Matter and Energy

1

Matter vs. Energy• Recall that:• Matter:

– Has mass and takes up space (volume)– Is usually a “thing”

• Energy:– Does not have mass or take up space– Is the ability to make something happen. Light,

heat, sound, motion, and electricity are all forms of energy.

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• Energy can NOT be observed

• You CAN observe the transformation of 1 form of energy into another

• Ex: Burning wood - chemical energy being converted to heat and light energy

Energy

3

Forms of Energy

• Heat

• Mechanical

• Electrical

• Chemical

• Nuclear

• Solar

• Light

• Sound

4

Types of Energy• Potential or kinetic

• Potential energy – stored energy – Used at a later time– Can’t be measured, only

CHANGES in PE can – Can be calculated – Examples are water behind a dam,

a compressed spring, chemical bonds in gasoline, coal, or food

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Types of Energy

• Kinetic energy-the energy of matter in motion – Examples are swimming,

water flowing over a dam, working out, burning gasoline

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Kinetic vs. Potential

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Let’s Do It!!!

• Number your paper from 1-5• Write whether these are kinetic energy (KE) or

potential energy (PE)– 1. Running to catch the bus– 2. Eating your lunch– 3. Holding your books in your arms– 4. The molecules of a mug of coffee– 5. An apple hanging from a branch

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Let’s Do It!!!

• Number your paper from 1-5• Write whether these are kinetic energy (KE) or

potential energy (PE)– 1. Running to catch the bus KE– 2. Eating your lunch KE– 3. Holding your books in your arms PE– 4. The molecules of a mug of coffee KE and

PE– 5. An apple hanging from a branch PE

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Temperature vs. Heat

• Temperature and heat are NOT the same thing!

• Temperature- a measure of the average amount of kinetic energy all the particles in a sample of matter– Symbol is T and the unit is

o F (British) and o C or K (metric)

– ↑Kinetic E = ↑Temperature

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– It’s only an indicator of energy– It’s impossible to measure the KE of all the individual

molecules so we use thermometers • Does NOT depend on AMOUNT of matter in the

sample, or mass, due to it being based on average KE

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Temperature

Heat, or Thermal Energy• Heat- total amount of energy (kinetic

and potential) transferred between objects due to a temperature difference– Symbol is q

– Metric units : calories or Joules

(1 cal = 4.184 J)– ***Note: a food Calorie is not

the same as this type of calorie

1 food Calorie = 1000 calories = 1 kcal

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• 1 kJ = 1000 J and 1kcal = 1000 cal

• English units: B.T.U. (British Thermal Unit)

• Does depend on mass of the sample

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Heat Transfer

Cup gets cooler while hand gets warmer

Ice gets warmer while hand gets cooler

• Heat energy flows from a warmer object to a colder object

• When heat is absorbed, part of it goes to storage as potential energy and part of it is used to make the particles move faster, raising the temperature.

• Energy lost (something cools) = energy gained (something warms)

.

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Think About It

• If you want to have a pool that can warm up quickly for summer, do you buy a big pool or a small pool?

• If you want to drink some hot chocolate right away, do you heat up a big pan of milk or a small mug?

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• Specific heat - the amount of heat energy required to raise the temp of 1 gram of a substance 1 o C• Symbol is Cp and the units are cal/g x o C OR J/g x o C

Specific Heat = Heat Energy change

(Mass ｘ Temperature change)• Rearrange this formula to calculate the energy change in a sample of matter

when the temperature of it changes:

Heat (cal or J)=mass(g) x T change (oC) x specific heat

OR q = m T Cp

(Tfinal – Tinitial)

Note: the units of specific heat must match the units on all other quantities so they can cancel

Specific Heat Capacity-Reference

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• One of the units of heat mentioned, the calorie, is a standard that is equal to the amount of energy required to change the temperature of 1 gram of WATER by 1 o C– So water’s specific heat is 1 cal/g o C (or 4.184 J/g oC

)– So it takes 1 calorie (or 4.184 Joules) to raise 1 gram

of water 1 oC

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Specific Heat Capacity of Water-Reference

Sample Problem #1-Reference (finding an energy change for water)

How much heat energy is required to raise the temperature of 125 g of water from 25.0 oC to 32.0 oC?

q = (m) (T) (Cp)

Need calories

q = (125 g) (32.0 oC-25.0 oC) (1 cal/g oC)

q = (125 g) (7.0 oC) (1 cal/g oC)

Cancel units!

q = 875 cal = 880 cal gained with proper sig figs

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This problem may also be done in the metric units of joules. The Cp for water in the metric system is 4.184 J/g oC.

q = (m) (T) (Cp)

Need joules

q = (125 g) (32.0 oC - 25.0 oC) (4.184 J/g oC)

q = (125 g) (7.0 oC) (4.184 J/g oC)

q = 3661 J = 3700 J gained with proper sig figs

**Keep numbers this way so are aware of sig figs

Sample Problem #1-Reference (finding an energy change for water)

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Sample Problem #2-Reference (finding an energy change for a metal)

How much heat energy (q) is lost when the temperature of 75.2 g of iron drops from 25.2 o C to 17.4 o C? The specific heat of iron is 0.107 cal/g o C.

q = (m) (T) (Cp) Need calories q = (75.2 g) (17.4o C-25.2 o C) (0.107 cal/g o C) q = (75.2 g) (7.8 o C) (0.107 cal/g o C)q = -62.76192 cal lost = - 63 cal lost with proper sig figs

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Sample Problem #3 -Reference (finding the mass of a metal)

When a piece of copper wire gains 45.0 J its temp changes from 26.5 o C to 29.7 o C. The specific heat of copper is 0.385 J/g o C. What is the mass of the sample?

q = (m) (T) (Cp) Need mass45.0J = (m) (29.7 o C-26.5 oC) (0.385 J/g o C) 45.0J = (m) (3.2 o C) (0.385 J/g o C) _______(45.0J)_______ = m

(3.2°C)(0.385J/g°C) 37 g = m with proper sig figs

***Note how the units cancel! 21

1. A 225g sample of hot tea cools from 74.6 o C to 22.4

o C. How much heat, in joules, is lost, assuming that tea has the same specific heat as water?

Let’s Do It!!!!!

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q = (m) (T) (Cp)

Need joules

q = (225 g) (22.4 °C – 74.6 °C) (4.184 J/g °C )

q = (225 g) (-52.2 °C) (4.184 J/g °C )

q = 225 g x -52.2 °C x 4.184 J = - 49.1 J lost g °C

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Let’s Do It!!!!!

Let’s Do It!!!!!

2. Aluminum is used to make kitchen utensils. What is the mass of an aluminum spatula if 3250 Joules of heat raises its temperature from 20.0 °C to 45.0 °C? Aluminum Cp = 0. 897 J/g °C.

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q = (m) (T) (Cp)

Need mass so can solve for m first (isolate it) since it’s embedded in the equation and then plug in numbers

___q___ = (m) (T) (Cp (T) (Cp) (T) (Cp)

___q_____ = m (T) (Cp)

_______3250 J_______________ =m = 145 g of Al

(45 °C - 20 °C) (0.897 J/g °C )

3250 J = (m) (25 °C) (0.897 J/g °C )

3250 J = (m) x 25 °C x 0.897 J

g °C

Solve for m

3250 J (g °C) = 145 g of Al

25.0 °C x 0.897 J

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Let’s Do It!!!!!

Review

• Number your paper from 1-5 and answer the following questions. Two will be cumulative review! – 1. Which of these has 4 significant figures?

• a. 3340• b. 3.340• c. 0.00334• d. 334.00

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Review• B• 2. Which of these is a quantitative observation?

• a. the sky is blue

• b. his arm is long

• c. she has 42 mosquito bites

• d. the candle was melting

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Review• C• 3. Which of these is kinetic energy?

• a. a balloon on a string

• b. blowing a bubble

• c. a match

• d. sitting on a horse

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Review• B• 4. Which of these is the correct specific heat for

water ?

• a. 1 cal/g °C

• b. 0.897 J/g °C

• c. 1 J/g °C

• d. 4.184 cal/g °C

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Review• A• 5. Substances’ specific heats are listed below.

Which is easiest to heat up?

• a. 1000 cal/g °C

• b. 100 cal/g °C

• c. 3 cal/g °C

• d. 4.184 J/g °C

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Review• D

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Substances Have Different Specific Heats (Cp)

• Some things heat up or cool down faster than others

Land heats up and cools down faster than water

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Different Specific Heats-ReferenceSubstance Specific Heat

J / kg oC

Water 4.184

Wood 1.700

Sand 0.830

Carbon (graphite) 0.710

Iron 0.450This is why land heats up quickly during the

day and cools quickly at night and why water takes longer

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Specific Heat• Why does water have such a high specific

heat?– Water molecules are polar– They form strong bonds with each other– It takes more heat energy to pry them apart

and get them moving (more kinetic E)• Other substances have weaker attractions and

do not need as much energy to pry them apart so their molecules get moving easier

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Specific Heat

Heat added to water must be used to overcome the attraction between the water molecules before they start moving faster (increase in temp)

Things like metals can have electrons travel freely so no strong attractions need to be overcome and they heat up right away 35

Specific Heat of Water

• A coolant is a substance that can absorb a great amount of thermal energy with little change in temperature

• Due to water’s specific heat, it make it a great coolant in things like car engines

36

The Law of Conservation of Energy

– The Law of Conservation of Energy- energy can’t be created nor destroyed by ordinary means, it can only be converted to another form

– The total energy of a system remains constant

– When something loses energy, something else must gain that SAME amount of energy

***Energy lost = Energy gained

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– Heat lost by the warmer object is equal and opposite to the heat energy gained by the colder object

– Both objects will come to one temperature somewhere in between the two and this is referred to as equilibrium

– q (gained by something) = -q (lost by something else) so that q of one substance = - q of the other

Determining Changes in Heat-Reference

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Transforming Energy• Some energy transformations are less

obvious because they do not result in visible motion, sound, heat or light– Green plants convert the radiant

energy from the Sun into the chemical energy stored in bonds in the plant, like in carbs (sugars)

– When we eat we transform the potential energy stored in the carb bonds and transform it into kinetic energy (movement)

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Law of Conservation of Energy Sample Problem #1-Reference

• The element copper has a specific heat of 0.0920 cal/g oC). If a sample of copper at 125.3 o C is added to 245 mL (245 g) of water at 25.8 oC and the resulting temperature of the water and copper is 37.4 oC, what mass of copper (in grams) was used?– The metal is losing energy (temp. goes from 125.3

oC to 37.4 oC)– The water is gaining energy (temp. goes from 25.8

oC to 37.4 oC)

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Need mass of metal

Energy gained by water (q) = Energy lost by metal (-q)

(m) (ΔT) (Cp) = -[ ( m ) (ΔT) (Cp) ]

(245 g) (37.4 oC-25.8 oC) (1 cal/g oC) =

-[( m ) (o C 37.4 - 125.3 oC)(0.0920 cal/g oC)]

(245 g)(11.6 oC)(1 cal/goC)=-[(m)(-87.9 oC)(0.0920 cal/goC)]

Negatives cancel!(245 g) (11.6 oC) (1 cal/g oC) = m = 351 g

(87.9 oC)(0.0920 cal/g oC)41

Law of Conservation of Energy Sample Problem #1-Reference

Law of Conservation of Energy Sample Problem #2-Reference

• A piece of unknown metal with mass of 23.8g is heated to 100.0 oC and dropped into 50.0 mL (50.0 g) of water at 24.0 oC. The final temperature of the water and metal is 32.5 oC. What is the specific heat of the metal?

42

Law of Conservation of Energy Sample Problem #2-Reference

Need specific heat of metal

Energy gained by water ( q) = Energy lost by metal (-q)

(m) (T) (Cp) = -[(m) (T) (Cp )]

(50.0 g) (32.5 o C-24.0 oC) (1 cal/g oC) =

-[(23.8 g) (32.5-100.0 oC) ( Cp )]

(50.0 g) (8.5 oC) (1 cal/g oC) = -[ (23.8 g) (-67.5 oC) (Cp )]

Negatives cancel!

(50.0 g) (8.5 oC) (1 cal/g oC) = Cp = 0.26 cal/g oC

(23.8 g) (67.5 oC)

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Law of Conservation of Energy Sample Problem #3- Reference

• If a piece of cadmium with mass 65.6 g and a temperature of 100.0 oC is dropped into 25.0 cm3 (25.0 g) of water at 23.0 oC, what will be the final temperature of the water and cadmium? The specific heat of cadmium is 0.05523 cal/g oC.

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Law of Conservation of Energy Sample Problem #3- Reference

Need final temp of the water AND cadmium

Energy gained by water (q) = Energy lost by metal (-q)

(m) (T) (Cp) = -[(m) (T) (Cp)]

(m) (Tfwater– Tiwater) (Cp) = -[(m)(Tfcad - Ticad)(Cp) ]

Remember: the final temperature of both will be the same, X!

(25.0 g)(X -23.0 oC)(1 cal/g oC) =

-[(65.6 g)(X -100.0 oC)(0.05523 cal/g oC)]

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Law of Conservation of Energy Sample Problem #3- Reference

(25.0 g)(X -23.0 oC)(1 cal/g oC) =

-[(65.6 g)(X -100.0 oC)(0.05523 cal/g oC)]

Use the distributive property with 65.6 g and cancel units

25.0 X - 575 = -[65.6 X – 6560.o C(0.05523 cal/g oC)]

Distribute again with 0.05523 cal/g oC and cancel units

25.0 X – 575 = -[3.62 X – 362.3] = -3.62X = 362.3

Combine like terms 28.6 X = 937

X= 32.8 oC = final temp for both water and cadmium

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Let’s Do It!!!! • Get with a partner and solve these two questions• 1. The element copper has a specific heat of 0.0920 cal/g oC.

If a sample of copper at 125.3 oC is added to 245 mL (245g) of water at 25.8 oC and the resulting temp of the water and copper is 37.4oC, what mass of copper (in grams) was used?

• 2. A piece of unknown metal with a mass of 23.8 g is heated to 100.0 oC and dropped into 50.0 mL (50.0g) of water at 24.0 oC. The final temp of the water and metal is 32.5 oC. What is the specific heat of the metal? Again the energy lost by the metal must equal the energy gained by the water.

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Solution #1

• Solving for mass of copper• Energy gained by water (q) = Energy lost by metal (-q)• (m) (ΔT) (Cp) = -[(m) (ΔT) (Cp)]

(245 g)(37.4 - 25.8)(1cal/goC) =

-[(m)(37.4-125.3)(0.0920cal/goC)]

(245g)(11.6oC)(1cal/goC) = -[(m) (-87.9oC) (0.0920 cal/goC)]

Cancel the negatives!

(245g)(11.6oC)(1cal/goC) = m = 351 g of metal

(87.9oC) (0.0920cal/goC)

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Solution #2

Solving for Cp of the metal

Energy gained by water (q) =Energy lost by metal (-q) (m)(ΔT) (Cp) = - [(m)(ΔT)(Cp)]

(50.0g)(32.5-24.0 o C)(1cal/goC) =

-[(23.8g)(32.5-100o C)(Cp)]

(50.0g)(8.5oC)(1cal/goC) = -[(23.8g)(-67.5oC)(Cp)]

Cancel the negatives!

(50.0g)(8.5oC)(1cal/goC) = Cp of metal = 0.26 cal/goC

(23.8g)(67.5oC)

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Review

• Number your paper from 1-5 and answer the following questions. Two of these will be cumulative review!

• 1. What does this conversion factor mean?

4.184 J

1 cal – a. There are calories that are Joules– b. There are 4.184 calories in a Joule – c. Calories and Joules are equal– d. There is 1 calorie in 4.184 Joules

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Review• D• 2. Which of these is potential energy?

• a. working out

• b. blowing a bubble

• c. a match

• d. chewing a pizza

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Review• C• 3. The specific heat of cadmium is 0.05523 cal/g oC. What

precisely does this mean?

– a. It takes 0.05523 calories per gram of cadmium to raise it 1 degree Kelvin

– b. It takes 0.05523 calories per gram of cadmium to raise it 1 degree Celsius

– c. It takes 0.05523 calories per kilogram of cadmium to raise it 1 degree Kelvin

– d. It takes 0.05523 degrees Celsius to raise 1 gram of cadmium one calorie

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Review

•B•4. Which of these is the correct specific heat for water ?

• a. 10 cal/g °C

• b. 0.897 J/g °C

• c. 1 J/g °C

• d. 4.184 J/g °C

53

Review

•D•5. What is the mass of an aluminum block if 3250 Joules of heat raises its temperature from 20.0 °C to 45.0 °C? Aluminum Cp = 0. 897 J/g °C?

• a. 145 g

• b. 0.897 J/g °C

• c. 10 grams

• d. 4.184 J/g °C 54

Review

• A

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Measuring Changes in Thermal Energy

• A calorimeter is used to help measure the thermal energy change

• Knowing its Cp value, its mass, and its T, its q (thermal energy change) can be calculated

56

Calorimeters and Heat Transfer-Reference

• To determine the amount of food Calories in a food, a Bomb Calorimeter can be used. Then the results are converted to calories (remember that 1 cal = 1 food Calorie

• The container (called a calorimeter) absorbs or releases energy depending on whether its temperature is going up or going down in the process

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Calorimeters-Reference

• Each calorimeter is different and has a calorimeter constant associated with it- the amount of heat absorbed/released by the calorimeter per degree Celsius.

• To find the energy change associated with any calorimeter use the following:

• Energy of calorimeter = (calorimeter constant)(ΔT)

q = (cc) (ΔT)

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Calorimeter Sample Problem #1- Reference

• Finding a Calorimeter Constant (easy way)

• If 225 calories of energy is added to a calorimeter with an initial temperature of 24.8 °C and the temperature rises to 33.8 °C, calculate the calorimeter constant for the calorimeter

• Need the calorimeter constant

q = (cc) (ΔT)

225 cal = (cc) (33.8°C - 24.8°C)

225 cal = (cc) (9.0°C)

25 cal/°C = cc

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• If a calorimeter has a calorimeter constant of 24.8 cal/°C, how much energy would need to be added to raise the temperature of the calorimeter from 24.8°C to 37.2°C?

• Need energy

q = (cc) (ΔT)

= (24.8 cal/°C) (37.2°C - 24.8°C)

= (24.8 cal/°C) (12.4°C)

q = 308 cal

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Calorimeter Sample Problem #2- Reference

Calorimeter Sample Problem #3- Reference

83.2 mL of water is put into a calorimeter. The starting temp of the water and calorimeter is 24.7 °C. 91.3 mL of water with a temperature of 39.5 °C is added to the calorimeter. The final temperature of the water and calorimeter is 31.5 °C. Find the calorimeter constant of the calorimeter.•The cool water (water starting at 24.7°C) is gaining energy (temp. going from 24.7°C to 31.5°C).•The warm water (water starting at 39.5°C) is losing energy (temp. going from 39.5°C to 31.5°C).•The calorimeter is gaining energy (temp. going from 24.7°C to 31.5°C). 61

Calorimeter Sample Problem #3- Reference

Need calorimeter constant

Energy gain by cool H2O (q) = Energy lost by warm H2O (-q) q of cool water + q of calorimeter = -q of warm water

(m) (ΔT) (Cp) + (cc) (ΔT) = -[(m) (ΔT) (Cp)]

(83.2g)(6.8°C)(1 cal/g °C) + (cc)(6.8°C)

= -[(91.3g)(-8.0°C)(1cal/g °C)]

Cancel the negatives!!!!

570 cal+ (cc) (6.8°C) = 730 cal

(cc) (6.8°C) = 160 cal

cc = 24 cal/°C62

Calorimeter Sample Problem #3- Reference

• 3.95 g of a solid is added to 78.3 mL of water in a calorimeter. The temperature of the water and calorimeter was originally 26.9 °C. After the solid dissolved, the temperature of the water and calorimeter was 24.0 °C. The calorimeter constant is 32.5 cal/°C. Calculate the amount of energy absorbed by the solid in cal/g as it dissolved.

• The solid is gaining energy (it says so in the problem).• The water is losing energy (temp. going from 26.9°C to

24.0°C).• The calorimeter is losing energy (temp. going from 26.9°C

to 24.0°C).63

Calorimeter Sample Problem #3- Reference

Need the amount of energy absorbed by the solid PER GRAM

Energy gained (q) = Energy lost (-q)

q of solid = -q of water + -q of calorimeter

q of solid = -[(m)(ΔT)(Cp)] + -[(cc)(ΔT)]

q of solid =

-[(78.3 g)(24.0 – 26.9 °C)(1 cal/g °C)] + -[(32.5 cal/°C) (24.0-26.9 °C)]

-[(78.3 g)(-2.9 °C)(1 cal/g °C)] + -[(32.5 cal/°C)(-2.9°C])

Cancel the negatives!!!

(78.3 g)(2.9 °C)(1 cal/g °C)] + (32.5 cal/°C)(2.9°C])

q of solid = 230 cal + 94 cal = 320 cal

q of solid PER GRAM = 320 cal/3.95 g of solid = 81 cal/g64

Let’s Do It!!!! Get with a partner and solve this problem. Bring up your papers when you’re done to check your work with the key.

98.7 mL of water is put into a calorimeter. The temperature of the water and calorimeter is 22.6 °C. A 82.3 g piece of metal is heated to 99.3°C and added to the water in the calorimeter. The final temperature of the water, calorimeter and metal is 28.5°C. The calorimeter constant is 29 cal/°C. Find the Cp of the metal.•The water is gaining energy (22.6°C to 28.5°C).•The calorimeter is gaining energy (22.6°C to 28.5°C).•The metal is losing energy (99.3°C to 28.5°C).

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SolutionNeed Cp of the metal

Energy gained water +calorimeter (q) = Energy lost (-q) metal (m) (ΔT) (Cp) + (cc) (ΔT) = -[(m) (ΔT) (Cp)]

(98.7 g)(28.5-22.6°C)(1 cal/g°C) +(29 cal/°C)(28.5-22.6°C) =

-[(82.3 g)(28.5-99.3°C)(Cp)]

(98.7 g)(5.9°C)(1cal/g°C) + (29cal/°C) (5.9°C) =

-[(82.3g)(70.8°C)(Cp)]

Cancel the negatives!!

580 cal + 170 cal = 5830 g°C (Cp)

750 cal = 5830 g°C (Cp)

Cp = .13 cal/g°C66

Review

• Number your paper from 1-5 and answer the following questions. Two of these will be cumulative review!

• 1. Which of these is written in proper scientific notation?– a. 4.5 X 32

– b. 45.6 X 1010

– c. 456 X 106

– d. 4.56 X 106

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Review

• D• 2. What’s the first step in setting up a dimensional

analysis problem?– a. set up the question mark format

– b. cancel the units

– c. write down all the conversion factors

– d. cry

68

Review

• A• 3. What’s the equation use for figuring out the

energy absorbed or released by an object?– a. q = cc x T– b. Cp = m ΔT cc

– c. q = m x ΔT x Cp– d. F = a x r x t

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Review

• C• 4. If I burn a sample of pizza in a calorimeter,

where does the energy released from the bonds of the pizza go as it burns?– a. the water

– b. the calorimeter

– c. the air

– d. the water and the calorimeter

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Review

• C• 5. If I have a bar of aluminum heated to 100 o C

and I put it in a calorimeter with and the water in the calorimeter at 25 o C, which of these is a possible and likely equilibrium temperature?– a. 15 o C

– b. 25 o C

– c. 43 o C – d. 99 o C

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Review

• C

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