Chapter 18 – Temperature, Heat and the First Law of Thermodynamics
This chapter deals with the concepts of temperature and heat.
Temperature
Qualitatively, temperature is a measure of how ‘hot’ or ‘cold’ an object is. If two objects
are in contact, then energy (heat) will flow from the hot to the cold object. If they are in
thermal equilibrium with each other, they are at the same temperature and there will be
no heat flow.
The zeroth law of thermodynamics says that if two objects (A and B) are each in thermal
equilibrium with a third object (C), then the two objects are in thermal equilibrium with
each other. While this may seem a bit abstract, it essentially says that we can use a
thermometer (object C) to decide if two objects (A and B) are at the same temperature.
A thermometer is a device that has some property that changes with temperature so that,
when properly calibrated, it can be used to measure temperature. One such device could
be a mercury-in-glass thermometer. As the temperature of the mercury increases the
mercury expands and this expansion in the glass is a measure of the temperature.
Another more fundamental thermometer is the pressure of a constant volume of a dilute
gas. In the Celsius scale (also called Centigrade) we define the temperature of an icewater mixture as 0oC and the temperature of boiling water as 100oC. Knowing the
pressure of the constant volume gas thermometer at these two points would allow one to
measure any temperature by extrapolation.
Absolute zero temperature
If the ice point is defined at 0oC and boiling water as 100oC, then the pressure of the gas
thermometer would extrapolate to zero at -273.15oC. This is referred to as absolute zero.
In the Kelvin scale, absolute zero is defined as 0 K, and the difference between the ice
point and boiling water is assumed to be 100 K. This makes the ice point 273.15 K and
boiling water 373.15 K.
In the Fahrenheit scale, the ice point is 32oF and boiling water is 212oF.
Conversions:
TC  TK  273.15
TF  9 TC  32
5
TC  TK , TF  9 TC
5
Question: What temperature is the same in the Celsius and Fahrenheit scales?
Thermal Expansion
Most objects expand with increase in temperature. For small temperature changes, the
change in length is
L   L0 T
where  is the coefficient of linear expansion.
Area expansion is given by
A   A0 T
and volume expansion is given by
V   V0 T
It can be shown that  = 2 and  = 3.
Example:
A steel railroad track consists of 4-m sections. The track reaches a low temperature of
-40oC in the winter and a high of 100oC in the summer. What gap should be allowed
between the sections? For steel,  = 11 x 10-6/oC.
L  (11x10 6 / o C )( 4m)(140 C )  6.2 x10 3 m  6.2 mm
Expansion of water
The thermal expansion of water is different from most materials. As the temperature is
increased above 4oC, it expands. But between 0oC and 4oC, water contracts with
increasing temperature. Water is most dense at 4oC. This is the reason why the top of a
pond freezes before the bottom.
Heat
Heat is the transfer of energy between a system and its environment because of a
temperature difference between them.
Internal energy is the energy associated with motions of the atoms and molecules of a
system and with the potential energy related to the bonding between atoms or molecules.
The molecular motions can be in the form of translation, rotation, and vibration.
If heat is transferred from one system to another, then the internal energy of one system
will increase and the other will decrease.
Units of heat
A calorie is the energy required to raise the temperature of 1 g of water from 14.5oC to
15.5oC.
1 cal = 4.186 J
One food calorie = 1 Calorie = 1000 calories. (Capital C for food calorie).
A Btu is the energy required to raise the temperature of 1 lb of water from 63oF to 64oF.
Example:
If you consume 2000 Calories of food in a day, then the energy you are consuming is
2000 Calories x 1000 cal/Cal x 4.186 J/cal = 8.37 x 106 J
Heat Capacity and Specific heat
The heat capacity, C, of a substance is the amount of energy, Q, required to raise the
temperature by 1 degree. Thus,
Q = CT
In the SI system, the units of C are J/oC. C is also often expressed in cal/oC.
Specific heat is defined as
c
Q
m T
Units are J/kg·oC or cal/g·oC. Whereas the heat capacity, C, depends on the amount of
the material, the specific heat, c, is an intrinsic property of the material and is
independent of the amount. By definition of the calorie, for water c = 1 cal/g·oC = 4186
J/kg·oC. Most substances have a smaller specific heat than that of water. From the above
equation, the heat required to raise the temperature of something by T is
Q  mcT
Example:
How much heat is required to raise the temperature of 100 g of water from 20oC to 50oC?
Q  (0.1kg)(4186 J / kg o C )(30o C )  12.6 x103 J
Example:
A 500-g block of aluminum initially at 200oC is placed in a bucket containing 1000 g of
water initially at 20oC. What would be the final equilibrium temperature of the block and
water if heat absorbed by the bucket and heat exchanged with atmosphere could be
neglected? The specific heat of aluminum is 0.215 cal/g·oC.
Q(lost by Al )  Q( gained by water)
m Al c Al (200 o C  T f )  m w c w (T f  20 o C )
(mw c w  m Al c Al )T f  m Al c Al 200 o C  mw c w 20 o C
Tf 

m Al c Al 200 o C  mw c w 20 o C
mw c w  m Al c Al
(500 g )(0.215cal / g o C )( 200 o C )  (1000 g )(1cal / g o C )( 20 o C )
(1000 g )(1cal / g o C )  (500 g )(0.215cal / g o C )
 37.5 o C
Latent heat
Latent heat refers to the energy required to cause the phase change of a substance without
changing the temperature. Examples of phase changes would be melting, vaporization,
sublimation, and change in crystal structure of a solid.
Latent heat of fusion is the energy per unit mass to melt a solid. Latent heat of
vaporization is the energy per unit mass required to boil a liquid.
L
Q
m
[cal/g, or J/kg]
The latent heat of fusion of water is about 80 cal/g. The latent heat of vaporization of
water is about 540 cal/g.
Example:
How much energy is required to convert 200 g of ice that is initially at 0oC to steam?
Q  Q(melt ice )  Q( warm water)  Q(boil water)
 mL f  mc(100 o C  0 o C )  mLv
 (200 g )(80cal / g )  (200 g )(1cal / g o C )(100 o C )  (200 g )(540cal / g )
 16000cal  20000cal  108000cal  144000 cal
Work and heat in thermodynamics processes
A gas in a cylinder can be compressed by pushing
on a piston, as shown to the right. The work done
on the gas by the external force F during an
infinitesimal volume change is given by
V
A
F
V
dWext   Fdx   PAdx   PdV
x
If the gas is compressed, then x and V are negative and the work done is positive.
(Note the negative sign in the above equation.) If the gas expands, then the work done on
the gas is negative. The work done by the gas during the volume change is the negative
of the work done on the gas and is given by
dW  PdV
The net work done by the gas during a finite volume change is
Vf
W   PdV
Vi
Isobaric process
In an isobaric process the pressure is constant. Thus, the work done by the substance is
W  PV  P(V f  Vi )
Isovolumeric process
In an isovolumeric (isochoric) process the volume is constant. Thus,
W  PV  0
If the pressure changes during the compression or expansion, then the work done by the
substance is the area under the P versus V curve.
Isothermal process
In an isothermal process the temperature is constant. We will see in the next chapter that
the relationship between the pressure, volume, and temperature of an ideal gas is given by
PV  nRT ,
where n is the number of moles and R is the ideal gas constant. Thus, the work done by
an ideal gas during an isothermal process is
Vf
Vf
Vf 
nRT
dV  nRT ln 

Vi V
 Vi 
W   PdV  
Vi
isobaric
P
i
isovolumetric
f
P
f
isothermal
P
i
f
i
V
V
V
First Law of Thermodynamics
The first law of thermodynamics relates the change in internal energy to the heat
absorbed and the work done by a substance. It is essentially a statement of conservation
of energy.
U  U f  Ui  Q  W
Q is positive if heat is absorbed and is negative if heat is lost by the system. (Note: Some
textbooks write the first law in terms of the work done on the system, in which case you
have U = Q + W.)
In an isolated system, Q = 0 and W = 0. Thus, U = 0 and the internal energy remains
constant.
The internal energy, U, depends on the state of the system. Thus, if a system goes
through a cycle and returns to its original state, then U = 0 and Q = W.
In the next chapter, we will see that for a monotonic gas the internal energy just depends
on T. Thus, for a monotonic gas undergoing an isothermal process, we also have U = 0
and Q = W.
An adiabatic process is one for which no heat flows into or out of a system. For an
adiabatic process, U = -W. Thus, in an adiabatic expansion the internal energy
decreases and in an adiabatic compression the internal energy increases.
Heat Transfer
Heat can be transferred from one system to another by
conduction, radiation, and convection.
Th
A
Heat transfer by conduction
Tc
Consider a slab of material with one face hot and one face
cold. The rate of energy transfer from the hot face to the
cold face is given by
P
x
(T  Tc )
Q
T
 kA
 kA h
t
x
L
The rate of energy transfer, P, is given in cal/s or J/s (watts). L is the thickness x.
The above equation assumes that the faces are at uniform temperature and that there is no
heat flow into or out of the sides.
The quantity k is the thermal conductivity and is a measure of how easily heat can flow
through a material. For example, k for copper would be much larger than for Styrofoam.
Example:
A room has a wall made of concrete with thickness 15 cm and areal dimensions 2.5 m x 3
m. The inside and outside temperatures are 25oC and 5oC. What is the rate of heat loss
through the wall? k(concrete) = 1.3 J/s·m·oC.
P  (1.3 J / s  m o C )( 2.5m x 3m)
(25 o C  5 o C )
 1300 W
0.15m
Heat flow through stacked materials
For layered materials, the rate of heat flow would be given by
PA
(Th  Tc )
 Li / k i
i
where Li and ki are the thickness and thermal conductivity of each layer and Th and Tc are
the temperatures of the outside surfaces of the stack of materials.
The quantity L/k for a material is called the “R” value. It is a measure of the resistance of
a material to the flow of heat. Units used for commercial materials for R are ft2·oF·h/Btu.
R depends on the material and its thickness. A good insulating material would have a
high R value. For a given thickness, fiberglass would have a much higher R value than
wood.
Heat transfer by radiation
All objects emit heat in the form of electromagnetic radiation. The rate depends on the
temperature of the object and the surface area. It also depends on the nature of the
surface. The rate of energy transfer (watts) is given by Stefan’s law:
P   AeT 4
 = 5.67 x 10-8 W/m2·K4 is a universal constant
A = area of object
T is the absolute (Kelvin) temperature.
e = emissivity. This depends on the nature of the surface and how well it radiates.
e is unitless and e  1.
An object will also absorb energy by radiation from its surroundings. If the surrounding
temperature is T0, then the net energy radiated will be given by
P   A e (T 4  T0 4 )
An object that is a good absorber is also a good radiator. A perfect radiator (and
adsorber) is called a ‘blackbody’ and has emissivity e = 1.
The electromagnetic radiation of an object occurs over a range of wavelengths. The
average wavelength decreases with increasing temperature. Much of the radiation of a
hot object may be in the infrared region.
Heat transfer by convection
An object can also lose or gain heat by the process of convection. This occurs because of
the movement of a liquid or gas in the vicinity of the object. For example, air can absorb
heat by direct contact with a radiator. This hot air will be expanded and will be buoyed,
thus rising and transferring heat elsewhere. You could also have forced convection by
blowing air onto a hot or cold object