Chapter 11: Energy in Thermal
Process
Example homework problems: 11,17,27,38,65
Heat and Internal Energy
 Terminology
• Internal energy U is the energy associated with microscopic
components of a system – the atoms and molecules of the system.
It includes kinetic and potential energy associated with the random
translational, and vibrational motion of the particles that make up
the system, and any potential energy bonding the particles together.
• Heat Q is the transfer of energy between a system and its
environment due to a temperature difference between them.
Heat and Internal Energy
 Units
of heat
• Calorie (cal)
The energy necessary to raise the temperature of 1 g of water
from 14.5o to 15.5oC. “Cal” used for energy content in food is kcal.
• British thermal unit (Btu)
The energy necessary to raise the temperature of 1 lb of water
from 63o to 64oF.
• SI unit : joule (J)
1 cal = 4.186 J
Heat and Internal Energy
 Example
• Example 11.1 : Working off breakfast
Breakfast : two bowls of cereal and milk, 3.20x102 Cal
Exercise : Do curls with a 25.0-kg barbell to consume all the
calories from the breakfast
Breakfast calories:
3

1
.
00

10
cal  4.186 J 
2

E  (3.20 10 Cal)

 1.00 Cal  cal 
 1.34 106 J
Work done by one lifting or lowering barbell:
W  KE  PE  0  (mgh  0)  mgh
Number of repetitions n needed:
n(2mgh)  E
n
E
 6.84 103 times
2mgh
But see the remark on p.354!
Heat and Internal Energy
 Specific
heat
• Definition of specific heat
Q
c
mT
SI unit : joule per kilo-gram
degree Celsius
J/(kg oC)
Q  mcT
T=Tf-Ti
• The energy required to raise the temp.
of 0.500 kg of water by 3.00oC.
Q  (0.500 kg)( 4186 J/(kg C))(3.00C)
 6.28 103 J
Heat and Internal Energy
 Specific
heat
• Example 11.2 : Stressing a strut
A steel strut near a ship’s furnace :
L0=2.00 m, ms=1.57 kg, A=1.00x10-4 m2, Q=2.50x105 J.
(a) Find the change in temp. of the strut.
Q  ms cs T  T 
Q
 355C
ms cs
cs=448 J/(kg oC)
(b) Find the change in length of the strut if it is allowed to expand.
L  L0 T  7.8 10 6 m
=11x10-6 oC-1
(c) Find the compressional stress in the strut.
F
L
Y
 7.8  108 Pa
A
L0
Y=2.00x1011 Pa
Calorimetry
 Isolated
system
• A system whose energy does not leave out of the system is
called isolated system.
• The principle of energy conservation for an isolated system
requires that the net result of all the energy transfer is zero.
If one part of the system loses energy, another part has to
gain the energy.
 Calorimeter
and calorimetry
• Imagine a vessel made of good insulating material and containing
cold water of known mass and temperature and the temperature of
the water can be measure. Such a system of the vessel and water
is called calorimeter. If the object is heated to a higher temperature
of known value before it is put into the water in the vessel, the specific
heat of the object can be measured by measuring the change in
temperature of the water when the system (the object, vessel, and
water) reaches thermal equilibrium. This measuring process is called
calorimetry.
Calorimetry
 Calorimeter
and calorimetry
• When a warm object, put into a calorimeter with cooler water
described in the previous page, becomes cooler while the water
becomes warmer.
Qcold  Qhot
Qcold (>0 ) is the heat transferred (energy change) to the cooler
object and Qhot (<0) is the heat transferred (energy change) to the
warmer object.
• In general, in an isolated system consisting of n objects :
Tf common to all objects in equilibrium.
n
Q
k 1
k
 0 Qk  mk ck [(Tk ) f  (Tk ) i ]
Calorimetry
 Calorimeter
and calorimetry
• Example 11.3 : Finding a specific heat
A 125-g block of an unknown substance with a temperature of 90.0oC
is placed in a Styrofoam cup containing 0.326 kg of water at 20.0oC.
The system reaches an equilibrium temperature of 22.4oC. What is
the specific heat, cx , of the unknown substance if the heat capacity of
the cup is ignored.
Qcold  Qhot
mwcw (T  Tw )  mx cx (T  Tx )
cx 
mwcw (T  Tw )
 388 J/(kg C)
mx (Tx  T )
Calorimetry
 Calorimeter
and calorimetry
• Example 11.4 : Calculate an equilibrium temperature
Suppose 0.400 kg of water initially at 40.0oC is poured into a 0.300-kg
glass beaker having a temperature of 25.0oC. A 0.500-kg block of
aluminum at 37.0oC is placed in the water, and the system is insulated.
Calculate the final equilibrium temperature of the system.
Qw  Qal  Qg  0
mwcw (T  Tw )  mal cal (T  Tal )  mg cg (T  Tg )  0
T
mwcwTw  mal calTal  mg c g Tg
mwcw  mal cal  mg c g
 37.9C
Calorimetry
 Latent
heat and phase change
• Phase change (change of molecular structure)
- Phases of material : solid, liquid, gas
- A change of phase : phase transition
- For any given pressure a phase change takes place at a
definite temperature, usually accompanied by absorption
or emission of heat and a change of volume and density
Heated added and phase changes of water
Calorimetry
 Latent
heat and phase change
• Latent heat
- To change structure of material, change in energy required.
- When a substance goes under phase transition, all the heat
transferred is used to change the phase.
Q   mL
L : latent heat (J/kg) , m : mass (kg)
+(-) when energy is absorbed (removed).
Heated added and phase changes of water
Calorimetry
 Latent
heat and phase change
• Latent heat
- The latent heat of fusion Lf is used when a phase change occurs
during melting or freezing.
- The latent heat of vaporization Lv is used when a phase change
occurs during boiling or condensing.
Calorimetry
 Latent
heat and phase change
• Water
Consider an addition of energy to a 1.00-g cube of ice at -30.0oC in
a container held at constant pressure. Suppose this input energy
turns ice to steam (water vapor) at 120.0oC.
A: Q  mcice T , T  30.0C
 62.7 J cice  2090 J/(kg C)
5
B:Q  mL f , L f  3.33 10 J / kg
 333 J
C: Q  mcwater T , T  100C
 4.19 J cwater  4190 J/(kg C)
6
D: Q  mLv , Lv  2.26 10 J / kg
 2.26 103 J
E: Q  mcsteamT , T  20.0C
 40.2 J csteam  2010 J/(kg C)
Calorimetry
 Latent
heat and phase change
• Example 11.5 : Boiling liquid helium
Liquid helium has a very low boiling point, 4.2 K, as well as a low latent
heat of vaporization, 2.09x104 J/kg. If energy is transferred to a container
of liquid helium at the boiling point from an immersed electric heater at
a rate of 10.0 W, how long does it take to boil away 2.00 kg of the liquid?
Q  mLv  (2.00 kg)( 2.09 10 4 J/kg)  4.18 10 4 J
Q mLv 4.18 104 J
t  

 4.18 104 s  69.7 min
P
P
10.0 W
Calorimetry
 Latent
heat and phase change
• Example 11.6 : Ice water
6.00 kg of ice at -5.00oC is added to a cooler holding 30 liters of water
at 20.0oC. What temperature of the water when it comes to equilibrium?
mwater   waterV  30.0 kg
Q
Qice  Qmelt  Qicewater  Qwater  0 Q
ice
T  3.03C
m(kg) c(J/(kgoC)) L (J/kg) Tf
6.00 2090
0
Ti
Exp.
-5.00 mcT
3.33x105 0
0
mLf
Qice-water 6.00 4190
T
0
mcT
Qwater
T
Qmelt
6.00
30.0
4190
20.0 mcT
Calorimetry
 Latent
heat and phase change
• Example 11.7 : Ice water
A 5.00-kg block of ice at 0oC is added to an insulated container partially
filled with 10.0 kg of water at 15.0oC.
(a) Find the final temperature (neglect the heat capacity of the container.)
The amount of energy needed to completely melt the ice:
Qmelt  mice L f  (5.00 kg)(3.33 105 J/kg)  1.67 106 J
The maximum energy that can be lost by the initial mass of liquid water
without freezing it :
Qwater  mwater cT  (10.0 kg)(4190 J/(kg C))(0 - 15.0C)
 6.29 105 J
Since this is less than half the energy needed to melt all the ice, so the
final state of the system is a mixture of water and ice at the freezing point.
(b) Find the mass of the ice melted.
Qwater  mL f  m  1.89 kg
Energy Transfer
 Mechanism
of heat transfer
• Thermal conduction
• Convection
• Radiation
 Thermal
conduction
• One form of heat transfer when there is temperature difference is
thermal conduction.
• This process is caused by an exchange of kinetic energy between
microscopic particles such as molecules, atoms, and electrons etc.
In the process, less energetic particles gain energy as they collide
with more energetic particles.
The rate of energy transfer P=Q/t is
proportional to cross-sectional area A,
the difference in temperature T=Th-Tc>0,
and inversely proportional to the thickness
of the slab.
P
Q
T
A
t
x
Energy Transfer
 Thermal
conduction
• Consider a substance is in the shape of a long, uniform rod of length
L and assume that the rod is insulated so that thermal energy does not
escape from the surface.
P
T T
Q
 kA h c
t
L
k : proportionality constant
thermal conductivity
[J/(s m oC)]
Th : higher temperature (oC)
Tc : lower temperature (oC)
A : area (m2)
L : length (m)
Q : heat transferred (J)
t : time interval (s)
Energy Transfer
 Thermal
conduction
• Thermal conductivities
• Example 11.9 : Energy transfer through a concrete
wall
Find the energy transferred in 1.00 h by conduction
through a concrete wall 2.0 m high, 3.65 m long,
and 0.20 m thick if one side of the wall is held at
20oC and the other at 5oC.
Q  Pt  kA
Th  Tc
t
L
 (1.3 J/(s m C))(7.3 m 2 )
 2.6  10 6 J
15C
(3600 s)
0.20 m
Energy Transfer
 Thermal
conduction
• Example : Two rods cases
k1,L,,A1
k2,L2
k1,L1
Tm
k2,L,,A2
 Th  Tm 
 Tm  Tc 
Q



 k1 A
 k2 A

t
 L1 
 L2 
Q
A(k1 / L1 )t
Q
Tm  Tc 
A(k 2 / L2 )t
Th  Tm 
Q A(Th  Tc )

L1 L2
t

k1 k 2
Q  L1 L2 
  
Th  Tc 
At  k1 k2 
T T
Q1
 k1 A1 h c
t
L
T T
Q2
 k 2 A2 h c
t
L
T T
Q Q1  Q2

 (k1  k 2 ) h c
t
t
L
Energy Transfer
 Thermal
conduction
• Thermal conduction through multiple layers
L1
k1
T T
Q
P L1 P
P
 k1 A h1 c1  Th1  Tc1 
 R1
t
L1
A k1 A
L2
k2
L3
k3
P L2 P
Th 2  Tc 2  Tc1  Tc 2 
 R2
A k2 A
P L3 P
Th 3  Tc 3  Tc 2  Tc 3 
 R3
A k3 A
Th1  Tcn 
n
Li
P

A i 1 ki
Th1  Tcn  Th  Tc
Th1
Th2 Th3 Th3
Th
Tc1 Tc2 Tc2
Q A(Th  Tc ) A(Th  Tc )
P


t  Li / ki
 Ri
i
i
Energy Transfer
 Thermal
conduction
• Example 11.11 : Staying warm in the arctic
An arctic explore builds a wooden shelter out of wooden planks that
are 1.0 cm thick. To improve the insulation, he covers the shelter with
a layer of ice 3.2 cm thick.
(a) Compute the R factors for the wooden planks and the ice.
Rwood  Lwood / k wood  0.10 m 2sC Rice  Lice / kice  0.020 m 2sC
(b) Find the rate of heat loss.
P
Q A(Th  Tc )

 830 W
t
 Ri
i
(c) Find the temperature in between the ice and the wood.
k
A(Th  Tc )
P  wood
 830 W, Th  5.00C , Tc  T
L
T  16C
Energy Transfer
 Convection
• The transfer of energy by the movement of a substance is called
convection.
Energy Transfer
 Radiation
• All objects radiate energy because of microscopic movements
(accelerations) of charges, which increase with temperature.
Heat current in radiation
Q
P
 A  T 4
t
Stefan' s Law
 : emissivity that depends on nature of surface
T in K
A :area
=5.670400(40) x 10-8 W/(m2 K4) :Stefan-Boltzman const.
• If an object is at temperature T and its surroundings are at
temperature T0, the net flow of heat radiation between the
object and its surroundings is
Heat transfer by radiation
Q
4
4
   A(T  T0 )
t
Energy Transfer
 Radiation
• Example 11.12 : Polar bear club
A member of the Polar Bear Club, dressed only in bathing trunks
of negligible size, prepares to plunge into the Baltic Sea from the
in St. Petersburg, Russia. The air is calm, with a temperature of
5oC. If the swimmer’s surface body temperature is 25oC, compute
the net rate of energy loss from his skin due to radiation. How much
energy is lost in 10.0 min? Assume his emissivity is 0.900 and his
surface area is 1.50 m2.
T5C  5  273  278 K T25C  25  273  298 K
Pnet  A (T 4  T04 )  146 W
Q  Pnet  t  Pnet  600 s  8.76 10 4 J