When energy is absorbed as heat by a solid or liquid, the temperature of the object does not necessarily rise.
The thermal energy may cause the mass to change from one phase, or state, to another.
The amount of energy per unit mass that must be transferred as heat when a mass undergoes a phase change is called the heat of transformation, L.
2. Change of State
Matter exists in 3 states:
•Solid
•Liquid
•Gas
water steam
fusion (at
melting point)
vaporization (takes
place at boiling point)
solidification condensation
Consider water:
at freezing point at boiling point
ice
3. Heats of
Transformation
When energy is absorbed as heat by a
solid or liquid, the temperature of the
object does not necessarily rise.
The thermal energy may cause the
mass to change from one phase, or
state, to another.
The amount of energy per unit mass
that must be transferred as heat when
a mass undergoes a phase change is
called the heat of transformation, L.
4. Heat of Fusion
To melt a solid means to change its
from the solid state to the liquid state.
This process requires energy because
the molecules of the solid must be
freed from their rigid structure.
To freeze a liquid to form a solid is the
reverse of melting and requires that
energy be removed from the liquid so
that the molecules can settle into a
rigid structure.
5. Heat of Fusion
When the phase change is from
solid to liquid, the sample must
absorb heat; when the phase
change is from a liquid to solid, the
sample must release heat.
The heat of transformation for these
phase changes is called the heat of
fusion, Lf.
Water: Lf = 334 J/g = 79.5 cal/g
fLmQ ⋅=
6. Heat of Vaporization
To vaporize a liquid means to change it from the liquid
state to the vapor or gas state. This process requires
energy because the molecules must be freed from the
liquid state.
Condensing a gas to a liquid is the reverse of
vaporizing; it requires that energy be removed from the
gas so that the molecules can cluster together instead
of flying away from each other.
The heat of transformation for these phase changes is
called the heat of vaporization, Lv.
Water: Lv = 2256 J/g = 539 cal/g
vLmQ ⋅=
9. SAMPLE PROBLEMS
1. In a water distilling store, 1.00 L
( 1.00kg ) of boiling water is to be
changed to steam. How much heat
will be needed to do that?
2. A freezer motor can remove 515
000J of heat from water at 0 °C. How
much water will turn to ice?
10. SUMMARY
Latent heat capacity – is the amount
of heat that will change the state of a
substance at a specific temperature.
When heat is added to a substance,
melting, vaporization or sublimation
may occur.
When heat is extracted from a
substance, freezing, condensation
and deposition may occur.
11.
12. Copyright 1999, PRENTICE
HALL
Chapter 1112
Phase ChangesPhase Changes
Energy Changes Accompanying PhaseEnergy Changes Accompanying Phase
ChangesChanges
19. vaporization
energy in
2260 kJ
fusion
energy in
334 kJ
condensation
energy out
2260
kJ
solidification
energy out
334 kJ
steam
(1 kg)
water
(1 kg)
ice
(1 kg)
Summary: Change of State
20. Energy involved in heating 1 kg of
water
ice
steam(ice and water) melting
(334) (420)
water
boiling
(water and stream)
(2260) energy / kJ
Summary: from ice to steam
0
100
21. Heat of Combustion (HC)
The amount of heat released per unit
mass or unit volume of a substance
when the substance is completely
burned.
Heats of combustion are used as a
basis for comparing the heating value
of fuels, since the fuel that produces
the greater amount of heat for a given
cost is the more economic.
Equation:
CHmQ ⋅=
22. Phase Change Example
Heat is added to 0.5 kg of water at 20° C.
How many joules of heat energy are required
to change the water to steam at 110° C?
Heat has to be added to the water to raise its
temperature from 20° C to the boiling point
100° C.
Heat has to be added to vaporize all of the
water that is at 100° C.
( )
JQ
CC
Ckg
J
kgTcmQ oo
o
167440
2010041865.0
=
−⋅
⋅
⋅=∆⋅⋅=
23. Phase Change Example
Heat has to be added to raise the
temperature of all of the steam from 100° C
to 110° C.
Total heat energy: Q = 167440 J +
1130000 J + 10050 J = 1307490 J
J1130000Q
kg
J
2260000kg5.0LvmQ
=
⋅=⋅=
( )
J10050Q
C100C110
Ckg
J
2010kg5.0TcmQ oo
o
=
−⋅
⋅
⋅=∆⋅⋅=
24. Final Temperature Example
A 0.3 kg piece of ice at 0° C is placed in 1 kg of
water at 40° C in an insulated container. If no
heat is lost to the container, what is the final
temperature of the water?
Q lost water = Q gained ice + Q gained ice water
(m·c·∆T)water = m·Lf + (m·c·∆T)ice water
( )
( )
FFo
o
Fo
F
o
o
T
kg
J
1255.8J100200T
C
J
4186J167440
C0T
Ckg
J
4186kg0.3
kg
J
334000kg0.3TC40
Ckg
J
4186kg1
⋅+=⋅−
−⋅
⋅
⋅+
⋅=−⋅
⋅
⋅