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 T_{1} & T_{2 }temperatures respectively. Let T_{1} > T_{2}, 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 }= ΔS_{A} + ΔS_{B} + ΔS_{C}

= Q / T_{1} + Q / T_{2} + -Q + Q / T

= Q ( T_{1} – T_{2 }/ T_{1} T_{2} + 0

= Q ( T_{1} – T_{2} / T_{1} T_{2}

Since T_{1} > T_{2}, ΔS _{univ} > 0 and the process is possible and irreversible. If T_{1} = T_{2}, 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 m_{1} m_{2}

specific heat c_{1} c_{2}

temperature T_{1 } T_{2}.

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 T_{1} be the final temperature. So, T_{2} < T_{1}< T_{1.} As there is no heat transfer through adiabatic wall, we have

Heat lost by A= Heat gained by B

m_{1}c_{1} (T_{1}- T_{1}) = m_{2}c_{2} (T_{F} T_{2})

T _{f }= m_{1} c_{1} T_{1} + m_{2 }c_{2} T_{2} / m_{1} c_{1} + m_{2} c_{2}

The change of entropy for subsystem A is

ΔS_{1} < 0 , since T _{f} < T_{1}.

The change of entropy for subsystem B is

T _{f} > T_{2} . ΔS_{1}, > 0

ΔS univ = ΔS_{1} + ΔS_{2}

= m1 c1 In T _{f} / T_{1} + m_{2} c_{2} T _{f }/ T_{2}

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 m_{1 }= m_{2} = m (say), and c_{1} = c_{2} = c ( say)

ΔS _{univ} = mc In Tf_{2} / T_{1} T_{2}

T_{F} = m_{1} c_{1} T_{1} + m_{1} c_{2} T_{2} / m_{1} c_{1} + m_{2} c_{2} = T_{1} + T_{2} / 2

Thus ΔS _{univ} = 2 mc In ( T_{1} + T_{2} / 2√ T_{1} T_{2})

We have ΔS _{univ} > 0 this is possible when

T_{1} + T_{2} > 2 √ T_{1} T_{2}

T_{1} + T_{2} / 2 > √ T_{1} T_{2}.

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 = T_{1} + T_{2} / 2

r2_{1} = T_{1} T_{2}

r_{1} = T_{1} T_{2}

r > r_{1}

T_{1} + T_{2} / 2 > T_{1} T_{2} proved.

The reverse of the statement i.e., if

T_{1} + T_{2} / 2 > T_{1} T_{2} the ΔS _{univ} > 0

**8.11.3 Maximum Work Out (from two Fixed Temperatures) by an Engine Working Reversibly**

** **ΔS univ = 0

ΔS_{1} + ΔS_{2} = 0

or In T _{f} / T_{1} + In T _{f} / T_{2} = 0

or In Tf_{2} / T_{1} T_{2} = 0 = In 1

T _{f }= T_{1} T_{2}

_{ }

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

= Q_{1} – Q_{2}

= mc _{p} ( T_{1} – T_{F}) – mc _{p} ( T_{F }– T_{2})

= mc _{p} [ ( T_{1} + T_{2}) – 2 T_{1} T_{2}]

= mc _{p }( T_{1} – T_{2})_{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=dU+W

- Q= 0+(-W)

dU = 0 for isothermal process

Q =W

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

1st law,

or dQ = dU + dW

0 = dU + dW

Or U_{2} – U_{1} = – ʃ d W = – W

ΔS _{univ} = ΔS _{surr} + ΔS _{sys}

= 0 + mc _{p} in T _{f }/ T_{i}

where c P is the specific heat at constant pressure.