11. HEAT
Key Words: Calorie (cal), British Thermal Unit, Mechanical Equivalent
of Heat, Heat, Internal Energy or Thermal Energy, Specific Heat
Capacity, Phase Changes, Latent Heat, Heat of Fusion, Heat of
Vaporization, Calorimetry, Internal Energy, Heat Ideal Gas, Internal
Energy of an Ideal Gas, Kinetic Theory, Heat Transfer, Conduction,
Convection, Radiation, Thermal Conductivity, Thermal Conductors,
Thermal Insulators, Forced Convection, Natural Convection or Free
Convection, Stefan-Boltzmann Law, Stefan-Boltzmann constant,
Thermography.
In everyday life we commonly speak of the flow of heat. Heat flows
spontaneously from an object at higher temperature to one at lower
temperature. This observation lead to an 18th century model of heat as
movement of a fluid substance called caloric. Caloric was not observed
experimentally. Nevertheless, the common unit of heat, still in use today, is
named after caloric. It is called the Calorie (cal).
1kilocalorie = 1kcal = 10 3 cal
In the British system of units the unit of heat is called British Thermal Unit
(Btu).
1Btu = 0.252kcal
The next crucial step in understanding the nature of heat was done in the 19th
century. The idea that heat is related to energy was confirmed. The English
physicist Joule experimentally found that a given amount of work done was
always equivalent to a particular amount of heat input. Quantitatively, 4.186
J of work is equivalent to 1 cal of heat. This is known as the Mechanical
Equivalent of Heat.
1cal = 4.186 J
(11.1)
Experiments demonstrated that heat is not a substance, and not exactly a
form of energy. Heat is energy transferred form one object to another
because of a difference in temperature. Therefore, in SI units, the unit of
heat, as for any form of energy or transfer of energy, is the joule.
11.2. Internal Energy
We know that work is the transfer of mechanical energy. What type of
energy is transferred by heat? This type of energy is called Internal Energy
or Thermal Energy. Internal Energy, U, refers to the total energy of all
the molecules in an object. We derive the expression for internal energy
for an ideal monatomic gas. In this case, atoms are separated; we neglect
their potential energy of interaction. This is the model of an Ideal Gas.
Therefore, considering the Internal Energy of an Ideal Gas we should take
into account only the kinetic energy of each of N atoms.
The Kinetic Theory used this model and yielded the following expressions
for the Internal Energy of an Ideal Gas. In microscopical terms:
U = NK av =
⎛1
N ⎜ mv
⎝2
2
⎞
⎟
⎠
= N ⎛⎜ KT ⎞⎟ → U = NkT
3
⎝2
⎠
3
2
(11-2)
In macroscopical terms:
U=
3
3
3 N
3
R
NkT = N
T=
RT → U = nRT
2
2 NA
2 NA
2
(11-3)
If the gas molecules contain more than one atom, then the rotational and
vibrational energy of the molecules must also be taken into account. The
results of the Kinetic Theory describe not only ideal gas but also the
properties of the real gases at high temperature and low pressure. The
internal energy of liquids and solids is quite complicated, for it includes
electrical potential energy associated with the forces (or “chemical” bonds)
between atoms and molecules.
EXAMPLE 11.1 Exercises against excess calories.
The energy value of food is usually specified in Calories (with a capital C):
1Calorie = 1kcal = 1 ⋅ 10 3 cal
Suppose you ate two excess chocolate bars (~500 calories). To compensate,
you want to do an equivalent amount of work climbing stairs or a mountain.
How much total height must you climb?
m = 60.0kg
1Calorie = 1kcal = 4.186 ⋅ 10 3
J
cal
U = 500kcal
m
g = 9.80 2
s
h=?
Actually, you need to perform mechanical work W . You should elevate
your body:
W = mgh
This mechanical work should be equal to the energy value of chocolate:
U = 500kcal .
mgh = U
h=
U
=
mg
500kcal
(60kg )(9.80
m
s
2
=
(500kcal )(4.186 ⋅ 10 3
)
(60kg )(9.80
m
s2
J
)
cal
)
h = 3600m
Even the human body does not transform energy with 100% efficiency, you
can deduce from this simple example. It is easier to avoid excess calories by
decreasing food intake than to work off these calories by exercising.
11.3. Specific Heat
How do you calculate the amount of heat needed to increase the temperature
of an object? Now we will write the corresponding formula. The quantity
of heat Q required to raise the temperature of an object with mass m by a
small amount ΔT is:
Q = mCΔT
(11-4)
where C is the Specific Heat Capacity. So the amount of heat Q is
proportional to the temperature change ΔT , the mass of an object being
heated, and the material from which the object is made. The SI unit of
specific heat capacity is J (kg ⋅ K ) .
The physical sense of specific heat, C is the amount of heat needed to raise
the temperature of a unit of mass of the sample per unit temperature change
(ΔT = 1K or ΔT = 1C ° ).
11.4 Phase Changes. Latent Heat.
Heat is needed to increase the temperature of an object. It is also involved in
phase changes, such as the melting of ice or boiling of water. The process
of a phase change occurs at constant temperature (melting or boiling
temperature correspondingly).
To change a mass m of a material to a different phase at the same
temperature requires the addition or subtraction of a quantity of heat Q given
by:
Q = ± mL
(11-5)
Where L depends on phase change, the heat of fusion, heat of vaporization,
or heat of sublimation, which are called latent heats. Sublimation is the
process in which a substance evaporates directly from the solid to the
gaseous state. It is a comparatively rare event and we will focus our
attention on the cases of fusion and vaporization.
The heat required to change 1.0 kg of a substance from the solid to the liquid
state is called the Heat of Fusion Lƒ. For water, L f = 3.33 ⋅ 10 5 J kg . In the
case of liquid-gas phase transition, the corresponding physical quantity is
called Heat of Vaporization, Lv . For water, Lv = 2.26 ⋅ 10 6 J kg . The heats
of fusion and vaporization also refer to the amount of heat released by a
substance when it changes from a gas to a liquid or from a liquid to a solid.
Thus, steam releases 2.26 106 J/kg when it changes to water and water
releases 3.33 105 J/kg when it becomes ice.
11.5 Calorimetry
“Calorimetry” means “measuring heat.” The basic principle in calorimetry
calculations is as follows: If we have a system of objects isolated from the
environment and these objects are initially at different temperatures.
Finally, all of them will have the same temperature. The amount of heat lost
by some objects will be equal to the amount of heat gained by other objects.
Heat is energy in transit, so this principle is just another form of
conservation of energy specified to the situation. We take each quantity of
heat added to an object as positive and each quantity leaving an object as
negative. Thus, the basic principle of calorimetry can be stated as follows:
The algebraic sum of the quantities of heat transferred to all objects of an
isolated system must be zero.
∑Q = 0
(11-6)
EXAMPLE 11.2 A temperature change with no phase changes.
An aluminum pan with mass mAl = 1.5 kg is heated on a stove to TiAl =
180oC, then plunged into a sink containing mw = 8.0kg of water at room
temperature Tiw = 20oC. Assuming that none of the water boils, and that no
heat is lost to the surroundings, what is the final temperature Tf of the water
and pan?
m Al = 1.5kg
m w = 8.0kg
TiAl = 180 °C
Tiw = 20 °C
C Al = J
(kg ⋅ K )
C w = 4186 J (kg ⋅ K )
Tf = ?
Because there is not any phase change in this problem, we can write
equation (11-6) for the case of two objects in the form
m Al C Al (T f − TiAl ) + m w C w (T f − Tiw ) = 0
Solving for T f gives
(11-7)
Tf =
m Al C Al TiAl + m w C wTiw
m Al C Al + m w C w
(11-8)
T f = 26 °C
EXAMPLE 11.3 Changes in both temperature and phase.
You want to cool 0.25kg of spring water initially at 25oC by adding ice
initially at –20oC. How much ice should you add so that the final
temperature will be 0oC with all the ice melted if the heat capacity of the
container may be neglected?
The spring water loses the amount of heat
Because T f < Tiw and Qw < 0 the heat Qice that is needed to warm the ice to
the melting temperature 0oC is:
Qice = mice Cice (0°C − Tice )
Because 0°C > Tice , Qice > 0 , additional heat, Qice → w , is needed to melt this
mass of ice.
Qice− w = mice L f
Qice− w > 0
According to (11-6), the sum of these three quantities must equal zero.
∑ Q = Qw + Qice + Qice−w (11-7)
Solving (11-7) for mice, we get mice = 0.070kg = 70g
So, three of four medium size ice cubes at temperature –20oC are needed to
cool spring water.
11.6 Heat Transfer: Conduction, Convection, and Radiation
Conduction is transfer of energy of molecular motion within materials
without bulk motion of the materials. The kinetic energy of thermal motion
is transferred by molecular collision along the object, from the warmer end
(greater molecular speeds) to the colder one (smaller molecular speeds).
The heat transfer per unit time
T −T
ΔQ
= kA H C
L
Δt
(11-9)
Where k is the Thermal Conductivity of material, A is the cross-section of
a sample, L is its length, TH is the temperature of the hot end and TC is the
temperature of the cold end. Substances for which k is large are called
Thermal Conductors. Most metals fall into this category. Substances for
which k is small are called Thermal Insulators.
EXAMPLE 11.4 Heat loss through windows.
Calculate the rate of heat flow through a glass window 2.0m 1.5m in area
and 3.2mm thick if TH=15.0oC and TC=14.0oC.
A = (2.0m)(1.5m) = 3.0m2
L = 3.2•10-3m
k = 0.84
J
S ⋅m⋅Co
TH = 15oC
TC = 14oC
H=
ΔQ
?
Δt
H=
o
o
T −T
J
ΔQ
2 (15 C − 14 C )
= kA H C = [0.84
)(3.0m
)
Δt
L
s ⋅ m⋅Co
3 ⋅ 2 ⋅ 10 −3 m
H=
ΔQ
J
= 790
Δt
s
Convection involves the motion of a mass of fluid from one region to
another. Familiar examples include hot air and hot water home heating
systems, the cooling system of an automobile engine, and the heating and
cooling the body by the flow of blood. If the fluid is circulated by means of
pumps, the process is called Forced Convection. If the flow is caused by
differences in density, due to thermal expansion such as hot air rising, the
process is called Natural Convection or Free Convection. This type of
convection plays a dominant role in determining weather patters and
convection in the oceans. These processes are very complicated and could
not be described by one simple formula.
Radiation is energy transfer through electromagnetic radiation. This type of
heat transfer occurs without any medium. For example, all life on Earth
depends on the transfer of energy from the Sum to the Earth through empty
space. Radiation is described by the Stefan-Boltzmann Law. It is named
after the discoverers of this law – Austrian physicists Stefan and Boltzmann.
The rate at which an object radiates energy has been found to be
proportional to the fourth power of the Kelvin temperature. It is also
proportional to the area A of the emitting object. Finally, the StefanBoltzmann Law can be written as follows:
H=
ΔQ
= eσAT4
ΔT
(11-10)
where Б is called the Stefan-Boltzmann constant.
σ = 5.67 ⋅ 10 −8
W
m2 K 4
(11-11)
e is emissivity. It is a number between 0 and 1 that is characteristic of the
surface of the radiating material. Emissivity is larger for dark surfaces than
for light ones.
An interesting application of thermal radiation to diagnostic medicine is
Thermography. A special instrument, the thermograph, scans the body
measuring the intensity of radiation from many points and forming a picture
that resemble and x-ray. Areas, such as tumors, can be detected on a
thermogram as a result of their higher temperature and consequent increased
radiation. One use of the thermograph specifically is for the early detection
of breast cancer. The temperature of a tumor will often be 1°C to 2°C higher
than that of normal tissue. This difference is not so large but because the
rate of radiation energy is proportional to T4 (see equation (11-10)), the
difference in temperatures between tumor and normal tissue becomes
detectable by thermographs.