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Lecture 17b- Water Curve Calcs
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Lecture 17b- Water Curve Calcs

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

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

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  • 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