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The amount of heat that must be supplied or

removed to change the phase of a mass of a

substance is called Latent Heat.

where:Q is heat, J

m is mass, kg

L is latent heat, J/kg

Solid & LiquidLatent Heat of Fusion, Lf

Liquid & GasLatent Heat of Vaporization, Lv

Solid & GasLatent Heat of Sublimation, Ls

+, heat gained

-, heat lost


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Heat of fusion (Q), solid TOliquid

Q = mLf

(Lf is latent heat of fusion)Lf (water) = 334 x10

3J/kg

Lf (mercury) = 11.8 x103

J/kg


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Heat of vaporization (Q), liquid TO gas

Q = mLv(Lv is latent heat of vaporization)Lv (water) = 2256 x10

3J/kg

Lv (mercury) = 272 x103J/kg

Heat of sublimation (Q), solid TO gas

Q = mLs(Ls is latent heat of sublimation)


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A plot of temperature versus energy added when 1.00 g of ice

initially at -30oC is converted to steam at 120oC.

cice= 2090 J/kg Co

csteam= 2010 J/kg Co

Lf= 3.33 x 105J/kg at 0 oC

Lv= 2.26 x 106

J/kg at 100o

C


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PROBLEMNO. 2A 7-kg glass bowl [c = 840 J/(kgCo)] contains 16-kg

punch at 25oC. Two-and-a-half kilograms of ice [c =

2090 J/(kgCo)] are added to the punch. The ice

has an initial temperature of -20oC. The punchmaybe treated as if it were water [c = 4186

J/(kgCo)] and that there is no heat flow from punch

bowl to its surroundings. The latent heat of fusion

for water is 3.33 x 105J/kg. When thermalequilibrium is reached, all ice has melted.

Determine the temperature at thermal equilibrium.


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ASSIGN P N 4


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ASSIGN: PROBLEMNO. 4 CROSSWISE

1. What mass of steam at 100oC must be added to

1.0-kg of ice at 0oC to yield liquid water at 20oC?

2. What mass of steam at 120oC must be added to

1.0-kg of ice at -3.0oC to yield liquid water at 20oC?

Substance Melting

Point(oC)

Lf

(J/kg)

Boiling

Point(oC)

LV

(J/kg)

H20 0 3.33 x 105 100 2.26 x 106

cice= 2090 J/kg Co

csteam= 2010 J/kg CocH20= 4186 J/kg C

o

PRACTICE SET (HEAT)


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PRACTICE SET (HEAT)1. A 250 g of water at 376.3 oC is mixed with 350.0 mL (D of

water is 1.00 g/mL) of water at 5.0 oC. Calculate the final

temperature of the mixture.2. If 40.0 kJ were absorbed by 500.0 g H2O at 10.0

oC, what

would be the final temperature of the water?

3. What mass of liquid water initially at 3oC is needed to

warm a 0.2-kg aluminum bar from -155oC to 0oC?4. A 1-kg solid mercury at its melting point of -39oC is placed in a

0.62-kg aluminum calorimeter with 0.4-kg water at 12.8oC.

Determine the latent heat of fusion of mercury if the resulting

equilibrium temperature of the mixed system is 5.06o

C5. What mass of steam (in g) at 100OC is needed to heat 500

g of water at 10OC in a 250-g aluminum container to a final

equilibrium temperature of 30OC?


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CONSTANTS

cice= 2090 J/kg Co

csteam= 2010 J/kg Co

cH20

= 4186 J/kg Co

cAl= 900 J/kg Co

cHg= 138 J/kg Co

Substance Melting

Point

(oC)

Lf

(J/kg)

Boiling

Point

(oC)

LV

(J/kg)

Al 660 3.97 x 105 2450 1.14 x 107

H20 0 3.33 x 105 100 2.26 x 106


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REFERENCES:

Padgett, Miles. 2013. Thermal Physics. Physics 9.

Available at [email protected]

[accessed on July 28, 2014]

No Author. 2013. 11 Heat. Available at

http://faculty.lacitycollege.edu[accessed on July 28,2014]
mailto:[email protected]://faculty.lacitycollege.edu/http://faculty.lacitycollege.edu/mailto:[email protected]

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