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
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
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.
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
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