Irreversibility

Physics 313Professor Lee

CarknerLecture 16

Exercise #15 Carnot Engine Power of engine

= 1 – QH/QL

= W/QH

Source temp

= 1 – TL/TH

Max refrigerator COP

For a Carnot refrigerator operating between the same temperatures:

Since K < KC (8.2<9.9), refrigerator is possible

Entropy Entropy (S) defined by heat and

temperature

Total entropy around a closed reversible path is zero

Can write heat in terms of entropy:dQ = T dS

General Irreversibility

Since S = Sf - Si

Sf > Si

This is true only for the sum of all entropies

Since only irreversible processes are possible, Entropy always increases

Reversible Processes Consider a heat exchange between a system

and reservoir at temperature T So:

dSs = +dQ/T

dSr = - dQ/T

For a reversible process the total entropy change of the universe is zero

Irreversible Processes

How do you compute the entropy change for an irreversible process?

What is the change in entropy for specific irreversible processes?

Isothermal W to U

Friction or stirring of a system in contact with a heat reservoir

The only change of entropy is heat Q (=W) absorbed by the reservoir

S = W/T

Adiabatic W to U Friction or stirring of insulated

substance

System will increase in temperature

S = dQ/T = CPdT/T = CPln (Tf/Ti)

Heat Transfer Transferring heat from high to low T

reservoir

For any heat reservoir S = Q/T S for cool reservoir = + Q/TC

Assumes no other changes in any other

system

Free Expansion Gas released into a vacuum

Replace with a reversible isothermal

expansion Thus, (dQ/T) = (nRdV/V)

Note: Entropy increases even though temperature

does not change

Entropy Change of Solids

Solids (and most liquids) are incompressible

We can thus write dQ as CdT and dS as (C/T)dT If we approximate C as being constant with T

Note:

If C is not constant with T, need to know (and be able to integrate) C(T)

General Entropy Changes For fluids that under go a change in T, P or V we

can find the entropy change of the system by finding dQ

For example ideal gas:

dQ = CPdT – VdP

dQ = CVdT + PdV

These hold true for any continuous process involving an ideal gas with constant C

Notes on Entropy Processes can only occur such that S

increases

Entropy is not conserved

The degree of entropy increase indicates the degree of departure from the reversible state

Use of Entropy

How can the second law be used?

Example: total entropy for a refrigerator S (reservoir) = (Q + W) /TH

The sum of all the entropy changes must be greater than zero:

Use of Entropy (cont.)

We can now find an expression for the work:

Thus the smallest value for the work is:

Thus for any substance we can look up S1-

S2 for a given Q and find out the minimum amount of work needed to cool it