Unit 14 - Solon City Schools

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H o n o r r s s C h e m i i s s t t r r y

U n i i t t 1 4

M a t t t t e r r & E n e r r g y

 Label a heating / cooling curve o Solid, liquid, gas o Evaporation, condensation, freezing, melting o Enthalpy of fusion, enthalpy of vaporization o Specific heat o Boiling point, melting point

 Separate mixtures based on physical properties. o boiling point (distillation), magnetism, density etc.

 Evaluate energy changes of matter (Specific Heat & Calorimetry) o Calculation of specific heat (q = mCΔT) o Calorimetry of various systems involving exothermic and endothermic heat exchange. o The calculation of energy released from a food substance using calorimetry.

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We are looking for:

1a. Identification of all phase changes and energy change values

1b. Evaporation, condensation, freezing, melting

1c. Enthalpy of fusion, enthalpy of vaporization

1d. Specific heat

1e. Boiling point, melting point

1f. Solid, liquid, gas

2. Physical properties such as boiling point, magnetism, density etc.

3a. Calculations of energy released/gained using specific heat (q = mCΔT)

3b. Calorimetry of various systems involving exothermic and endothermic heat exchange.

3c. The calculation of energy released from a food substance using calorimetry.

Matter

Objectives:

Distinguish between a mixture and a pure substance

Describe the states of matter in terms of particles

Building Blocks of Matter

Atom – smallest unit of an element that maintains the properties of that element. Neutral charge

Element – a pure substance made of only one kind of atom

Compound – a substance that is made of the atoms from two or more elements that are chemically combined

Classification of Matter

Pure substance – composed of one kind of atom or molecule

 has a fixed composition

 has the same physical and chemical properties throughout

Mixture – a blend of two or more kinds of matter

Can be physically separated .

Methods include: centrifugation, distillation, filtration, and magnets . o Homogeneous – uniform in composition throughout

Solution - one or more substances (the solutes) dissolved in another substance (the solvent)

Colloid - Particles suspended in a liquid that do not settle out

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o Heterogeneous – NOT uniform throughout and will show Tyndall Effect

(scattering of light due to particles in a solution)

Suspension -

Particles are suspended in a fluid that will settle out

States of Matter:

1.

Solid – has definite volume and definite shape.

(same shape regardless of its container)

2.

Liquid - has definite volume but an indefinite shape (assumes the shape of its container)

3.

Gas – has neither a definite volume nor definite shape.

(expands to fill any container)

4.

Plasma – high temperature state in which atoms lose their electrons .

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Heating and Cooling Curve Definitions

Specific Heat –

Solid –

Liquid-

Gas –

Plasma –

Heating Curve -

Enthalpy of Fusion/ Molar heat of fusion–

Melting–

Melting Point –

Enthalpy of Vaporization/Molar heat of Vaporization–

Evaporation–

Boiling Point –

Sublimation-

Cooling Curve –

Condensation –

Condensation Point –

Freezing –

Freezing Point –

Deposition -

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The graph below shows the relationship between heat (energy) added, in calories (cal), and temperature for 1 g of water. A student applied heat to 1 g of ice that had been cooled to -40⁰C and measured the rise in temperature.

Read and fill-in the notes below and on the following pages and label the steps/regions A, B, C, D, E on the graph.

Step A:Solid Water (Ice) Rises in Temperature (Keep in mind the graph is for water!)

 If the __________________ is not at 0 o C, it will rise as heat is ____________to get there. (Kinetic energy is _________________)

 Each gram of water requires a constant amount of energy to increase 1 o = specific

heat

 IMPORTANT – the ice has not________________ yet!

Step B: Solid Water (Ice) Melts

 By ______________energy the ice begins to _____________.

 Temperature does not ___________ as more energy is being ______________

(Kinetic energy is _____________________ but potential energy is ____________)

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 Each mole of water requires a given amount of energy to melt = molar heat of fusion

(∆ H fus

) in kJ / mole.

 Energy is overcoming water molecules attraction for each other so it can be converted from a solid to liquid.

 How many calories of energy did it take to completely change the 1 gram of solid water (ice) at 0

C to liquid water?________________________

Step C: Liquid Water Rises in Temperature

 Now the ice is completely _________ and the water temperature begins to

_________________ as heat is ________________. (specific heat)

 Kinetic energy is ______________________.

 The water has not started to____________ yet.

 How many calories of energy did it take to make the 1 gram of liquid water to change temperature from 0

C to 100

C (just beginning to boil)?____________

Step D: Liquid Water Boils

 As we __________ energy the temperature does not change.

 Each mole of water will require a constant amount of energy to boil = molar heat of vaporization (∆H vap

) KJ/mole.

 The energy is being used to overcome water's attraction to each other to convert the liquid to a gas (kinetic energy _________________ but potential energy is

_________________).

 How many calories of energy did it take to make the 1 gram of liquid water to completely turn to steam once it hit 100

C?________________________

Step E: Steam Rises in Temperature

 Temperature ___________ again when all water is turned to steam

 Each gram of water requires a constant amount of energy to rise 1 o = specific heat.

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Specific Heat Capacity “C”

The amount of energy required to be absorbed to warm 1 gram of a substance by 1 o C (or 1 K) or the amount of energy required to be released to cool 1 gram of a substance by 1 o C (or 1 K).

-or-

How easily things warm up & cool down.

Energy Calculations Involving Specific Heat: q = mC∆T

where: q = Heat Energy

+ q means heat/energy is being absorbed (endothermic process)

- q means heat/energy is being released (exothermic process). m = mass in grams c = specific heat capacity (also “s”)

∆T = change in temperature (temperature final – temperature initial)

Energy Units:

Heat energy (q) is in joules(J), kilojoules (kJ) or calories (cal).

1 calorie = 4.184 joules

Mass (m) is in grams or kilograms

Specific heat capacity, c, is in J/g o C or kJ/kg o C

Water (L) = 4.184 J/g o C

Water (s) = 2.03 J/g o C

Water (g) = 2.0 J/g o C

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Temperature , T, is usually in o C (temperature can be in K)

Metals have low specific heat values

Aluminum 0.900 J/g o C

Iron 0.450 J/g o C

Gold 0.126 J/g o C

Doesn’t take much heat to heat them up and they don’t hold the heat well!!! (better conductors of heat/energy)

Water and organic materials hold heat much better – have higher specific heats also takes more energy to heat them up. (better insulators of heat/energy)

Water = 4.184 J/g o C

Wood = 1.76 J/g o C

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Specific heats and molar heat capacities for various substances at 20

C

Substance

Aluminum

Bismuth

Copper

Brass

Gold

J/g o C

cal/g K or Molar C

Btu/lb F J/mol K

0.900 0.215 24.3

0.123 0.0294 25.7

0.386 0.0923 24.5

0.380 0.092 ...

0.126 0.0301 25.6

Lead

Silver

Tungsten

Zinc

0.128

0.233

0.134

0.387

0.0305

0.0558

0.0321

0.0925

Mercury 0.140 0.033

Alcohol(ethyl) 2.4 0.58

26.4

24.9

24.8

25.2

28.3

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Water

Ice (-10 C)

Granite

Glass

4.184 1.00

2.05 0.49

.790 0.19

.84 0.20

75.2

36.9

...

...

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Name _____________________________________

Specific Heat Examples

1.

How much heat energy does a copper sample absorb if its specific heat is 0.386 J/g o C, its mass is 12.5 g and it is heated from 25.0 o C to 40.0 o C?

2.

How much heat energy is released by 10.0 g of gold, when it is cooled from 35.0 o o C? The specific heat of gold is 0.129 J/g o C.

C to 25.0

3.

A 4.00 kg sample of iron was heated from 0.0 o C to 20.0 as heat. What is the specific heat of this piece of iron? o C. It absorbed 35.2 kJ of energy

4.

42.6 J of energy is needed to heat 2.00 grams of carbon from 50.0 o C to what final temperature? The specific heat of carbon is 0.790 J/g o C.

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Name _____________________________________________

Specific Heat Problems 2

1. What amount of heat is required to raise the temperature of 85.9 g of water by 7.0

 C?

2. When 1045 joules are absorbed by a certain mass of water, the temperature of the water increases from 45.0 ºC to 50.0 ºC. What is the mass of the water sample?

3. How many joules are required to heat 38.0 grams of gold from 60.0 ºC to 260.0 ºC? The specific heat of gold is 0.126 J/(g·ºC).

4. Iron has a specific heat of 0.450 J/(g·ºC). If 1400. joules are absorbed by a chunk of iron that weighs 40.0 grams, how much does the temperature of the iron increase?

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Name _____________________________________

Practice Specific Heat Problems

** Pay attention to units AND sig figs **

1.

What is the specific heat value of a sample of unknown material, if it weighs 36.359 grams and 59.912 J of heat raise its temperature 152.0 o C?

2.

What would be the final temperature of a 73.174 g sample of cobalt with an initial temperature of 102.0 o C, after it loses 800 J? (The specific heat of cobalt is 0.4210 J/g o C)

3.

What mass of iron would release 0.1854 kJ when its temperature changed from 1550.0 o to 75.0 o C? (The specific heat of iron is 0.450 J/g o C)

C

4.

The specific heat of mercury is 0.0335 cal/g o C. If 152.00 g of mercury at 75.0 o C are cooled to 23.5 o C, what is the value of q in Joules?

5.

Kelly has 2.00 kg of water at 80.0 o C and wants it to cool to 45

20.9 kJ of energy every minute, how long will it take to cool? o C. If the water releases

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Calorimetry

From the point of view of the system

Endothermic

Feels cold

Surroundings lose heat (energy)

System gains energy

(+) Energy term

Energy is absorbed

Measured in Joules

Exothermic

Feels hot

Surroundings gain heat (energy)

System loses energy

(-) Energy term

Energy is released

Measured in Joules

To convert between Joules and Calories:

1 calorie = 4.184 Joules

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Calorimeter

Q

water

= -Q

system

Mass

H2O

x C

H2O

x ∆T

H2O

= Mass

sys

x C

sys

x ∆T

sys

Mass

H2O

x C

H2O

x ∆T

H2O

= Mass

sys

x C

sys

x ∆T

sys

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Name:________________________________

Calorimetry Problems

1) A 2.8 kg sample of metal with a specific heat of 0.43 kJ/kg°C is heated to 100.0°C and then placed in a 50.0g sample of water at 30.0°C. What is the final temperature of the water and the metal?

2) The specific heat of mercury is 0.033 cal/g°C. If 152g of mercury at 75.0°C is placed in 145g of water at 23.5°C, what will be the final temperature of the water?

3) A 37.7 g piece of metal is heated to 100.0

 C and placed into 75.0 g of water in a coffeecup calorimeter. Initially, the temperature of the water in the calorimeter was 23.1

 C.

After the metal was added to the water the temperature of the water increased until its temperature and the temperature of the metal were 30.6

 C. a. What is the specific heat of the metal? b. What kind of metal was added to the water in the calorimeter?

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4) A 440.00 g sample of mercury (specific heat = 0.140 J/g o C, initial temperature of 22.00

o C) is placed into 134.00 g of water (initial temperature of 35.00

o C). Find the final temperature of the system.

5) Abbey is testing her baby’s bath water and finds that it is too cool, so she adds some hot water from kettle on the stove. If Abbey adds 2.00 kg of water at 80.0°C to 20.0 kg of water at 27.0°C, what is the final temperature of the bath water?

6) Jason is emptying the dishwasher. He removes a 0.200 kg glass that has a temperature of 30.0°C.

Into the glass, he pours 0.100 kg of diet soda (mostly water) which comes out of the refrigerator with a temperature of 5.00°C. Assuming no external heat loss, what will be the final equilibrium temperature of the glass of diet soda (no ice was added)? (c for glass =0.84 J/g°C).

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Name ___________________________________________________________________

More Calorimetry Problems

1) 45.3 g of a shiny metal, with a specific heat of 0.561 cal/g ⁰C, is placed into a water bath that has a temperature of

99.7 ⁰C. It is then placed into a calorimeter that has 54.7 mL of water. If the water and the metal end up with a temperature of 17.2 ⁰C, what was the initial temperature of the water in the calorimeter?

2) A metal with a mass of 97.4 g is heated to a temperature of 81.4 ⁰C. It is then placed into a calorimeter containing

0.246 kg of benzene, which has a specific heat of 1.74 J/g ⁰C. The temperature of the benzene rises from 15.5 ⁰C to 32.5 ⁰C. What is the specific heat of the metal in calories?

3) A metal with a specific heat of 0.126 cal/g ⁰C is placed into a water bath with a temperature of 94.5 ⁰C. The metal is then placed into a calorimeter containing 86.5 g of acetic acid at a temperature of 20.6 ⁰C. The acetic acid and metal have a final temperature of 35.5 ⁰C. The acetic acid has a specific heat of 2.05 J/g ⁰C. What is the mass of the metal?

4) A metal with a specific heat of 2.03 J/g ⁰C and a mass of 68.5 g is placed into a hot water bath with a temperature of 74.5 ⁰C. The metal is then placed into a calorimeter containing acetic acid at a temperature of 14.5 ⁰C. The final temperature of the acetic acid and metal is 45.5 ⁰C. The density of acetic acid is 1.04 g/mL and a specific heat of

0.49 cal/g ⁰C. What is the volume of acetic acid in the calorimeter?

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Distillation Apparatus:

Separating a mixture of liquids based upon the boiling point of each liquid.

Using the graphed data and the table of compounds and their boiling point temeperature, which compound(s) is definitely not in the mixture?_______________________________

Which compound(s) are definitely in the mixture?_____________________________

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