Download 20. Heat and the First Law of Thermodynamics

Physics 121
Chapter 20
Two bodies brought in thermal contact will change their temperature until they are at the
same temperature. In the process of reaching thermal equilibrium, heat is transferred from one
body to the other.
Suppose we have a system of interest at temperature TS surrounded by an environment
with temperature TE. If TS > TE heat flows from the system to the environment. If TS < TE
heat flows from the environment into the system. Heat, presented by the symbol Q and unit
Joule, is chosen to be positive when heat flows into the system, and negative if heat flows out of
the system (see Figure 20.1). Heat flow is a results of a temperature difference between two
bodies, and the flow of heat is zero if TS = TE.
Heat is not the only way in which energy can be transferred between a system and its
environment. Energy can also be transferred between a system and its environment by means of
work (W). The unit of work is the Joule.
Another commonly used unit is the calorie. The calorie is defined as the amount of heat
that would raise the temperature of 1 g of water from 14.5 C to 15.5 C. The Joule and the
calorie are related as follows: 1 cal = 4.1860 J
When heat is added to an object, its temperature increases.
Te
Te
Ts
Q<0
Te
Ts
Q=0
Ts > Te
The change in the
Ts
Q>0
Ts = Te
Figure 20.1. Heat Flow.
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Ts < Te
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Chapter 20
temperature is proportional to the amount of heat added
Q = C T f -T i
The constant C is called the heat capacity of the object. The heat capacity of an object depends
on its mass and the type of material of which it is made. The heat capacity of an object is
proportional to its mass, and the heat capacity per unit mass, c, is commonly used. In that case
Q = c m Tf - Ti
where m is the mass of the object. The molar heat capacity is the heat capacity per mole of
material. For most materials the molar heat capacity is 25 J/mol K.
In order to determine the heat capacity of a substance we not only need to know how
much heat is added, but also the conditions under which the heat transfer took place. For gases,
adding heat under constant pressure and under constant temperature will lead to very different
values of the specific heat capacity.
When heat is added to a solid or a liquid, the temperature of the sample does not
necessarily rise. During a phase change (melting, boiling) heat is added to the sample without
an increase in temperature. The amount of heat transferred per mass unit during a phase change
is called the heat of transformation (symbol L) for the process. The amount of heat
needed/released is given by
Q = Lm
where m is the mass of the sample.
Problem 28P
What mass of steam of 100C must be mixed with 150 g of ice at 0C, in a thermally
insulated container, to produce liquid water at 50C ?
We start with calculating the heat required to transform 150 g of ice at 0C to 150 g of
liquid at 0C. The heat of transformation of water is 333 kJ/kg (see Table 20-2, page 555). The
transformation of ice into water therefore requires a total heat given by
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Physics 121
Chapter 20
Q = L m = 50 kJ
The heat required to change the temperature of 150 g of water from 0C to 50C is given by
Q = c m T f - Ti = 31.5 kJ
The total heat that needs to be added to the system is therefore equal to 81.5 kJ. This heat must
be supplied by the steam. Heat will be released when the steam is transformed into liquid, The
heat of transformation for this process is 2260 kJ/kg. Suppose the mass of the steam is m. The
total heat released in the conversion of steam into water is given by
Q = L m = 2260 mkJ
The heat released when the steam cools down from 100C to 50C is given by
Q = c m T f - Ti = 210 m kJ
The total heat released by the cooling of the steam is therefore equal to 2470 m kJ. The total
heat required is 81.5 kJ, and we therefore conclude that the mass of the steam must be equal to
33 g.
Suppose a system starts from an initial state described by a pressure pi, a volume Vi, and a
temperature Ti. The final state of the system is described by a pressure pf, a volume Vf, and a
temperature Tf. The transformation from the initial state to the final state can be achieved in a
variety of ways (see for example Figure 20.2). In Figure 20.2a both pressure and volume change
simultaneously. In Figure 20.2b the pressure of the system is first lowered while keeping the
volume constant (this can for example be achieved by cooling the sample) and subsequently, the
volume is increased while keeping the pressure constant (this can be achieved by heating the gas
while increasing the volume).
If the pressure of a gas increases it can move a piston (this happens in an engine). In this
case, work is done by the system as the expanding gas lifts the piston. On the other hand, if we
increase the weight of the piston, work will be done on the system as the piston falls. The force
exerted by the gas on the piston is equal to p A, where A is the area of the piston and p is the gas
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Physics 121
Chapter 20
pressure. If the piston is displaced by a distance ds, the amount of work done can be calculated
as follows:
 = F  ds = (p A) ds = p dV
dW
The total work done during a finite displacement of the piston is now easy to calculate
W =
dW =
Vf
p dV
Vi
If W is positive, work was done by the system (for example, the expanding gas lifts the piston).
A negative value of W tells you that work was done on the system (the piston is pressed down in
order to compress the gas).
The amount of work done is equal to the area under the curve in the pV diagrams shown
in Figure 20.2. Clearly, the amount of work done depends on the path chosen. The work W for
the path shown in Figure 20.2a is significantly more than the work W for the path shown in
Figure 20.2b. Any change in the system in which the volume does not change will not
produce/cost any work. The work done for the paths shown in Figure 20.2 can be calculated
easily
Pressure
Pressure
(a)
(b)
Pi
Pi
Pf
Pf
Volume
Vi
Vf
Volume
Vi
Vf
Figure 20.2. Two possible ways to get from the initial state to the final state.
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Physics 121
W 2a =
Chapter 20
Vf
Vi
Vf
p dV =
Vi
pi - pf
Vi - Vf
V - Vf + p f dV =
1
p + p f V f - Vi
2 i
No work is done for the path shown in Figure 20.2b between (pi, Vi) and (pf, Vi) since there is no
change in volume. The work done to move from (pf, Vi) to (pf, Vf) is calculated easily
W2b =
Vf
Vi
p dV =
Vf
Vi
p f dV = p f Vf - Vi
Clearly, W2b is always less then W2a, and we can make the amount of work done as small or as
large as we want. For example no work would be done if the transition follows the following
path:
(pi, Vi) (0, Vi) (0, Vf) (pf, Vi)
A system can be taken from a given initial state to a given final state by an infinite number of
processes. In general, the work W and also the heat Q will have different values for each of
these processes. We say that heat and work are path-dependent quantities.
From the previous discussion neither Q nor W represents a change in some intrinsic
properties of the system. Experimentally, however, it is observed that the quantity Q - W is the
same for all processes. It depends only on the initial and final states and it does not matter at
what path is followed to get from one to the other. The quantity Q - W is called the change in
the internal energy U of the system:
∆U = Uf - Ui = Q - W
This equation is called the first law of thermodynamics. For small changes the first law of
thermodynamics can be rewritten as
dU = dQ - dW
If a system is well insulated, no transfer of heat will occur between it and its environment.
This means that Q = 0, and the first law of thermodynamics shows that
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∆U = - W
If work is done by the system (positive W) its internal energy decreases. Conversely, if work is
done on the system (negative W) its internal energy will increase. For gases, the internal energy
is related to the temperature: a higher internal energy means a higher temperature. Adiabatic
expansion of a gas will lower its temperature; adiabatic compression of a gas will increase
its temperature.
If the volume of a system is held constant, the system can do no work (W = 0 J). The
first law of thermodynamics then shows that
∆U = Q
If heat is added to the system its internal energy will increase; if heat is removed from the
system its internal energy will decrease.
Processes which, after certain interchanges of heat and work, are restored to their initial
state are called cyclical processes. In this case, no intrinsic properties of the system are changed,
and therefore ∆U = 0. The first law of thermodynamics now immediately yields
Q = W
Free expansion is an adiabatic process in which no work is done on or by the system.
This means that Q = W = 0 J, and the first law of thermodynamics now requires that
∆U = 0 J
The transfer of heat between a system and its environment can occur in a variety of ways.
Three different mechanisms of heat transfer will now be discussed: conduction, convection, and
radiation.
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Chapter 20
Consider the slab of material shown in Figure 20.3. The left end of the beam is
maintained at a temperature TH; the right end of the beam is maintained at a temperature TC. As
a result of the temperature difference heat will flow through the slab, from its hot end to its cold
end. Experimentally it is shown that the rate of heat transfer (Q/t) is proportional to the crosssectional area of the slab, proportional to the temperature difference, and inversely proportional
to the length of the slab
H =
T H - TC
Q
= kA
t
L
Here, k is the thermal conductivity, which is a constant that depends on the type of material.
Large values of k define good heat conductors. The thermal resistance R is related to the
thermal conductivity k in the following manner
R =
L
k
Thus, the lower the thermal conductivity of the material, the higher the thermal resistance R.
From the definition of R it immediately follows that
H =A
T H -T C
R
Figure 20.3. Conduction.
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Physics 121
Chapter 20
Figure 20.4. Heat transport through a composite slab.
Consider a composite slab is made up out of two different materials, with length L1 and L2, and
with thermal conductivity k1 and k2, is placed between two heat baths (see Figure 20.4).
Suppose that the temperature of the interface between the two slabs is equal to Tx. The amount
of heat flowing from TH to Tx is given by
H Hx = k 1 A
TH - Tx
L1
The amount of heat flowing from Tx to TC is given by
H xC = k 2 A
T x - TC
L2
Of course, the heat flowing through slab 1 must equal the heat flowing through slab 2. Thus
k1 A
TH - Tx
L1
= k2A
Tx -T C
L2
This equation can be used to obtain the temperature at the interface between slab 1 and slab 2:
Tx =
k 1 L 2 T C + k 2 L 1 TH
k1 L2 + k2 L 1
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Physics 121
Chapter 20
The heat flowing through the slab can now be calculated easily
H = k1A
TH - Tx
L1
= A
T H - TC
L1
k1
+
L2
=A
TH - TC
•R
k2
A composite slab therefore has a thermal resistance equal to the sum of the thermal resistance of
each of the individual slabs.
Heat transfer by convection occurs when a fluid, such as air or water, is in contact with an
object whose temperature is higher than the temperature of its surroundings. The temperature of
the fluid increases and (in most cases) the fluid expands. Being less dense than the surrounding
cooler fluid, it rises because of buoyant forces. The surrounding cooler fluid falls to take the
place of the rising warmer fluid and a convective circulation is set up.
Every object emits electromagnetic radiation. The energy spectrum of the radiation
emitted depends on the temperature of the object; the average energy increases with increasing
temperature.
Problem 57P
A container of water has been outdoors in cold weather until a 5.0 cm-thick slab of ice
has formed on its surface. The air above the ice is at -10C. Calculate the rate of formation of
ice (in centimeters per hour) on the bottom surface of the ice slab. Take the thermal
conductivity of ice to be 0.0040 cal/s . cm . C and the density to be 0.92 g/cm3.
The water at the boundary between the water and the ice will be at a temperature of 0C.
The heat transported through 5 cm of ice is equal to
T H - TC
Q
H =
= kA
= 0.0080 Acal/s
t
L
This heat is released when water is transformed into ice. The heat of transformation of this
process is 79.5 cal/g. Suppose a mass m of water is transformed each second into ice. This
produces a total heat equal to
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Physics 121
Chapter 20
H = 79.5 m cal/s
This must be equal to the heat flow through the ice:
79.5 m = 0.0080 A
A mass m of ice (covering an area A) will have a thickness d, where d is given by
d =
m
0.92 A
Combining the last two expression we obtain for the rate of ice formation:
d =
m
-4
= 1 x 10 cm/s = 0.36 cm/h
0.92 A
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Physics 121
Chapter 20
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