Chapter 14: Heat
•Internal Energy
•Heat
•Heat Capacity
•Specific Heat
•Phase Transitions
•Thermal Conduction
•Thermal Convection
•Thermal Radiation
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§14.1 Internal Energy
The internal energy of a system is the sum total of all the
energy of all the molecules in the system. It does not
include macroscopic kinetic energy nor energy due to
external interactions (potential energy).
2
Internal energy includes:
•Translational and rotational kinetic energy of the
particles due to their individual motions.
•Vibrational kinetic and potential energy
•Potential energy due to interactions between particles in
the system.
•Chemical and nuclear energy (binding energies)
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Internal energy does not include:
•The kinetic energy of the molecules due to translations,
rotations, and vibrations of the whole or large fraction of
the system.
•Potential energy due to the interactions of the molecules
of the system with bodies outside of the system (external
interactions).
4
Example (text problem 14.5): A child of mass 15 kg climbs
to the top of a slide that is 1.7 m above a horizontal run that
extends for 0.5 m at the base of the slide. After sliding
down, the child comes to rest just before reaching the very
end of the horizontal portion of the slide. How much internal
energy was generated during this process?
U = mgh
KE = 0
1.7 m
U=0
KE = 0
5
Example continued:
The change in mechanical energy of the child is
ΔE= Ef – Ei = -mgh = -250 J. This is the increase in
internal energy and is distributed between the child, the
slide, and the air.
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§14.2 Heat
Heat is energy in transit between two systems at different
temperatures. Heat flows from the system at high
temperature to the system at low temperature.
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An experiment by Joule showed that the quantity of work
done on a system or the same quantity of heat flowing into
a system causes the same increase in the system’s internal
energy.
Heat is measured in joules or calories. 1 cal = 4.186 J
(the mechanical equivalent of heat); 1 Calorie (used on
food packaging) = 1 kcal.
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§14.3 Heat Capacity and Specific
Heat
For many substances, under normal circumstances ΔT∝Q.
Or Q = CΔT where C is the heat capacity.
The specific heat capacity, or just specific heat, of a
substance is the heat capacity per unit mass.
C
Q
c= =
m mΔT
or
Q = mcΔT
9
Example (text problem 14.12): If 125.6 kJ of heat are
supplied to 5.00×102 g of water at 22 °C, what is the final
temperature of the water?
Q = mcΔT = mc(Tf − Ti )
Q
Tf = Ti +
mc
125.6 kJ
= 22°C +
= 82°C
(0.5 kg )(4.186 kJ/kg °C)
10
Example (text problem 14.19): A 0.400 kg aluminum
teakettle contains 2.00 kg of water at 15.0 °C. How much
heat is required to raise the temperature of the water (and
kettle) to 100 °C?
The heat needed to raise the temperature of the water to
Tf is
Qw = mw cw ΔTw = (2 kg )(4.186 kJ/kg °C )(85 °C ) = 712 kJ.
The heat needed to raise the temperature of the aluminum
to Tf is
QAl = mAlcAl ΔTAl = (0.4 kg )(0.900 kJ/kg °C )(85 °C ) = 30.6 kJ.
Then Qtotal= Qw + QAl = 732 kJ.
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§14.4 Specific Heat of Ideal Gases
The average kinetic energy of a molecule in an ideal gas is
K tr
3
= kT .
2
And the total kinetic energy of the gas is
3
3
K tr = NkT = nRT .
2
2
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Define the molar specific heat at constant volume; this
is the heat capacity per mole.
Q
CV =
nΔT
Heat is allowed to flow into a gas, but the gas is not
allowed to expand. If the gas is ideal and monatomic, the
heat goes into increasing the average kinetic energy of
the particles.
13
3
The added heat is ΔK tr = Q = nRΔT .
2
3
nRΔT
Q
3
2
CV =
= R = 12.5 J/K/mol
=
nΔT
nΔT
2
5
If the gas is diatomic: CV = R = 20.8 J/K/mol
2
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Internal energy will be distributed equally among all possible
degrees of freedom (equipartition of energy). Each degree
of freedom contributes ½kT of energy per molecule and ½R
to the molar specific heat at constant volume.
Rotational motions of a 2-atom molecule:
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Example (text problem 14.28): A container of nitrogen gas
(N2) at 23 °C contains 425 L at a pressure of 3.5 atm. If
26.6 kJ of heat are added to the container, what will be the
new temperature of the gas?
For a diatomic gas, Q = nCV ΔT .
PiVi
.
The number of moles n is given by the ideal gas law n =
RTi
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Example continued:
The change in temperature is
⎛ Q ⎞⎛ RTi ⎞
⎟⎟⎜⎜
⎟⎟
ΔT = ⎜⎜
⎝ CV ⎠⎝ P iVi ⎠
⎛ 26.6 ×103 J ⎞
R(296 K )
⎟⎟
= ⎜⎜
5
2
−3
3
(
)
(
)
2
.
5
3
.
5
atm
1
.
013
×
10
N/m
/atm
425
L
10
m
/L
R
⎝
⎠
= 21 K
(
)
(
The final temperature of the gas is Tf = Ti + ΔT = 317 K
= 44 °C.
17
)
§14.5 Phase Transitions
A phase transition occurs whenever a substance changes
from one phase (solid, liquid, or gas) to another.
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Latent heat is the amount of heat per unit mass required
to change the phase of a substance.
The latent heat of fusion (Lf) is the heat per unit
mass needed to produce the solid-liquid phase
transition.
The latent heat of vaporization (Lv) is the heat
per unit mass needed to produce the liquid-gas
phase transition.
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Example (text problem 14.37): A 75 g cube of ice at
-10.0 °C is placed in 0.500 kg of water at 50.0 °C in an
insulating container so that no heat is lost to the
environment. Will the ice melt completely? What will be
the final temperature of this system?
The heat required to completely melt the ice is
Qice = mice cice ΔTice + mice Lf
= (0.075 kg )(2.1 kJ/kg °C )(10°C ) + (0.075 kg )(333.7 kJ/kg )
= 27 kJ
The heat required to cool the water to the freezing point is
Qw = mw cw ΔTw
= (0.5 kg )(4.186 kJ/kg °C )(50°C )
= 105 kJ
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Example continued:
Since Qice < Qwater the ice will completely melt.
To find the final temperature of the system, note that no
heat is lost to the environment; the heat lost by the water is
gained by the ice.
0 = Qice + Qw
0 = mice cice ΔT + mice Lf + mice cw (Tf − Tice,i ) + mw cw (Tf − Tw,i )
0 = mice cice ΔT + mice Lf + (mice + mw )cwTf − mw cwTi
0 = 27 kJ + (mice + mw )cwTf − 105 kJ
Tf = 32.4 °C
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Example (text problem 14.43): Compute the heat of fusion of
a substance from these data: 31.15 kJ will change 0.500 kg
of the solid at 21 °C to liquid at 327 °C, the melting point.
The specific heat of the solid is 0.129 kJ/kg K.
Q = mcΔT + mLf
Q − mcΔT
Lf =
= 22.8 kJ/kg
m
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On a phase diagram, the
triple point is the set of
P and T where all three
phases can coexist in
equilibrium.
Sublimation is the process by which a solid phase
transitions into a gas (and gas → solid).
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The critical point marks the end of the vapor pressure
curve. A path around this point (i.e. the path does not cross
the curve) does not result in a phase transition. Past the
critical point it is not possible to distinguish between the
liquid and gas phases.
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§14.6 Thermal Conduction
Through direct contact, heat can be conducted from regions
of high temperature to regions of low temperature. Energy
is transferred by collisions between neighboring atoms or
molecules.
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ΔT
The rate of energy transfer by conduction is P = κA
d
where κ is the thermal conductivity, A is the crosssectional area, and ΔT/d is the temperature gradient.
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d
= PR
Also ΔT = P
κA
where R is the thermal resistance.
This is convenient when heat is conducted through multiple
layers because
n
ΔTtotal = P ∑ Ri .
i =1
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Example (text problem 14.50): A metal rod with a diameter
of 2.30 cm and a length of 1.10 m has one end immersed in
ice at 0 °C and the other end in boiling water at 100 °C. If
the ice melts at a rate of 1.32 grams every 175 s, what is
the thermal conductivity of the metal? Assume no heat loss
to the surrounding air.
Qc
ΔT
= κA
.
Heat is conducted to the ice at a rate of P =
d
Δt
The heat conducted to the ice in a time period Δt is
ΔT
Qc = κA
Δt.
d
The heat needed to melt a given mass of ice is Q = mice Lf .
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Example continued:
Since all the heat conducted by the rod is absorbed by
the ice,
ΔT
Δt = mice Lf
κA
d
(
mice Lf
0.00132 kg )(1.10 m ) 333.7 × 103 J/kg
=
κ=
2
−2
ΔT
π 1.15 ×10 m (100 K )(175 s )
A
Δt
d
= 66.6 W/m K.
(
)
(
)
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Example (text problem 14.55): For a temperature difference
of 20 °C, one slab of material conducts 10.0 W/m2; another
of the same shape conducts 20.0 W/m2. What is the rate of
heat flow per m2 of surface when the slabs are placed side
by side with a total temperature difference of 20 °C?
For each slab, the thermal resistance per square meter is
R1 =
ΔT
20.0 K
2
=
=
2
.
00
K/W/m
P1
10.0 W/m 2
A
ΔT
20.0 K
2
R2 =
=
=
1
.
00
K/W/m
.
2
P2
20.0 W/m
A
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Example continued:
When the materials oneplaced in series, the rate of heat
flow is
P=
ΔT
2
∑R
i =1
ΔT
=
= 6.67 W/m 2 .
R1 + R2
i
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§14.7 Thermal Convection
Convection is the movement of heat by fluid currents.
Material is transported from one place to another.
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The rate of energy transport by convection is
P = hAΔT
where h is the coefficient of convection, A is a surface
area, and ΔT is the temperature difference between the
surface and convecting fluid.
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§14.8 Thermal Radiation
All bodies emit electromagnetic (EM) radiation. The perfect
absorber and emitter of EM radiation is called a blackbody.
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The rate of energy emission by a blackbody is (Stefan’s
Law)
P = AσT 4
where A is the surface area of the emitting body, T is its
temperature, and σ = 5.670×10-8 W/m2 K4 is the StefanBoltzmann constant.
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Since an ideal blackbody does not exist, Stefan’s law is
written as
P = eAσT 4
where e is the emissivity; e = 0 for a perfect reflector of
EM radiation and e = 1 for perfect blackbody.
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A spectrum shows the
amount of radiation emitted
at a particular wavelength.
For a blackbody, the peak of
the spectrum is determined
only by its temperature.
Wien’s law
λmaxT = 2.898 ×10 m K.
−3
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The net energy gained or lost by a blackbody at a
temperature T is
(
P = eAσT 4 − eAσTs4 = eAσ T 4 − Ts4
)
where Ts is the temperature of the surroundings.
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Example (text problem 14.69): A sphere with a diameter of
80 cm is held at a temperature of 250 °C and is radiating
energy. If the intensity of the radiation detected at a
distance of 2.0 m from the sphere’s center is 102 W/m2,
what is the emissivity of the sphere?
The power emitted by a point source is P = 4πd 2 I .
The emissivity is
P
4πd 2 I
e=
=
4
AσT
4πR 2σT 4
102 W/m 2 (2.0 m )2
=
(0.40 m )2 5.67 ×10 −8 W/m 2 K 4 (523 K )4
= 0.60.
(
(
)
)
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Summary
•Definition of Internal Energy
•Heat Capacity
•Specific Heat
•Phase Transitions
•Latent Heat
•Phase Diagrams
•Energy Transport by Conduction, Convection, and Radiation
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