CalorimetryCalorimetryHow to use math to describe How to use math to describe the movement of heat energythe movement of heat energy

Temperature Change Problems

Phase Change ProblemsUnit 7Honors Chemistry

Energy ConversionsEnergy Conversions

Heat is a specific type of Heat is a specific type of energyenergy that can be measured in different that can be measured in different ways.ways.

The SI unit for heat is The SI unit for heat is JoulesJoules– 4.184 Joules = 1 calorie (this will be 4.184 Joules = 1 calorie (this will be

given)given)– 1000 calories = 1 kilocalorie1000 calories = 1 kilocalorie– 1000 Joules = 1 kiloJoule1000 Joules = 1 kiloJoule

Heat ConversionsHeat Conversions

How many joules are in 130 calories?How many joules are in 130 calories?

How many calories are in 50 Joules?How many calories are in 50 Joules?

130 calories

1 calorie

4.184 Joules

50 Joules

4.184 Joules

1 calorie

= 543.92 Joules

(= 540 J (Sig figs!)

= 11.95 calories

Heat ConversionsHeat Conversions

How many kilojoules are in 130 How many kilojoules are in 130 Calories?Calories?130 Calories 4.184 Joules

1 Calorie

= 0.54 KiloJoules

1000J

1 kJ

CalorimetryCalorimetry Allows us to calculate the amount of Allows us to calculate the amount of

energy required to heat up a substance energy required to heat up a substance or to make a substance change states.or to make a substance change states.

Molar Heat of FusionMolar Heat of Fusion (H (Hff))—— The heat The heat absorbedabsorbed by one mole of a substance by one mole of a substance when changing from a when changing from a solidsolid to a to a liquidliquid. .

For water, it = For water, it = 6.0 kiloJoules/mole6.0 kiloJoules/mole– or or 334 Joules/gram 334 Joules/gram (specific heat of (specific heat of

fusion)fusion) Heat of solidificationHeat of solidification is opposite of is opposite of

heat of fusion (heat is heat of fusion (heat is releasedreleased).).

Molar Heat of VaporizationMolar Heat of Vaporization (H (Hvv))—— The heat absorbed by The heat absorbed by one mole of a substance when changing from a one mole of a substance when changing from a liquidliquid to a to a gas.gas.

For water, it = 40.7 kiloJoules/mole.For water, it = 40.7 kiloJoules/mole.

or or 2260 Joules/gram 2260 Joules/gram (specific heat of vaporization(specific heat of vaporization

Heat of condensationHeat of condensation is the opposite of heat of is the opposite of heat of vaporization (heat is vaporization (heat is releasedreleased))

Every pure substance will have a unique Molar heat of fusion (HEvery pure substance will have a unique Molar heat of fusion (H ff) or ) or

vaporization (Hvaporization (Hvv))

Heat Required For a Phase ChangeHeat Required For a Phase Change Heat Absorbed or Released = qHeat Absorbed or Released = q

For Melting or Freezing use the following:For Melting or Freezing use the following:

For Vaporization or Condensation use the For Vaporization or Condensation use the following:following:

q = (moles) x Molar Heat Fusion

q = (moles) x Molar Heat vaporization

Calculating Heat Required To Calculating Heat Required To Change StateChange State

Example #1Example #1: How much heat is needed to : How much heat is needed to melt 56.0 grams of ice into liquid (the molar melt 56.0 grams of ice into liquid (the molar heat of fusion for ice is 6.0 kJ/mol)?heat of fusion for ice is 6.0 kJ/mol)?

56.0 g 1 mole H56.0 g 1 mole H22O 6.0 kJO 6.0 kJ = = 18.0 g18.0 g 1 mole 1 mole

= 18.7 kJ will be absorbed= 18.7 kJ will be absorbed

q = (moles) x (Hf)

Example #2Example #2 How much heat energy in kJ will be How much heat energy in kJ will be

released when 200grams steam released when 200grams steam condenses back to a liquid water?condenses back to a liquid water?

HHvv = 40.7kJ/mol = 40.7kJ/mol

q = (moles) x (Hv)

200gram 1 mole 40.7 kJ

18gram 1 mole= 452 kJ released

Or -452kJ

Heating a Substance with Heating a Substance with No Phase ChangeNo Phase Change

Specific Heat CapacitySpecific Heat Capacity--The amount --The amount of energy required to raise one of energy required to raise one gram of a substance one degree gram of a substance one degree Celcius.Celcius.

Water’s Specific Heat (as a liquid) Water’s Specific Heat (as a liquid)

CCpp= 4.184 Joules/gram = 4.184 Joules/gram ooCC

**Every pure substance will have its own unique Every pure substance will have its own unique specific heat for every phase!specific heat for every phase!

Heating a Substance with Heating a Substance with No Phase ChangeNo Phase Change

When you see an increase in the When you see an increase in the temperature of a sample, the heat is temperature of a sample, the heat is being added to being added to raise the temperatureraise the temperature

How much the temperature increases is How much the temperature increases is based upon the based upon the heat capacity (Cheat capacity (Cpp)) and and the the massmass of your sampleof your sample

The The higherhigher the heat capacity the heat capacity number, the number, the longerlonger it takes to heat it takes to heat a substance up and the a substance up and the longerlonger the the substance holds on to the heatsubstance holds on to the heat..

Energy to Change Energy to Change TemperatureTemperature

q = (mass) ( Cq = (mass) ( Cpp) ( T) ( T ) )

HeatMeasured in Joules

Mass In grams

Specific Heat Capacity

Change in TemperatureTfinal – Tinitial

In OCelcius

Example #3Example #3How much energy is needed to How much energy is needed to heatheat 80 80

g of water from 10 g of water from 10 ooC to 55 C to 55 ooC?C?

q = mCpΔT = m Cp (Tfinal – Tinitial )

(55oC – 10oC)

q = 15062 joules

AbsorbedAbsorbed, because temperature in increasingincreasing

m

Tinitial

Tfinal

= (80g) ( 4.184 J/g C)

Is the energy absorbed or released?

Final Answer: 15,062 J = 15.06 kJ 15,062 J = 15.06 kJ

absorbed/ endothermicabsorbed/ endothermic

Example #4Example #4How much energy is needed to How much energy is needed to coolcool

150 g of ice from -2 150 g of ice from -2 ooC to -55 C to -55 ooC?C?

q = mCpΔT = m Cp (Tfinal – Tinitial )

(-55oC – -2oC)

q = - 16377 joules

Released, because temperature in decreasing

m

Tinitial

Tfinal

= (150g)

( 2.06 J/g C)

Is the energy absorbed or released?

Final Answer: - 16377 J = -16.3 kJ released/exothermic

Heat Problem Road MapHeat Problem Road Map

q = mCq = mCppΔΔTTSolidHeats

LiquidHeats

Gas Heats

q = (q = (moles)moles)HHvv

q = (q = (moles)moles)HHff

Melting or Freezing

Vaporization or Condensation

* Add each individual energies (in kJ) together for total heat energy required for multistep problems (up to 5 steps max!)

Example #5Example #5 -How much energy in kJ is needed to -How much energy in kJ is needed to change 150grams of ice from 0change 150grams of ice from 0ooC to 50C to 50ooC?C?

This problem requires two steps. Since water is solid ice at 0oC, we need to melt the ice and then heat it up to 50oC.

Step 1 – Calculate heat required to melt 150grams ice

Step 2 - Calculate heat required to heat liquid water from 0oC to 50oC

q = mC T = (150g)(4.184 J/goC)(50oC)

= 31380 J convert to kJ = 31.38kJ

150g 1 mole 6.0 kJ = 50 kJ 18grams 1 mole

*Add both heat values together for your final answer

50 kJ + 31.38kJ = 81.38 kJ heat absorbed.

Calorimetry Formula Calorimetry Formula SummarySummary

Phase ChangePhase ChangeUse Molar Heat constantsUse Molar Heat constants

Melting use q = (moles) x (HMelting use q = (moles) x (H fusionfusion))

Vaporize use q = (moles) x (HVaporize use q = (moles) x (HVaporizationVaporization))

No Phase ChangeNo Phase ChangeUse specific heat capacityUse specific heat capacity

q = q = (mass) ( C(mass) ( Cpp ) ( ) ( ΔΔTT ) )