Chapter 20 – First Law of Thermodynamics
This chapter deals with the concept of heat and the First Law of Thermodynamics.
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 = CΔT
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
mΔT
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 DT is
Q = mcΔT
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 )(30 o C ) = 12.6 x10 3 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 )
(m w c w + m Al c Al )T f = m Al c Al 200 o C + m w c w 20 o C
m Al c Al 200 o C + m w c w 20 o C
Tf =
m w 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
during an infinitesimal volume change is given by
A
ΔV
F
V
dW = − Fdx = − PAdx = − PdV
Δx
If the gas is compressed, then ΔV is 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. (Work is done by the gas.)
The net work done during a finite volume change is
Vf
W = − ∫ PdV
Vi
Isobaric process
In an isobaric process the pressure is constant. Thus, the work done is
W = − PΔV = P(Vi − V f )
Isovolumeric process
In an isovolumeric (isochoric) process the volume is constant. Thus,
W = − PΔV = 0
If the pressure changes during the compression or expansion, then the work done on the
substance is the area under the P versus V curve.
Isothermal process
The work done on a gas during an isothermal (constant temperature) process is
Vf
⎛V
nRT
dV =nRT ln⎜ i
⎜V f
Vi V
⎝
Vf
W = − ∫ PdV = − ∫
Vi
isobaric
P
f
⎞
⎟
⎟
⎠
isovolumetric
i
P
i
isothermal
P
f
i
f
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 on a substance. It is essentially a statement of conservation of energy.
ΔU = U f − U i = Q + W
Q is positive if heat is absorbed and is negative if heat is lost by the system.
In an isolated system, Q = 0 and W = 0. Thus, DU = 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 DU = 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 DU = 0 and Q = -W.
An adiabatic process is one for which no heat flows into or out of a system. For an adiabatic
process, ΔU = 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
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=
Tc
Dx
(T − Tc )
ΔT
Q
= kA
= kA h
Δt
Δx
L
The rate of energy transfer, P, is given in cal/s or J/s (watts). L is the thickness Dx.
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 = (1.3J / s ⋅ m ⋅o C )(2.5m x 3m)
(25 o C − 5 o C )
= 1300W
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 and 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 = σ A eT 4
σ = 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 = σ 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