ThermodynamicsTopic #21 of 35

Second Law and Entropy

Entropy, heat engines, refrigerators, and the Carnot cycle.

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.

ΔStotal0\Delta S_{\text{total}} \geq 0

Entropy

Definition

For a reversible process:

dS=dQrevTdS = \frac{dQ_{\text{rev}}}{T} ΔS=dQrevT\Delta S = \int \frac{dQ_{\text{rev}}}{T}
  • Unit: J/K
  • State function (path independent)

For Isothermal Process

ΔS=QT\Delta S = \frac{Q}{T}

For Temperature Change

ΔS=mcln(T2T1)=nCln(T2T1)\Delta S = mc \ln\left(\frac{T_2}{T_1}\right) = nC \ln\left(\frac{T_2}{T_1}\right)

For Ideal Gas

ΔS=nCvln(T2T1)+nRln(V2V1)\Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)

Or:

ΔS=nCpln(T2T1)nRln(P2P1)\Delta S = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right)

Heat Engines

Efficiency

η=WQH=QHQCQH=1QCQH\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}

Where:

  • QHQ_H = heat absorbed from hot reservoir
  • QCQ_C = heat rejected to cold reservoir
  • WW = net work done

Carnot Engine (Maximum Efficiency)

ηCarnot=1TCTH\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

No real engine can exceed Carnot efficiency!

Carnot Cycle

  1. Isothermal expansion at THT_H
  2. Adiabatic expansion (THTCT_H \to T_C)
  3. Isothermal compression at TCT_C
  4. Adiabatic compression (TCTHT_C \to T_H)

Refrigerators and Heat Pumps

Coefficient of Performance (COP)

Refrigerator:

COPref=QCW=QCQHQC\text{COP}_{\text{ref}} = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}

Heat Pump:

COPhp=QHW=QHQHQC\text{COP}_{\text{hp}} = \frac{Q_H}{W} = \frac{Q_H}{Q_H - Q_C}

Carnot COP

COPref,Carnot=TCTHTC\text{COP}_{\text{ref,Carnot}} = \frac{T_C}{T_H - T_C} COPhp,Carnot=THTHTC\text{COP}_{\text{hp,Carnot}} = \frac{T_H}{T_H - T_C}

Note: COPhp=COPref+1\text{COP}_{\text{hp}} = \text{COP}_{\text{ref}} + 1

Entropy Changes

Heat Transfer Between Reservoirs

When heat QQ flows from THT_H to TCT_C:

ΔStotal=QTCQTH>0\Delta S_{\text{total}} = \frac{Q}{T_C} - \frac{Q}{T_H} > 0

Irreversible Process

ΔStotal>0(entropy increases)\Delta S_{\text{total}} > 0 \quad \text{(entropy increases)}

Reversible Process

ΔStotal=0(entropy unchanged)\Delta S_{\text{total}} = 0 \quad \text{(entropy unchanged)}

For the Universe

Natural processes always increase total entropy.

Statistical Interpretation

Boltzmann Entropy

S=kBln(Ω)S = k_B \ln(\Omega)

Where:

  • kBk_B = Boltzmann constant = 1.38×10231.38 \times 10^{-23} J/K
  • Ω\Omega = 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=1TCTH=1300600=0.5=50%\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.5 = 50\%

Example 2: Real Engine

An engine takes in 1000 J of heat and does 300 J of work.

η=WQH=3001000=30%\eta = \frac{W}{Q_H} = \frac{300}{1000} = 30\% QC=QHW=700 JQ_C = Q_H - W = 700 \text{ J}

Example 3: Entropy of Melting

1 kg of ice melts at 0°C. Find entropy change.

ΔS=QT=mLfT=1×334000273=1223 J/K\Delta S = \frac{Q}{T} = \frac{mL_f}{T} = \frac{1 \times 334000}{273} = 1223 \text{ 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(TfTi)=1×4186×ln(323353)=373 J/K\Delta S_{\text{hot}} = mc \ln\left(\frac{T_f}{T_i}\right) = 1 \times 4186 \times \ln\left(\frac{323}{353}\right) = -373 \text{ J/K} ΔScold=mcln(TfTi)=1×4186×ln(323293)=+417 J/K\Delta S_{\text{cold}} = mc \ln\left(\frac{T_f}{T_i}\right) = 1 \times 4186 \times \ln\left(\frac{323}{293}\right) = +417 \text{ J/K} ΔStotal=373+417=+44 J/K>0\Delta S_{\text{total}} = -373 + 417 = +44 \text{ J/K} > 0 \checkmark

Example 5: Refrigerator

A refrigerator removes 400 J from food at 5°C, rejecting heat at 35°C.

Carnot COP:

COP=TCTHTC=278308278=9.3\text{COP} = \frac{T_C}{T_H - T_C} = \frac{278}{308 - 278} = 9.3

Minimum work required:

Wmin=QCCOP=4009.3=43 JW_{\min} = \frac{Q_C}{\text{COP}} = \frac{400}{9.3} = 43 \text{ 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=THTHTC=29320=14.65\text{COP}_{\text{hp}} = \frac{T_H}{T_H - T_C} = \frac{293}{20} = 14.65 QH=COP×W=14.65×1000=14,650 JQ_H = \text{COP} \times W = 14.65 \times 1000 = 14{,}650 \text{ J}

Practical Implications

  • No perpetual motion machines (second kind)
  • All real engines have η<ηCarnot\eta < \eta_{\text{Carnot}}
  • Waste heat is unavoidable
  • Energy quality decreases in all processes
  • Heat death of the universe (maximum entropy)