1. Unit 2- Matter and Energy
Boiling
Condensing
Endothermic
Energy
Evaporating
Exothermic
Freezing
Heat
Heat of fusion
Heat of vaporization
Hypothesis
Kinetic energy
Law
Law of conservation of energy
Law of conservation of mass
Melting
Potential energy
Scientific notation
Significant figure
Specific heat
Sublimation
Temperature
Theory

3. Matter/Energy Relationship
Energy- the capacity to do work
2 changes in matter:
Physical- affects physical properties, ice melts, water boils
Chemical- occurs when a new substance is made, H and O combining to produce water
*every change in matter requires a change in energy
Endothermic- change in matter that requires E (E is absorbed)
Melting ice, boiling water
Exothermic- change in matter that releases E
Freezing water, water vapor condensing
**energy is released when a bond is formed
**energy is absorbed when a bond is broken
If you have a lot of energy- you can get a lot done
If you don’t have a lot of energy you cant get much (work) done
Pg 40 of text barium hydroxide and ammonium nitrate (beaker frozen to block)
Physical- crumple paper
Chemical- burn paper

5. Law of conservation of mass- mass of products = mass of reactants
Law of conservation of energy – during any change (phys or chem) the total amount of E remains constant
E can’t be created or destroyed; it’s transferred
Endothermic- transfer from surrounding to system
Exothermic- transfer from system to surrounding
Law of conservation of mass- mass of products = mass of reactants

6. E transfer can happen in different forms:
Ex: photosynthesis: light E  chem E
Firefly/glo stick is the opposite
Radiant
Electrical
Chemical
Thermal
Nuclear
Magnetic
Sound
Mechanical
Photosyn actually electromagnetic to chem
Solar E radiant to electical
Motor- electromagnetic to mechanical
In a battery, chemical energy is converted into electromagnetic energy.
The mechanical energy of a waterfall is converted to electrical energy in a generator.
In an automobile engine, fuel is burned to convert chemical energy into heat energy. The heat energy is then changed into mechanical energy.

7. Heat vs. Temperature
Heat- E transferred between objects of 2 different temperatures
Physical- ice cube into water
Chemical-most Rx’s involve heat
E released as heat- E can be released from E stored in a chemical
Ex: explosions
E absorbed as heat- E (heat) is absorbed by chemicals
Ex: baking
Temperature- measurement of the average kinetic E of the particles in a substance
So… heat is what’s transferred, temperature is a measure of it
Kinetic Energy- E of motion
Potential Energy- stored E
Heat is the total kinetic energy of moving molecules.
Temperature is the average kinetic energy of moving molecules.
If you heat a cup and a pool with a match which one’s temp will increase more?

8. Expressing temperature
F- Fahrenheit; C- Celcius; K-Kelvin
t(oC)= T(K)-273 =
T(K) – t(oC) =
**transfer of heat doesn’t always = change in temp
Heating curve
Where melting/freezing is
Where evaporating/condensing is

9. Specific heat Different substances are affected by heat differently
Specific heat- the quantity of heat required to raise 1g of a substance 1 oC
Q=mc∆T
Q= heat (joules)
M=mass (grams)
C= specific heat (joules/goC)
∆T= final temp – initial temp (oC)
Metals- low specific heat- get hot quick
Water- high specific heat- takes

10. Label everything you can:

12. Reading a Heating Curve
The heating curve below shows what happens when a 20.0 g sample of a substance absorbs 60 J of heat per minute.
During which lettered intervals is the phase changing?
What is the boiling point?
How much heat is absorbed heating the liquid from its melting
point to its boiling point?
What is the heat of fusion?
BC and DE
110°C
(10 min – 5 min) × 60 J/min = 300 J
= 9 J/g
(5 min – 2 min) × 60 J/min
20.0 g

13. Sample Problem 1 Step1: List the known variables
How much heat is needed to raise the temperature of 500. g of water by 15°C?
Step1: List the known variables
m = 500. g
C = 4.2 J/g°C
ΔT = 15°C
Step 2: Determine the product
q = mCΔT
q = (500. g)(4.2 J/g°C)(15°C)
q = 31,500 J ≈ 32,000 J

14. Sample Problem 2 Step 1: Determine ΔT (ΔT = Tf − Ti)
How much heat is needed to raise the temperature of 25 g of water from 27°C to 47°C?
(In this problem, the initial temperature [Ti] and the final temperature [Tf] are given instead of the temperature change [ΔT].)
Step 1: Determine ΔT (ΔT = Tf − Ti)
ΔT = 47°C − 27°C = 20.°C
Step 2: List the known variables
m = 25 g
C = 4.2 J/g°C
ΔT = 20.°C
Step 3: Determine the product
q = mCΔT
q = (25 g)(4.2 J/g°C)(20°C)
q = 2100 J

15. Sample Problem 3
What is the final temperature when 84 joules of heat are added to 2.0 gram of water at 15°C?
(In this problem, the amount of heat [q] and the initial temperature [Ti] are given. The final temperature [Tf] and the temperature change [ΔT] are the unknowns.)
Step 1: List the known variables
q = 84 J
m = 2.0 g
C = 4.2 J/g°C
Ti = 15°C
Step 2: Determine ΔT (q = mCΔT)
84 J = (2.0 g)(4.2 J/g°C)(ΔT)
ΔT = 10°C
Step 3: Determine Tf (ΔT = Tf − Ti)
10°C = Tf − 15°C
Tf = 25°C

16. Other Specific Heats
Consider two frying pans, one with a metal handle, and the other with a wood handle:
Which one is more comfortable to handle with the bare hands after it has been on a hot flame?
Why are they different?
Wood has a higher specific heat than metal. Wood is more resistant to temperature change. The wood is cooler even though it absorbed as much heat as the metal.
It is possible to do calculations with specific heats other than that of water (4.2 J/g°C). It is also possible for specific heat to be the unknown.
The wood handle

17. Sample Problem 4 Step 1: List the known variables
The specific heat of gold is J/g°C. How many joules will it take to make the temperature of a 20.0 g nugget go up 10.0°C?
(In this problem, the specific heat of gold is used instead of the specific heat of water.)
Step 1: List the known variables
m = 20.0 g
C = J/g°C
Ti = 10°C
Step 2: Determine the product
q = mCΔT
q = (20.0 g)(0.134 J/g°C)(10°C)
q = 26.8 J

18. Scientific Method Series of steps followed to solve problems
Hypothesis- theory or explanation based on observations that can be tested
Variable- a factor that could affect the results of an experiment
**only 1 is changed at a time
Control- the variable that is kept constant; the procedure that is kept constant is the control experiment
Theory- a well-tested explanation of observations
Can be disproved
Attempts to explain the cause of an event
Law- a summary of many experimental results and observations; tells how things work
**a hypothesis predicts an event, a theory explains it and a law describes it
Ex: law of conservation of mass- states mass of products=mass of reactants (DOESN’T explain why)

19. Measurements and Calculations
MUST measure using right equipment
Accuracy- how close measurement is to the actual value
Precision- how exact repeated measurements are
Let’s use a golf analogy…

20. Accurate? No
Precise?
Yes 20

21. Accurate?
Yes
Precise?
Yes 21

22. Precise? No
Accurate?
Maybe? 22

23. Why need or use significant figures? 1st: Estimation
Examine the object being measured below:
It appears to be between 3.2 cm and 3.3 cm long.
Perhaps it is 3.27 cm or 3.28 cm long.
The last number is estimated.
It makes no sense to say it is cm long. Since the 7 is already estimated, the 5 is nonsense.
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CENTIMETERS
Click to show object. Ask how long it is. Take several suggestions.
Point out that the object is between 3.2 and 3.3 cm. Click. Ask for a better answer than 3.2 or 3.3.
Click for estimate. Click. Point out the last number is estimated.
Ask if it is possible to get a better estimate. Click. Point out that it’s not possible.
Click for next slide. 23

24. The Nature of Measurements
The measured length, 3.27 cm, consists of a measured portion and an estimated portion
There can never be more than one estimated digit in a measurement. It is always the last digit.
Convert 3.27 cm to meters
3.27 cm = m
There are still only two measured digits and one estimated digit. The zeros are placeholders.
3.27
measured estimated
Using three clicks go through the fact that - a number consists of a measured and estimated portion - only one number can be estimated - it is always the last digit
Click to present conversion problem.
Ask for answer. Click to confirm.
Ask which numbers are measured and which are estimated. Point out there are no more measured numbers than there were before the conversion. Ask what the zeros are. Click to confirm.
Click for next slide. 24

25. The Nature of Significant Figures
All measured and estimated values are significant figures.
3.27 cm has 3 significant figures.
m has the same 3 significant figures. The zeros are place holders.
Place holders are not significant figures.
All nonzero digits are significant.
Only zeros can be place holders.
Not all zeros are place holders, however. Some zeros are significant.
Go through the points on the slide, clicking as needed.
Click for next slide. 25

26. Identifying Place Holders
The significance of a zero depends on where it is compared to the nonzero digits and the decimal
Decimal present
All zeros to the right of the first nonzero digit are significant.
Leading zeros between the decimal and the first nonzero digit are not significant.
Decimal absent
Trailing zeros, zeros following the nonzero digits, are not significant.
Describe how to tell when zeros are significant, clicking as needed.
Click for next slide. 26

27. Atlantic Pacific Rule
Review and explain Atlantic Pacific rule, clicking as needed.
Ask the number of sig. figs. in based on AP rule. Click to confirm.
Ask the number of sig. figs. in 1000 based on AP rule. Click to confirm.
Click for next slide. 27

28. Calculations with Significant Figures
Consider the following problem, 2.17 × 3.2:
The last digit of each number is estimated.
Anything multiplied by an estimated value is also estimated.
This results in an answer with 3 estimated digits. It must be rounded off to have only one.
The resulting answer, 6.9, has 2 significant figures like the smaller of the two multipliers.
2.17
× 3.2
434
651__
6.944
ESTIMATED
ESTIMATED
ESTIMATED
Click to show problem.
Ask which digits are estimated. Click to confirm.
Ask what kind of result you get when you multiply by an estimated value. Click to confirm.
Ask how many estimated digits are in the answer. Click to confirm.
Ask what it should be rounded to in order to have only 1 estimated digit. Ask how many significant figures the rounded value has. Ask how it compares, WRT significant figures, to the original multipliers. Click to confirm.
Ask for a suggested rule for multiplication.
Click for next slide. 28

29. Rules for Calculating with Significant Figures
Rules for Multiplication and Division
The number of significant figures in a product or quotient is the same as the number of significant figures in the measurement with the smaller number of significant figures.
Example: × 2.25 = Correct number of Significant Figures = 3 Solution: 7.07
Rules for Addition and Subtraction
The number of decimal places in the sum or difference is equal to the number of decimal places in the measured quantity with the smallest number of decimal places.
Example: = Correct number of Decimal Places = 1 Solution: 11.8
Click as needed.
Go over rules for multiplication and division.
Discuss and explain example.
Go over rules for addition and subtraction. 29