View
216
Download
1
Category
Preview:
Citation preview
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
Recommended