Thermal Properties
Ch 16 & 17: Temperature & Heat
Joseph Black (1728-1799)
• English Chemist
• Founder of calorimetry technique
• Proposed a theory of heat that did
not reference phlogiston, including:
– The heat equation
– Specific heat capacity & Latent heat
• Widely renowned as a great teacher:
"Many were induced, by the report of his students, to
attend his courses, without having any particular
relish for chemical knowledge."
Heat & Internal Energy
• Heat is energy that flows from a high temperature
object to a lower temperature object
– When something absorbs heat its internal energy (or the
energy of its atoms/molecules) increases
– When something releases heat its internal energy
decreases
• The SI units of heat are joules (J)
• Two things occur when an object absorbs or
releases heat energy (Q):
– The temperature will change (which is why they
expand/contract, due to changes in molecular motion)
– The object (or part of it) will change phase (solid,
liquid, gas)
Heat & Temperature Change
• When an solid (or liquid) of mass, m, absorbs heat
and NO phase change occurs:
Q = m.c. DT {Black’s Heat Equation}
Q is heat absorbed/lost,
DT is the temperature change
c is called the specific heat capacity
Example – Temp. Change
• When a car brakes, an
amount of heat equal to
112,500 J is generated in the
brake drums. If the mass of
the brake drum is 28 kg and
their specific heat capacity is
460.5 J kg-1 K-1, what is the
change in their temperature.
Q  mc DT
Q
DT 
mc
112500J

28x 460.5
o
 8.7 C
Heat & Temperature Change
Specific Heat Capacity:
– A physical property that relates how energy absorbed
reflects changes in temperature for a given substance
– In SI terms: the amount of energy required to change the
temperature of 1 kg of a substance by 1oC (or 1 K)
– SI units for specific heat capacity are J kg-1K-1
– Metals tend to have low specific heat capacities (which is
one reason they make great cooking vessels)
See Table 2.1
– Non-metal substances tend to have higher specific heat
capacities
– Water has an unusually high specific heat capacity
Heat & Temperature Change
• Specific Heat Capacity:
– Metals tend to have low specific heat capacities
(a reason they make great cooking vessels) See Table 2.1
– Non-metal tend to have higher specific heat capacities
– Water has an unusually high specific heat capacity
• Why do substances have different sp heat cap?
• Every 1 kg of material have different numbers of
molecules. If you provided the same heat to the
same number of moles the temperature change
would be the same!
Table 16-2
Specific Heats at Atmospheric Pressures
Substance
Specific Heat, c [J/(kg•K)]
Water
4186
Ice
2090
Steam
2010
Beryllium
1820
Air
1004
Aluminum
900
Glass
837
Silicon
703
Iron (steel)
448
Copper
387
Sliver
234
Gold
129
Lead
128
Heat Capacity
The amount a given substance heats up is characterized
by it’s “heat capacity”:
The concept of heat capacity is useful when a body
consists of several parts of different specific heat
capacities.

Heat capacity --- heat energy required to raise the
temperature of a sample by 1K (=1ºC)

Depends on amount of material
Units: J / K
Q  CDT
C  mc
Calorimetry
• A method to measure specific heat
capacity that is based on:
– The principle of conservation of energy
– The known specific heat capacity for water
(1000 cal/kg.oC or 4186 J/kg.oC )
• The Process:
• A heated object is placed into a thermally isolated container
containing a known amount of water
Qnet = Qgained by water + Qgained by object = 0
• Object and liquid reach thermal equilibrium:
Qgained by water = Qlost by object {= - Qgained by object}
• The final temperature of the object/water is used to determine
the specific heat of the object
Calorimetry
• A method to measure specific heat
capacity that is based on:
– The principle of conservation of energy
– The known specific heat capacity for water
(1000 cal/kg.oC or 4186 J/kg.oC )
• The Process:
• A heated object is placed into a thermally isolated container
containing a known amount of water
Qnet = Qgained by water + Qgained by object = 0
• Object and liquid reach thermal equilibrium:
Qgained by water = Qlost by object {= - Qgained by object}
• The final temperature of the object/water is used to determine
the specific heat of the object
Calorimetry Problems
• A piece of iron of mass 200g and temperature 300 oC is
dropped into 1.00 kg of water at 20 oC. What will be the
temperature of the water when it reaches thermal
equilibrium?
– Use c for iron as 470 J kg-1 K-1 and for water as 4200 J Kg-1K-1
• By conservation of energy Heat lost by the iron must
equal the heat gained by the water.
mironciron(300-T) = mwatercwater (T-20)
0.2 kg(470 J kg-1 K-1)(300-T) = 1kg(4200 J Kg-1K-1)(T-20)
T = 26.1 oC
Example: Cooking Tips
While cooking a turkey in a microwave oven that
puts out 500 W of power, you notice that the
temperature probe in the turkey shows a 1°C
temperature increase every 30 seconds. If you
assume that the turkey has roughly the same
specific heat as water (c = 4184 J/kg-K), what is
your estimate for the mass of the turkey?
What Equations? Q = m.c. DT ?
Turkey Time!
While cooking a turkey in a microwave oven that puts out 500 W of
power, you notice that the temperature probe in the turkey shows a 1°C
temperature increase every 30 seconds. If you assume that the turkey
has roughly the same specific heat as water (c = 4184 J/kg-K), what is
your estimate for the mass of the turkey?
DQ  Cturkey T f  Ti   mturkeycturkey T f  Ti 
DQ
mturkey 
cturkey T f  Ti 
In 30 seconds, DQ = (500 W)x(30 s) = 15,000 J of heat is put out by oven
mturkey 
15000 J
4184 J / kg  K 1C 
 3.6kg
Example: Cooking Tips
You place a copper ladle of mass mL= 0.15 kg (cL =
386 J/kg-K) - initially at ‘room’ temperature Tla= 20 C
- into a large pot containing 0.6 kg of hot cider (cC =
4184 J/kg-K) at a temperature of 90 C. If you forget
about the ladle while watching a football game on TV,
roughly what is its temperature when you try to pick it
up after a few minutes?
Example: Cooking Tips
You place a copper ladle of mass mL= 0.15 kg (cL = 386 J/kg-K) - initially
at ‘room’ temperature Tla= 20 C - into a large pot containing 0.6 kg of
hot cider (cC = 4184 J/kg-K) at a temperature of 90 C. If you forget
about the ladle while watching a football game on TV, roughly what is its
temperature when you try to pick it up after a few minutes?
DQcider  DQladle
mciderccider T f  Tco   mladlecladleT f  TLo 
mcider cciderTco  mladlecladleTLo
Tf 
mcider ccider  mladlecladle

0.6kg 4184 J / kg  K 363K   0.15kg 386 J / kg  K 293K 

0.6kg4184 J / kg  K   0.15kg386 J / kg  K 
= 361 K = 88 C !
Phase (State) Changes & Latent Heat
• When a substance absorbs/loses heat energy its temperature
will change until:
– the substance reaches its “critical” temperature it will no longer
change temperature
– gain/loss of additional heat energy will result in the phase
transformation of the matter from one phase to another:
Solid
 liquid
gas
.
Phase (State) Changes & Latent Heat
• Why do substances change phase or states?
• Why does temperature not change during a phase
change?
When the kinetic energy of the molecules become comparable to the
energy required for separation the molecules change there position
and separate (PE increase). This is a phase transition
Phase (State) Changes & Latent Heat
Solid  liquid  gas
• The relationship that describes heat energy (Q) gained/lost
to mass of substance (m) that undergoes a phase change is:
Q = m.Lf or Q = m.Lv
L is called the specific latent heat (the SI units are J kg-1)
• Def: Specific Latent Heat is heat (energy) required to
change the phase of a unit of mass (1 kg) of material at its
melting or vaporization point.
• Latent heat is a physical property that describes how much
energy is required to transform the mass of a substance from
one phase to another (e.g. latent heat of fusion or vaporization,
etc.)
Heat vs. Temperature Graph (for water)
Examples of
Phase Change
Question:
So why do you feel so “cold”
when you step out of the
shower soaking wet?
Figure 17-22
Heat Required for a Given Change in Temperature
Table 17-4
Latent Heats for Various Materials
Material
Latent heat of fusion,
Lf (J/kg)
Latent heat of
vaporization, Lv (J/kg)
Water
33.5 x 104
22.6 x 105
Ammonia
33.2 x 104
13.7 x 105
Copper
20.7 x 104
47.3 x 105
Benzene
12.6 x 104
3.94 x 105
Ethyl alcohol
10.8 x 104
8.55 x 105
Gold
6.28 x 104
17.2 x 105
Nitrogen
2.57 x 104
2.00 x 105
Lead
2.32 x 104
8.59 x 105
Oxygen
1.39 x 104
2.13 x 105
Examples: Phase Changes
• An ice cube of mass 25.0 g and a temperature of -10
◦C is dropped into a bowl of punch of mass 300.0 g
and temperature 20.0 ◦C.
What is the temperature eventually?
(Specific heat of fusion of ice = 2218 J kg-1K-1; latent heat of fusion of ice = 334 kJ
kg-1.)
Examples: Phase Changes
Game Plan!
• Increase temp of ice from -10 to 0.
– Mct
• Melt the ice cube in to water
– MLf
• Increase the temperature of the former Ice cube form 0 to Tf
– mct
Thus,
T = 11.9 degrees
Figure 17-14
A Liquid in Equilibrium with its Vapor
Evaporation
• Factors effecting evaporation rates
– Temperature
– Surface Area
– Vapor Pressure
• Air Flow
• Molecule concentrations
• What happens to the liquid when it evaporates?
• Is evaporation the same as boiling?
Drinking Bird
• Explain the
Thermodynamic
processes which
occur during one
cycle of the
drinking bird.