What is the melting point of this substance?

50˚C

100˚CThe boiling point?

Just to review before we start…

Topic: Calculating Topic: Calculating Energy Changes at Energy Changes at Phase Changes (HPhase Changes (Hvv

and Hand Hff))

It takes energy to heat stuff It takes energy to heat stuff up!up!

for pure substance in for pure substance in single phasesingle phase - can - can calculate how much E needed using:calculate how much E needed using:

Q = mCQ = mCT T Q = energy in JoulesQ = energy in Joules m = mass in gramsm = mass in grams C = specific heat capacityC = specific heat capacity T = change in temperature = TT = change in temperature = Tff - T - Tii

on other hand, when something cools down, on other hand, when something cools down, energy is released!energy is released!

Tem

pera

ture

Time

Q = mCliquidT

Q = mCsolidT

Q = mCgasT

C = specific heat capacity

(amount heat required to raise temp of 1g of pure substance by 1C)

C is a physical constant

unique for every pure substance

CAN YOU FIND THE SPECIFIC HEAT OF WATER ON YOUR REFERENCE TABLE?

I II III IV V

Why can’t I use Q = mCT for II and IV??T

em

pera

ture

Time

I II III IV V

Because T = 0, temp isn’t changing!!!!

So, how do we calculate the amount of energy required during a phase change?

• HF = Heat of Fusion (Q = mHF)

• HV = Heat of Vaporization (Q=mHV)

• We use one of these two constants instead of specific heat and delta T

Q = mCT

HHf f == Heat Heat of Fusionof Fusion

is amount energy required to change 1 is amount energy required to change 1 gram gram

of pure substance from of pure substance from solid to liquid solid to liquid at at its MP (meaning you aren’t changing its MP (meaning you aren’t changing the temperature)the temperature)

Is a physical constantIs a physical constant Check out Reference Table B, what is the heat of Check out Reference Table B, what is the heat of

fusion for water?fusion for water?

The Equation Q = mHThe Equation Q = mHff

How much heat is absorbed How much heat is absorbed when 10 grams of ice melts when 10 grams of ice melts at 0at 0ooC?C?

Heat absorbed = Heat absorbed =

mass of substance x heat of fusionmass of substance x heat of fusion of of substancesubstance

Q = mHQ = mHff = (10 g)(334 J/g) = 3340 J = (10 g)(334 J/g) = 3340 J

Where does this energy go?Where does this energy go? Particles must overcome forces of attraction to Particles must overcome forces of attraction to

move farther apart during phase change (s → move farther apart during phase change (s → l)l)

HHVV = Heat of = Heat of VaporizationVaporization

is the amount energy required to is the amount energy required to convert 1 gram convert 1 gram

of pure substance from of pure substance from liquid to gas liquid to gas at at its BP (meaning you aren’t changing its BP (meaning you aren’t changing the temperature)the temperature)

Is a physical constantIs a physical constant Check out Reference Table B, what is the heat of Check out Reference Table B, what is the heat of

vaporization for water?vaporization for water?

The Equation Q = mHThe Equation Q = mHvv

How much energy does it How much energy does it take to vaporize 10 g of take to vaporize 10 g of

water?water? Q = mHQ = mHvv

Q = (10 g)(2260 J/g) = 22600 JQ = (10 g)(2260 J/g) = 22600 J

H2O changing from solid to liquid requires 3,340J/g

It takes a lot more energy to go from liquid to gas than from solid to liquid. Why?

H2O changing from liquid to gas requires 22,600J/g

* greater energy required to change from liquid to gas because particles are spreading farther apart!

Tem

pera

ture

Time

Q = mCsolidT

I II III IV V

Q = mHFQ = mHV

Q = mCLIQUIDT Q = mCgasT

Heating curve of H2O

3 equations for Q3 equations for Q

1. Q = mC1. Q = mCTT

2. Q = mH2. Q = mHff

3. Q = mH3. Q = mHvv

figure out which to use figure out which to use depends on section of heating depends on section of heating

curvecurve look for hints in word problemlook for hints in word problem

Q = mCQ = mCTT

Temperature Temperature changedchanged

T T ↑↑ T ↓T ↓ Initial Initial

temperaturetemperature Start Start

temperaturetemperature

Final Final temperaturetemperature

Ending Ending temperaturetemperature

From __ ˚C to __ From __ ˚C to __ ˚C ˚C

WaterWater

Q = mHQ = mHff

IceIce FreezingFreezing MeltingMelting Occurs at 0Occurs at 0C (for HC (for H22O)O) At constant temperatureAt constant temperature

Q = mHQ = mHvv

SteamSteam BoilingBoiling CondensationCondensation Occurs at 100Occurs at 100C (for HC (for H22O)O) At constant temperatureAt constant temperature

heating rate = 150 J/min

If the substance takes 4 minutes to melt, how much heat energy was used to melt it?150J/min x 4min = 600J