If the process is more irreversible, the more will be the increase of entropy of the universe. The increase of entropy quantifies the extent of irreversibility of the process. The entropy principle is a quantitative statement of 2nd law. We illustrate some applications of entropy principle in the following.
8.11.1 Heat Transfer Through a Finite Temperature Difference
Fig. 8.14 shows two reservoirs A and B with T1 & T2 temperatures respectively. Let T1 > T2, so heat (Q) will flow from A to B. So for the isolated system comprising of two reservoirs A & B and the rod, we can write
ΔS univ = ΔSA + ΔSB + ΔSC
= Q / T1 + Q / T2 + -Q + Q / T
= Q ( T1 – T2 / T1 T2 + 0
= Q ( T1 – T2 / T1 T2
Since T1 > T2, ΔS univ > 0 and the process is possible and irreversible. If T1 = T2, ΔS univ = 0, the process is reversible.
If T 2 > T 1, ΔS univ < 0 and the process is impossible.
8.11.2 Mixing of two Fluids
Subsystem A Subsystem B
mass m1 m2
specific heat c1 c2
temperature T1 T2.
There are two subsystems A and B as shown in Fig. 8.15. When the partition is removed, the two fluids mix together and at equilibrium let T1 be the final temperature. So, T2 < T1< T1. As there is no heat transfer through adiabatic wall, we have
Heat lost by A= Heat gained by B
m1c1 (T1- T1) = m2c2 (TF T2)
T f = m1 c1 T1 + m2 c2 T2 / m1 c1 + m2 c2
The change of entropy for subsystem A is
ΔS1 < 0 , since T f < T1.
The change of entropy for subsystem B is
T f > T2 . ΔS1, > 0
ΔS univ = ΔS1 + ΔS2
= m1 c1 In T f / T1 + m2 c2 T f / T2
In actual calculation the RHS of equation (8.21) is positive,
Thus ΔS univ > 0 and the mixing process
is irreversible. For integration the irreversible path is replaced by reversible paths.
For m1 = m2 = m (say), and c1 = c2 = c ( say)
ΔS univ = mc In Tf2 / T1 T2
TF = m1 c1 T1 + m1 c2 T2 / m1 c1 + m2 c2 = T1 + T2 / 2
Thus ΔS univ = 2 mc In ( T1 + T2 / 2√ T1 T2)
We have ΔS univ > 0 this is possible when
T1 + T2 > 2 √ T1 T2
T1 + T2 / 2 > √ T1 T2.
This is already known that arithmetic mean of any two numbers is greater than the geometric mean. The above statement can be proved geometrically as
r = T1 + T2 / 2
r21 = T1 T2
r1 = T1 T2
r > r1
T1 + T2 / 2 > T1 T2 proved.
The reverse of the statement i.e., if
T1 + T2 / 2 > T1 T2 the ΔS univ > 0
8.11.3 Maximum Work Out (from two Fixed Temperatures) by an Engine Working Reversibly
ΔS univ = 0
ΔS1 + ΔS2 = 0
or In T f / T1 + In T f / T2 = 0
or In Tf2 / T1 T2 = 0 = In 1
T f = T1 T2
When the system reaches TF the work done by the engine during that period is given by
W max= Heat lost by source- Heat gained by sink
= Q1 – Q2
= mc p ( T1 – TF) – mc p ( TF – T2)
= mc p [ ( T1 + T2) – 2 T1 T2]
= mc p ( T1 – T2)2
c p is the specific heat at constant pressure in (KJ/kg K).
8.11.4 External Mechanical Irreversibility
(i) Isothermal dissipation of work
Let I current flows through a resistor in
contact with reservoir. At steady state,
dU = 0,
- Q= 0+(-W)
dU = 0 for isothermal process
ΔS surr = Q / T = W / T
At steady state ΔS system = 0
ΔS univ = ΔS sys + ΔS surr
= O + ΔS surr
= ΔS surr = W / T
ΔS univ > 0 as W > 0.
Thus there is increase of entropy of the universe from this irreversible process.
(ii) Adiabatic dissipation of work
Let W is the work transfer to the system, according to
or dQ = dU + dW
0 = dU + dW
Or U2 – U1 = – ʃ d W = – W
ΔS univ = ΔS surr + ΔS sys
= 0 + mc p in T f / Ti
where c P is the specific heat at constant pressure.