1. Calculations associated with Heating and Cooling Curves

2. PROBLEM:
Suppose we have 100 g of ice starting at minus 10.0 ｰC and that the pressure is always one atmosphere. We will calculate the amount of energy needed to convert this to steam at ｰC.

3. Five major steps occur and the energy for each must be calculated
Five major steps occur and the energy for each must be calculated. Here they are:
1) the ice rises in temperature from to ｰC. 2) the ice melts at 0.00 ｰC. 3) the liquid water then rises in temperature from zero to ｰC. 4) the liquid water then boils at ｰC. 5) the steam then rises in temperature from to ｰC

4. Step One: SOLID ICE RISES IN TEMPERATURE
100 grams of ice (no liquid water yet!) has changed 10.0 ｰC. We need to calculate the energy needed to do this.
What information do we need? (Specific Heat of ICE = 2.06 Joules per gram-degree Celsius)
You do the Calculation ?

5. Step Two: SOLID ICE MELTS
Now, we continue to add energy and the ice begins to melt.
However, the temperature DOES NOT CHANGE. It remains at zero during the time the ice melts.
Each mole of water will require
A constant amount of energy to
melt. That amount is named
the molar heat of fusion and
its symbol is ΔHfus. The molar
heat of fusion is the energy
required to melt one mole of a
substance at its normal melting
point. Water's molar heat of
fusion is 6.02 kJ/mol.
Expressed per gram, it is J/g.
During this time, the energy is being used to overcome water
molecules' attraction for each other, destroying the three-dimensional
structure of the ice.
YOU do the Calculation ?

6. Step Three: LIQUID WATER RISES IN TEMPERATURE
Once the ice is totally melted, the temperature can now begin to rise again.
It continues to go up until it reaches its normal boiling point of ｰC.
Since the temperature went from zero to 100, the Δt is 100.
You do the Calculation!
Once the ice is totally melted,
the temperature can now
begin to rise again.
It continues to go up until it
reaches its normal boiling point
of ｰC.
Since the temperature went
from zero to 100, the Δt is 100.
You do the Calculation!

7. Step Four: LIQUID WATER BOILS
Each mole of water will require
a constant amount of energy to boil.
That amount is named
the molar heat of vaporization
and its symbol is ΔHvap.
The molar heat of vaporization is
the energy required to boil one
mole of a substance at its
normal boiling point. Water's
molar heat of vaporization is
40.7 kJ/mol. Expressed per gram,
it is 2261 J/g or 2.26 kJ/g.
You do the Calculation!

8. Step Five: Steam RISES IN TEMPERATURE
Once the ice is totally melted, the temperature can now begin to rise again.
It continues to go up until it reaches its normal boiling point of ｰC.
Since the temperature went from zero to 100, the Δt is 100.
You do the Calculation!
Once the steam is vaporized,
the temperature can now
begin to rise again.
It continues to go up until it
reaches ｰC.
Since the temperature went
From 100 to 120, the Δt is 20.
The Specific heat capacity of water vapor is J/-gK
You do the Calculation!

9. Sum Total: Step One: 2,060 J Step Two: 33,416 J Step Three: 41,840 J
Step Four: 226,100 J
Step Five: 3,992 J
Total- 307,408 J needed to heat 100 g f ice at -20oC to 120oC