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

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

• 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

Radiant Electrical Chemical

Thermal NuclearMagnetic

Sound

Mechanical

• E transfer can happen in different forms:

– Ex: photosynthesis: light E chem E– Firefly/glo stick is the opposite

Heat vs. Temperature• Heat- E transferred between objects of 2

different temperatures– Physical- ice cube into water– Chemical-most Rx’s involve heatE released as heat- E can be released from E stored in a

chemicalEx: explosions

E absorbed as heat- E (heat) is absorbed by chemicalsEx: 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

• Expressing temperature– F- Fahrenheit; C- Celcius; K-Kelvin– t(oC)= T(K)-273 =– T(K) – t(oC) + 273 =

**transfer of heat doesn’t always = change in temp

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

Label everything you can:

• During which letteredintervals 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?

Reading a Heating CurveThe heating curve below shows what happens when a 20.0 g sample of a substance absorbs 60 J of heat per minute.

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

• Step1: List the known variablesom = 500. go C = 4.2 J/g°C

o ΔT = 15°C• Step 2: Determine the product

o q = mCΔTo q = (500. g)(4.2 J/g°C)(15°C)o q = 31,500 J ≈ 32,000 J

Sample Problem 1 How much heat is needed to raise the temperature of 500. g of water by 15°C?

• Step 1: Determine ΔT (ΔT = Tf − Ti)o ΔT = 47°C − 27°C = 20.°C

• Step 2: List the known variableso m = 25 go C = 4.2 J/g°Co ΔT = 20.°C

• Step 3: Determine the producto q = mCΔTo q = (25 g)(4.2 J/g°C)(20°C)o q = 2100 J

Sample Problem 2How 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: List the known variableso q = 84 Jo m = 2.0 go C = 4.2 J/g°Co Ti = 15°C• Step 2: Determine ΔT (q = mCΔT)o 84 J = (2.0 g)(4.2 J/g°C)(ΔT)o ΔT = 10°C

• Step 3: Determine Tf (ΔT = Tf − Ti)o 10°C = Tf − 15°C o Tf = 25°C

Sample Problem 3What 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.)

• Consider two frying pans, onewith a metal handle, and theother with a wood handle:

• Which one is morecomfortable to handlewith the bare handsafter it has been on a hotflame?

• 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.

Other Specific Heats

The wood handle

• Step 1: List the known variableso m = 20.0 go C = 0.134 J/g°C

o Ti = 10°C• Step 2: Determine the product

o q = mCΔTo q = (20.0 g)(0.134 J/g°C)(10°C)o q = 26.8 J

Sample Problem 4The specific heat of gold is 0.134 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.)

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 itEx: law of conservation of mass- states mass of products=mass of reactants (DOESN’T explain why)

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…

Accurate? No

Precise? Yes

Accurate? Yes

Precise? Yes

Precise? No

Accurate?Maybe?

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 3.275 cm long. Since the 7 is already estimated, the 5 is nonsense.

|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||| 1 2 3 4 5 6 7 CENTIMETERS

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 = 0.0327 m•There are still only two measured digits and one estimated digit. The zeros are placeholders.

3.27

measured estimated

The Nature of Significant Figures

• All measured and estimated values are significant figures.•3.27 cm has 3 significant figures.•0.0327 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.

Identifying Place Holders

• 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.

The significance of a zero depends on where it is compared to the nonzero digits and the decimal

Atlantic Pacific Rule

Calculations with Significant Figures

• Consider the following problem, 2.17 × 3.2:•The last digit of each numberis estimated.•Anything multiplied by anestimated value is alsoestimated.•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

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: 3.1415 × 2.25 = 7.068375Correct number of Significant Figures = 3Solution: 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: 6.357 + 5.4 = 11.757Correct number of Decimal Places = 1Solution: 11.8