Spark 09/22/2011

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Spark 09/22/2011 On a very warm day at the beach, which heats up faster, the water or the sand? Once night rolls in, which cools faster? If some materials heat up faster than others, although exposed to the same energy, then where does the energy go if something is not getting hotter? Why might some materials heat up faster than others?

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Spark 09/22/2011. On a very warm day at the beach, which heats up faster, the water or the sand? Once night rolls in, which cools faster? If some materials heat up faster than others, although exposed to the same energy, then where does the energy go if something is not getting hotter? - PowerPoint PPT Presentation

Transcript of Spark 09/22/2011

Spark 09/22/2011On a very warm day at the beach,

which heats up faster, the water or the sand?

Once night rolls in, which cools faster?If some materials heat up faster than

others, although exposed to the same energy, then where does the energy go if something is not getting hotter? Why might some materials heat up faster

than others?

Reading QuestionsRead the Matter document up to

“Temperatures in Kelvin”What is the KMT?What is the difference between heat and

temperature?Why is the Kelvin temperature scale

used?How does the K scale convert to Celcius?

http://mutuslab.cs.uwindsor.ca/schurko/animations/waterphases/status_water.htm

Enthalpy and Heat

The study of energy and its transformations is known as thermodynamics. Energy is commonly defined as the capacity to do work or to transfer heat. Work is energy used to cause an object with mass to move. Heat is energy used to cause the temperature of an object to increase.

Enthalpy and Heat

Kinetic energy is the energy of motion, represented by the equation:

Ek = 1/2 mv2

Atoms and molecules have mass and are in motion, therefore they have kinetic energy.

Enthalpy and Heat

The SI unit for energy is the joule:   J = 1 kg-m2/s2

A 2 kg mass moving at 1 m/s possesses a kinetic energy of 1 J:

Ek = 1/2 mv2   = 1/2 (2 kg) (1 m/s)2  

= 1 kg-m2/s2   = 1 J Since a joule is a small amount of energy,

the unit kilojoules, kJ, is often used when describing chemical reactions.

Enthalpy and Heat

Traditionally, energy changes in chemical reactions have been expressed in calories, a non-SI unit.

A calorie was originally defined as the amount of heat energy required to raise the temperature of 1 g of water from 14.5 oC to 15.5 oC. It is now defined in terms of the joule:

1 cal = 4.184 J (exactly)

Enthalpy and Heat

A related energy unit used in nutrition is the nutritional Calorie (note the capital letter).

1 Cal = 1000 cal = 1 kcal

E N T H A L P Y - HE N T H A L P Y - HE N T H A L P Y - HE N T H A L P Y - H

heat content in matter:

E N T H A L P Y - HE N T H A L P Y - HE N T H A L P Y - HE N T H A L P Y - H

heat content in matter:

natural systems tend to go from a state of high energy to a state of low energy.

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

The phases of matterrepresent "classes" of molecular motion found at different temperatures.

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

The phases of matterrepresent "classes" of molecular motion found at different temperatures.

XKinetic Energ

y

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

The phases of matterrepresent "classes" of molecular motion found at different temperatures.

XHEAT

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E Thermal E Internal E Chem

Hot Coffee

Thermal E = 4

Internal E = 2

Solids = 1

Liquids = 2

Gases = 3

Chemical = 0

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Coffee Before Coffee After

Total Energy = 7 units

Total Energy = 5 units

Missing 2 units of E: Where did they go?

How can we diagram energy flow when a cup of coffee cools?

Energy Flow Diagram Is energy flowing in or

out of the system? 2 units of energy

leave causing the coffee to cool.

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Cup of Hot Coffee

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7 units - 2 units = 5 units

Spark

Does hot water or cold water freeze faster?Explain using the KMT.

Are you aware of the Mpemba Effect?What are the four conditions that make the Mpemba Effect possible?(p2-3)

Do you have an explanation for the Mpemba Effect?

HMMMM…

Can water boil if it’s cold?Explain using the KMT.

http://youtu.be/8oCjj8iDB9I

http://youtu.be/QGSo7zcg_qA

http://youtu.be/OHY9fFQhX68

http://youtu.be/RPlCO3AITV4

http://youtu.be/9q5gEZGoBnk

Heat Matter does not contain heat, it contains energy

(kinetic, thermal, internal energy, chemical energy).

Heat is energy that is transferred from hot to cold Heat is measured in calories or Joules 1 calorie = 4.185 J Measure average kinetic energy with a thermometer

and call it temperature How can we measure heat if things don’t heat and cool

the same way?

Specific Heat Capacity Think of it as heating ability… The amount of heat energy that is required

to raise the temperature of one gram of a substance by one degree Celsius is called the specific heat capacity, or simply the specific heat, of that substance.

Water, for instance, has a specific heat of 1.0 calorie per gram degree Celsius [1.0 cal / (g x o C)].

Water becomes our standard for studying energy.

Specific Heat Capacity Water

So:It takes 1 calorie of heat to raise 1 gram of

water 1 degree

How many calories are released if 1 gram of water cools 1 degree?

How many calories are needed to heat 1 gram of water at OoC to boiling (100oC)?

1 calorie

100 cal

Lab Burners and BTUs

BTU Inquiry Lab Tips

Rate of BTU production be lab burner.

1 BTU = Nrg to increase 1 lb of water @39.1oF, 1oF

Convert measurable metric masses and temperatures into BTU’s and predict the time to get a hypothetical rate (BTU/sec)

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

q = U + P Vq = heat gained or lostU = internal energyP = PressureV = volumeSee Energy notes

Spark 09/27/2011

How much energy is needed to melt a 10 lb bag of party ice at 0o C? (heat of fusion = 334 J/g or 80 cal/g) There are 2.2 lbs/1 kgHfus - heat of fusion

The amount of heat needed to change one

gram of a solid into a liquid at its melting point.

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

Hfus - heat of fusion

As energy is added to a solid at its melting point, ALL the energy is used to increase the kinetic energy of the molecules during the phase change.

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

Hfus - heat of fusion

The amount of heat needed to change one gram of a solid into a liquid at its melting point.

Every phase change isaccompanied by a changein the energy of the system.

As the temperature of a solidincreases, its particle vibration increases.

When a solid melts, itsparticles are freed to move with respect to one another.

This melting process is called fusion.

The increased freedom ofmotion of the molecules orions comes at a price...

measured by the heat of fusion, or enthalpy of fusion, Hfus

solid/liquid phase change

q = m Hfus

q = heat gained or lostm = massHfus = heat of fusion

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

Hfus H2O = 334 J/g

Hfus lead = 24.72 J/g

As the temperature of a liquidincreases, the molecules movewith increasing energy.

When increasing temperature,the concentration of gas-phasemolecules just above thesurface of the liquid increases.

These gas-phase moleculesexert a pressure on theliquid called vapor pressure.

This vapor pressure increases with temperature until it equals the atmospheric pressure.

This is the boilingpoint of the liquid.

The energy required tocause this transition..

is called the heat of vaporization or enthalpy of vaporization, Hvap

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

As energy is added to a liquid at its boiling point, ALL the energy is used to increase the kinetic energy of the molecules during the phase change.

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

Hvap - heat of vaporization

The amount of heat needed to change one gram of a liquid into a gas at its boiling point.

liquid/gas phase change

q = m Hvap

q = heat gained or lostm = massHvap = heat of vaporization

E N T H A L P YE N T H A L P YE N T H A L P YE N T H A L P Y

Hvap H2O = 2260 J/g

is used to measure

heat change

Heat of Fusion of Ice

Spark q = m x ΔT x s

50.0 grams of hot water (65.0oC) are added to a calorimeter that has 25.0 grams of water at 5.0oC. After mixing the final temperature reached was 44.0oC.1 cal = 4.185 J = E to raise 1 g H2O 1oC

How much energy was released from the hot water?How much was absorbed by the cold water?Is energy being conserved? Does the nrg released = the nrg absobed?

Calorimeter Constant

Heat of Solutions and Reactions

Spark

With your group, calculate your mean calorimeter constant.

Send a representative to me with your value for spreadsheet credit.

Be sure that all group members know what the calorimeter constant reflects!

What is

How “well” a substance heats up…How much energy is needed to heat something up…

Specific Heat Specific Heat CapacityCapacity

Specific Heat Specific Heat CapacityCapacity

The heat required to raise

the temperature of onegram of a substance byone Celsius degree.

change temp within change temp within phasephase

q = m Cp

q = heat gained or lostm = mass= “change in temperature”Cp = specific heat capacity

Specific Heat CapacityThe amount of heat energy involved in

changing the temperature of a sample of a particular substance depends on three parameters -- the specific heat of the substance, the mass of the sample, and the magnitude of the temperature change.

The Greek letter delta (Δ) is used to indicate a change. Δ T = temperature final - temperature

initial

The General Idea Behind Specific Heat Labs

Both ice or a heated sample will be placed into a crude calorimeter, consisting of water contained in a plastic foam cup. Shortly after mixing, the sample will have come to the same temperature as the water or the water temperature will increase or decrease depending on the change. Because Styrofoam is a good insulator, heat cannot easily escape from the calorimeter to the surroundings.

Specific Heat LabsThe heat lost or gained by the sample should

be equal to the heat gained or lost by the water.

The amount of heat energy gained by the water will be calculated in the following manner:

heat gained water = specific heat water X mass water X ΔTwater

The heat lost by the sample is given by a similar equation:

heat lostsample = specific heat sample X mass sample X ΔTsample

Because the heat gained must equal the heat lost, a third equation can be written:

specific heat water X mass water X ΔT water = specific heat sample X mass sample X ΔT sample

The specific heat of water is known. The temperature changes of the

water, and of the sample, can be measured, as can the mass of the water and the mass of the sample.

Using this data, the specific heat of the sample can be calculated.

Calorimetry values are all determined based on the temperature changes of the water sample.

Which has the greater?

Cu H2Oor

Specific Heat Specific Heat CapacityCapacity

Specific Heat Specific Heat CapacityCapacity

Cp of ice = 2.06 J/g.Co

Cp of water = 4.18 J/g.Co

Cp of steam = 2.02 J/g.Co

Cp of copper = .385 J/g.Co

q=mHvap change phase

q=mHFUS- change phase

q=mCp - change temp

q=mCp - change temp

q=mCp - change temp

How much heat is needed toconvert 250 grams of ice at -30 oC to vapor at 150 oC?

1. Change temp of ice2. Melt ice3. Change temp of liquid water4. Vaporize water5. Change temp of vapor6. Add total heat used

Step 1 raise temp of ice

q = m T Cp

q = (250g) (30 Co) (2.06 J/g. Co)

q = 15,450 Joules

Step 2 melt ice

q = m Hfus

q = (250g) (334 J/g)

q = 83,500 Joules

Step 3 raise temp of liquid water

q = m T Cp

q = (250g) (100 Co) (4.18 J/g. Co)

q = 104,500 Joules

Step 4 vaporize liquid water

q = m Hvap

q = (250g) (2260 J/g)

q = 565,000 Joules

Step 5 raise temp of water vapor

q = m T Cp

q = (250g) (50 Co) (2.02 J/g. Co)

q = 25,250 Joules

Step 6

q = 15,450 83,500 104,500 565,000 25,250

q = 794,000 Joules

Specific Heat of a Metal

Heat of Combustion of Wax

Spark

Get your materials for the specific heat lab.

3 groups may use burners otherwise use a hot plate.

Use 2 foam cups instead of oneCalorimeter water should be cold.I will give you your metals once you have everything else.

P h a s e D i a g r a mP h a s e D i a g r a m

A g r a p h s h o w i n g A g r a p h s h o w i n g t h e p h a s e s o f t h e p h a s e s o f m a t t e r a s t h e m a t t e r a s t h e y r e l a t e t o y r e l a t e t o t e m p e r a t e m p e r a t u r e a n d t u r e a n d p r e s s u r e . p r e s s u r e .

M a j o r p o i n t s o n a p h a s e d i a g r a m

M a j o r p o i n t s o n a p h a s e d i a g r a m

Normal Atmospheric PressureNormal Atmospheric Pressure

Triple pointTriple point

Boiling PointBoiling Point

Melting pointMelting point

Critical temperatureCritical temperature

101.325 kPanormal atmospheric pressure

TP Triple Point

Tb

Normal boiling point

Tm

Normalmelting point

Tc

Criticaltemperature

V a n d e r W a l l s A t t r a c t i o n

V a n d e r W a l l s A t t r a c t i o n

T h e f o r c e o f T h e f o r c e o f a t t r a c t i a t t r a c t i o n o n b e t w e e n b e t w e e n

m o l e c u l e sm o l e c u l e s

Calories in a Peanut