Overview
The second law of thermodynamics establishes the direction of natural processes and introduces entropy, a measure of disorder in a system.
Statements of the Second Law
Clausius Statement
Heat cannot spontaneously flow from a colder body to a hotter body.
Kelvin-Planck Statement
No cyclic process can convert heat completely into work without other effects.
Entropy Statement
The total entropy of an isolated system never decreases.
ΔStotal≥0
Entropy
Definition
For a reversible process:
dS=TdQrev
ΔS=∫TdQrev
- Unit: J/K
- State function (path independent)
For Isothermal Process
ΔS=TQ
For Temperature Change
ΔS=mcln(T1T2)=nCln(T1T2)
For Ideal Gas
ΔS=nCvln(T1T2)+nRln(V1V2)
Or:
ΔS=nCpln(T1T2)−nRln(P1P2)
Heat Engines
Efficiency
η=QHW=QHQH−QC=1−QHQC
Where:
- QH = heat absorbed from hot reservoir
- QC = heat rejected to cold reservoir
- W = net work done
Carnot Engine (Maximum Efficiency)
ηCarnot=1−THTC
No real engine can exceed Carnot efficiency!
Carnot Cycle
- Isothermal expansion at TH
- Adiabatic expansion (TH→TC)
- Isothermal compression at TC
- Adiabatic compression (TC→TH)
Refrigerators and Heat Pumps
Coefficient of Performance (COP)
Refrigerator:
COPref=WQC=QH−QCQC
Heat Pump:
COPhp=WQH=QH−QCQH
Carnot COP
COPref,Carnot=TH−TCTC
COPhp,Carnot=TH−TCTH
Note: COPhp=COPref+1
Entropy Changes
Heat Transfer Between Reservoirs
When heat Q flows from TH to TC:
ΔStotal=TCQ−THQ>0
Irreversible Process
ΔStotal>0(entropy increases)
Reversible Process
ΔStotal=0(entropy unchanged)
For the Universe
Natural processes always increase total entropy.
Statistical Interpretation
Boltzmann Entropy
S=kBln(Ω)
Where:
- kB = Boltzmann constant = 1.38×10−23 J/K
- Ω = number of microstates
Higher entropy = more disorder = more possible arrangements
Examples
Example 1: Carnot Efficiency
A Carnot engine operates between 600 K and 300 K.
η=1−THTC=1−600300=0.5=50%
Example 2: Real Engine
An engine takes in 1000 J of heat and does 300 J of work.
η=QHW=1000300=30%
QC=QH−W=700 J
Example 3: Entropy of Melting
1 kg of ice melts at 0°C. Find entropy change.
ΔS=TQ=TmLf=2731×334000=1223 J/K
Example 4: Mixing Water
1 kg water at 80°C mixes with 1 kg water at 20°C.
Final temperature = 50°C = 323 K
ΔShot=mcln(TiTf)=1×4186×ln(353323)=−373 J/K
ΔScold=mcln(TiTf)=1×4186×ln(293323)=+417 J/K
ΔStotal=−373+417=+44 J/K>0✓
Example 5: Refrigerator
A refrigerator removes 400 J from food at 5°C, rejecting heat at 35°C.
Carnot COP:
COP=TH−TCTC=308−278278=9.3
Minimum work required:
Wmin=COPQC=9.3400=43 J
Example 6: Heat Pump
How much heat can a heat pump deliver using 1 kJ of work between 0°C and 20°C?
COPhp=TH−TCTH=20293=14.65
QH=COP×W=14.65×1000=14,650 J
Practical Implications
- No perpetual motion machines (second kind)
- All real engines have η<ηCarnot
- Waste heat is unavoidable
- Energy quality decreases in all processes
- Heat death of the universe (maximum entropy)