The second lecture for chapter 17 (honors and prep chemistry) covers the water curve, specific heat, and the enthalpy of phase changes.

1. BELLWORK‐ heat flow in pool
When you enter a swimming
pool, the water may feel quite
cold. ABer a while, though,
your body “gets used to it,” and
the water no longer feels so
cold. Use the concept of heat to
explain what is going on.

2. The specific heat capacity (c) of any
substance is the amount of heat
required to raise the temperature of
1gram of that substance by 1°C.
• Every substance has its own
specific heat.

3. Water has a high specific heat.
c= 4.184 J/(g·°C)
Water can absorb or release a
lot of heat before changing
temperature

4. The heat absorbed during a change in
temperature is calculated using the
equation
Specific Mass of Change
Heat
heat sample in temp
(J)
capacity (g) (°C)
J/(g·°C)
ΔT= Tfinal-Tinitial

5. PracIce
How much energy is required to raise 50g
of water from 25°C to 40°C?
q = c  m  ΔT
q = unknown
c = 4.18J/(g°C) for water
m = 50g
ΔT = 15°C q = 4.18 x 50 x 15
= 3135 J of energy

6. ∆Hfus = the quanIty of heat absorbed
during melIng
= the quanIty of heat released
during freezing.
For water
H2O(s)  H2O(l) ∆Hfus = 6.01kJ/mol
H2O(l)  H2O(s) ∆Hfus = -6.01kJ/mol

7. How many grams of ice at 0°C will melt
if 2.25kJ of heat are added?
Known
• iniIal and final temps are 0 °C
• ΔHfus = 6.01kJ/mol
• ΔH = 2.25kJ
2.25 kJ x 1mol = 0.374 moles would
6.01kJ be melted

8. How much energy is needed to melt 50g of ice?
Known The heat of
• ΔHfus = 6.01kJ/mol fusion is PER
• 50g MOLE so
convert grams
to moles
50g H2O x 1mole 2.8 moles x 6.01 kJ
=
18g 1mole
=16.8 kJ

9. The quanIty of heat absorbed by a
evaporaIng liquid is exactly the same as
the quanIty of heat released when the
vapor condenses = ∆Hvap
For water
H2O(l)  H2O(g) ∆Hvap = 40.7kJ/mol
H2O(g)  H2O(l) ∆Hvap= -40.7kJ/mol

10. Phase change occurs at
a specific temperature
(aka the melIng point
and boiling point)
Temperature is constant during a phase change

11. Temperature is constant during a phase
change
During a phase change heat is
used to overcome
intermolecular a`racIons.

12. The ΔHvap is always
larger than ΔHfus
∆Hfus is the energy required to overcome some
intermolecular a`racIons.
∆Hvap is the energy required to overcome all
intermolecular a`racIons.

13. IN YOUR NOTES--
Include a labeled drawing of
the water curve and all of
the notes in black and red
on the slides to follow.

14. To calculate energy changes (ΔH) for a state change
moles x ΔHfus OR moles x ΔHvap

15. To calculate energy required to increase or decrease
temperature within a single state
q= c m ΔT
Use
the
specific
Use the heat of
specific heat Energy
steam
of liquid
water
Use the
specific heat
of ice

16. To calculate the energy change over more
than one secIon of the curve…
Calculate the energy for each step and
add them together.

17. Calculate the ΔH associated with changing 100g
of steam at 125°C to ice at ‐50 °C.
125°C
1.Determine if a phase change occurs
within the temperature range.
• water will condense at 100°C
• water will freeze at 0 °C
‐50°C

18. Calculate the ΔH associated with changing 100g
of steam at 125°C to ice at ‐50 °C.
125°C
2. Split problem into steps.
125 °C steam  100 °C steam
steam  water
100 °C water  0 °C water
‐50°C water  ice
0 °C ice  ‐50 °C ice

19. 3. Calculate the enthalpy change for each step.
125 °C steam  100 °C steam
q = 1.89 J/(g°C) x 100g x ‐25°C = ‐4725J
steam  water
100g x 1 mol/18g = 5.55mol x 40.7kJ/mol = ‐226kJ
100 °C water  0 °C water
q = 4.18 J/(g°C) x 100g x ‐100°C = ‐41,800J
water  ice
5.55mol x 6.01 kJ/mol = ‐33.4kJ
0 °C ice  ‐50 °C ice
q = 2.10 J/(g°C) x 100g x ‐50°C = ‐10,500J

20. 4. Add the values for each step to get
the total energy change
‐4725J = ‐4.725 kJ
+ ‐226kJ
+ ‐41,800J = ‐41.8 kJ
+ ‐33.4kJ
+ ‐10,500J = ‐10.5 kJ
Whoops! The units aren’t the same!!

21. 4. Add the values for each step to get
the total energy change
‐4.725 kJ
+ ‐226 kJ
The total enthalpy
+ ‐41.8 kJ
change and the
+ ‐33.4 kJ
changes in each step
+ ‐10.5 kJ
are all negaIve
ΔHtotal = ‐316.4 kJ values because
cooling water is an
exothermic process

22. EXOTHERMIC STATE CHANGES
Heat comes out of substance(system)
∆H is negative
Happens in a freezer
Freezing(solidification) & condensation
ENDOTHERMIC STATE CHANGES
Heat goes into substance
∆H is positive
Happens on a stove
Melting(fusion), vaporization, sublimation

23. The insulated device used to
measure heat changes in chemical
or physical processes is called a
calorimeter.
qwater = - qrxn
The heat change for
the water in the
calorimeter is equal
to the heat change
of the reaction but
opposite in sign.

24. CalculaIons for Honors Only