Unit 7
Honors
Chemistry
Temperature Change
Problems
Phase Change Problems
Calorimetry
How to use math to describe the
movement of heat energy
Energy Conversions
Heat is a specific type of energy that can
be measured in different ways.
 The SI unit for heat is Joules

– 4.184 Joules = 1 calorie (this will be given)
– 1000 calories = 1 kilocalorie
– 1000 Joules = 1 kiloJoule
Heat Conversions

How many joules are in 130 calories?
130 calories
4.184 Joules
1 calorie

= 543.92 Joules
(= 540 J (Sig figs!)
How many calories are in 50 Joules?
50 Joules
1 calorie
4.184 Joules
= 11.95 calories
Heat Conversions

How many kilojoules are in 130 Calories?
130 Calories
1 kJ
4.184 Joules
1 Calorie
1000J
= 0.54 KiloJoules
Calorimetry
Allows us to calculate the amount of energy
required to heat up a substance or to make a
substance change states.
Molar Heat of Fusion (Hf)— The heat
absorbed by one mole of a substance when
changing from a solid to a liquid.
For water, it = 6.0 kiloJoules/mole



–

or 334 Joules/gram (specific heat of fusion)
Heat of solidification is opposite of heat of
fusion (heat is released).

Molar Heat of Vaporization (Hv)— The heat
absorbed by one mole of a substance when
changing from a liquid to a gas.

For water, it = 40.7 kiloJoules/mole.
or 2260 Joules/gram (specific heat of vaporization

Heat of condensation is the opposite of heat
of vaporization (heat is released)

Every pure substance will have a unique Molar
heat of fusion (Hf) or vaporization (Hv)
Heat Required For a Phase Change

Heat Absorbed or Released = q

For Melting or Freezing use the following:
q = (moles) x Molar Heat

Fusion
For Vaporization or Condensation use the
following:
q = (moles) x Molar Heat
vaporization
Calculating Heat Required To
Change State

Example #1: How much heat is needed to melt
56.0 grams of ice into liquid (the molar heat of
fusion for ice is 6.0 kJ/mol)?
q = (moles) x (Hf)

56.0 g

1 mole H2O
6.0 kJ =
18.0 g
1 mole
= 18.7 kJ will be absorbed
Example #2

How much heat energy in kJ will be
released when 200grams steam
condenses back to a liquid water?

Hv = 40.7kJ/mol
q = (moles) x (Hv)
200gram
1 mole
18gram
40.7 kJ
1 mole
= 452 kJ released
Or -452kJ
Heating a Substance with
No Phase Change

Specific Heat Capacity--The amount
of energy required to raise one
gram of a substance one degree
Celcius.
Water’s Specific Heat (as a liquid)
Cp= 4.184 Joules/gram oC
*Every pure substance will have its own unique
specific heat for every phase!
Heating a Substance with
No Phase Change

When you see an increase in the temperature
of a sample, the heat is being added to raise
the temperature

How much the temperature increases is based
upon the heat capacity (Cp) and the mass
of your sample

The higher the heat capacity number, the
longer it takes to heat a substance up
and the longer the substance holds on to
the heat.
Energy to Change Temperature
Heat
Measured in Joules
Change in Temperature
Tfinal – Tinitial
In OCelcius
q = (mass) ( Cp) ( T )
Mass
In grams
Specific Heat
Capacity
Example #3
m
How much energy is needed to heat 80 g of
water from 10 oC to 55 oC? Tfinal
Tinitial
q = mCpΔT
= m Cp (Tfinal – Tinitial )
= (80g) ( 4.184 J/g C) (55oC – 10oC)
q = 15062 joules
Is the energy absorbed or released?
Absorbed, because temperature in increasing
Final Answer: 15,062 J = 15.06 kJ
absorbed/ endothermic
Example #4
m
How much energy is needed to cool 150 g
of ice from -2 oC to -55 oC?
Tinitial
q = mCpΔT
Tfinal
= m Cp (Tfinal – Tinitial )
= (150g) ( 2.06 J/g C) (-55oC – -2oC)
q = - 16377 joules
Is the energy absorbed or released?
Released, because temperature in decreasing
Final Answer: - 16377 J = -16.3 kJ
released/exothermic
Heat Problem Road Map
q = (moles)Hv
q = (moles)Hf
Solid
Heats
Melting
or
Freezing
Vaporization or
Condensation
Gas
Heats
Liquid
Heats
q = mCpΔT
* Add each individual energies (in kJ) together for total
heat energy required for multistep problems (up to 5
steps max!)

Example #5 -How much energy in kJ is needed
to change 150grams of ice from 0oC to 50oC?
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
150g
1 mole
18grams
6.0 kJ
1 mole
=
50 kJ
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
*Add both heat values together for your final answer
50 kJ + 31.38kJ = 81.38 kJ heat absorbed.
Calorimetry Formula Summary
Phase Change
 Use Molar Heat constants
Melting use q = (moles) x (Hfusion)
Vaporize use q = (moles) x (HVaporization)
No Phase Change
 Use specific heat capacity
q = (mass) ( Cp ) ( ΔT )