Chapter 5
The First Law of
Thermodynamics
Introduction

THE FIRST LAW OF THERMODYNAMICS
– FOR A CONTROL MASS UNDERGOING A CYCLE
– FOR A CHANGE IN STATE OF A CONTROL MASS

INTERNAL ENERGY
– A THERMODYNAMIC PROPERTY

ENTHALPY
– THE THERMODYNAMIC PROPERTY

SPECIFIC HEATS
– CONSTANT-VOLUME
– CONSTANT-PRESSURE

INTERNAL ENERGY,ENTHALPY, AND
SPECIFIC HEAT OF IDEAL GASES

THE FIRST LAW AS A RATE EQUATION
CONSERVATION OF MASS
THE FIRST LAW OF THERMODYNAMICS
FOR A CONTROL MASS UNDERGOING
A CYCLE
– The First Law is often called the
conservation of energy law.
– The first law of thermodynamics states
that during any cycle a system (control
mass) undergoes, the cyclic integral of
the heat is proportional to the cyclic
integral of the work.
THE FIRST LAW OF THERMODYNAMICS FOR
A CHANGE IN STATE OF A CONTROL MASS
Three observations should be
made regarding this equation

The first is that the property E, the
energy of the control mass, was found to
exist.
– However, rather than deal with this property
E, we find it more convenient to consider the
internal energy and the kinetic and potential
energies of the mass.

The second is that Eqs. 5.10 and 5.11 are in
effect a statement of the conservation of
energy.
– The net change of the energy of the control mass
is always equal to
the net transfer of energy across the boundary as
heat and work.
– There are two ways in which energy can cross the
boundary of a control mass—either as heat or as
work.

The third is that Eqs. 5.10 and 5.11 can
give only changes in energy.

We can learn nothing about absolute values of these
quantities from these equations.
– If we wish to assign values to internal energy, kinetic energy,
and potential energy, we must assume reference states and
assign a value to the quantity in this reference state.
• The kinetic energy of a body with zero velocity relative to the
earth is assumed to be zero.
• Similarly, the value of the potential energy is assumed to be
zero when the body is at some reference elevation.
• With internal energy, therefore, we must also have a
reference state if we wish to assign values of this property.
5.3 INTERNAL ENERGY





Internal energy, kinetic and potential
energies are extensive properties,
because they depends on the mass of
the system.
U designates the internal energy of a
given mass of a substance, and u as
the specific internal energy.
The values are given in relation to an
arbitrarily assumed reference state,
which, for water in the steam tables, uf
is taken as zero for saturated liquid at
the triple-point temperature, 0.010C.
u = ( 1 –x ) uf + x ug
u = uf + x ufg
Example 5.3
5.4 PROBLEM ANALYSIS AND SOLUTION
TECHNIQUE




What is the control mass or control
volume?
What do we know about the initial state
(i.e., which properties are known)?
What do we know about the final state?
What do we know about the process that
takes place? Is anything constant or
zero? Is there some known functional
relation between two properties?




Is it helpful to draw a diagram of the
information in steps 2 to 4 (for example, a
T .or P–v diagram)?
What is our thermodynamic model for the
behavior of the substance (for example,
steam tables, ideal gas, and so on)?
What is our analysis of the problem (i.e., do
we examine control surfaces for various
work modes, use the first law or
conservation of mass)?
What is our solution technique? In other
words, from what we have done so far in
steps 1–7, how do we proceed to find
whatever it is that is desired? Is a trial-anderror solution necessary?
Example 5.4
5.5 ENTHALPY
Let us consider a control mass undergoing a quasiequilibrium constant-pressure process, as shown in Fig.
5.6.
[ specific enthalpy, h, and total enthalpy,
H. ]
(The significance and use of enthalpy is not restricted to the
special process just described. Other cases in which this same
combination of properties u +Pv appear will be developed later,
notably in Chapter 6 in which we discuss control volume
analyses.)
Students often become confused about the validity of this
calculation when analyzing system processes that do not
occur at constant pressure, for which enthalpy has no physical
significance.
We must keep in mind that enthalpy, being a property, is a
state or point function, and its use in calculating internal
energy at the same state is not related to, or dependent
on, any process that may be taking place.
Example 5.5
(Table at page 398, P2=400kPa,T 2=300oC)
5.6 THE CONSTANT-VOLUME AND CONSTANT-PRESSURE
SPECIFIC HEATS
5.7 THE INTERNAL ENERGY, ENTHALPY,
and SPECIFIC HEAT of IDEAL GASES
For a low-density gas, however, u depends primarily on T
and much less on the second property, P or v.
For example,

As gas density becomes so low that the idealgas model is appropriate, internal energy does
not depend on pressure at all, but is a function
only of temperature.
That is, for an ideal gas,
P v = R T and
u =f (T )
only
where the subscript 0 denotes the specific heat of
an ideal gas.
Calculate the change of enthalpy
------ as 1 kg of ideal gas is heated from T1 to T2 K
(1)
(2)
Empirical
Equation
(3)
Ideal-gas Entropy
(to integrate at constant
pressure of 0.1 MPa)
Example 5.6
Calculate the change of enthalpy as 1 kg of oxygen is
heated from 300 to 1500 K. Assume ideal-gas behavior.
Let us solve this problem in these ways and compare
the answers.
(3)
Our most accurate answer for the ideal-gas enthalpy
change for oxygen between 300 and 1500 K would be
from the ideal-gas tables, Table A.8. This result is,
using Eq. 5.29,
h2 - h1 = 1540.2 – 273.2 = 1267.0 kJ/ kg
Example
5.7
5.8 The First Law as a Rate Equation
( Dividing the first law equation by δt )
5.9 Conservation of Mass
E = mc2
Ex.
2900 KJ = m x ( 2.9979x10-8 )2
m = 3.23 x 10 –11Kg