Entropy Change — Physica

Equivalent forms

dS = \frac{\delta Q_{\text{rev}}}{T}

\Delta S = nC \ln\frac{T_f}{T_i}

\Delta S \geq \frac{Q}{T}

A quantity that never decreases in isolation — the arrow of time itself written as a mathematical inequality.

Symbol	Name	SI unit	Dimension	Range
$Delta_S$	Entropy changeoutput Change in entropy of the system	J/K	M·L²·T⁻²·Θ⁻¹	-10 – 10
$Q$	Heat transferred Heat energy transferred reversibly	J	M·L²·T⁻²	0 – 10000
$T$	Temperature Absolute temperature at which heat transfer occurs	K	Θ	50 – 2000

Unit systems

Symbol	SI	CGS	Imperial
$Delta_S$	J/K	erg/K	BTU/°R
$Q$	J	erg	BTU
$T$	K	K	°R

Where it holds

The integral form applies to reversible processes. For irreversible processes,

\Delta S

> Q/T (Clausius inequality). Statistical mechanics generalizes via

S = k_B \ln \Omega

.

Dimensional analysis

$[\Delta S] = [Q]/[T] \to M\cdot L^{2}\cdot T^{-2}\cdot \Theta ^{-1} = (M\cdot L^{2}\cdot T^{-2})/(\Theta ) = M\cdot L^{2}\cdot T^{-2}\cdot \Theta ^{-1} \checkmark$

Discovery

Rudolf Clausius · 1865

Clausius coined the word 'entropy' from the Greek τροπή (transformation). He stated: 'The entropy of the universe tends to a maximum' — one of the most profound sentences in physics.

Try this

Why can't you unscramble an egg — even though no energy is lost?

500 J of heat flows reversibly from a reservoir at 400 K to one at 300 K. Find the total entropy change of the universe.

Research status: stable

Real-world applications

Second law efficiency analysis of engines and refrigerators
Chemical equilibrium prediction (Gibbs free energy $G = H - TS)$
Information theory — Shannon entropy is mathematically identical to Boltzmann entropy
Black hole thermodynamics (Bekenstein-Hawking entropy)

Common misconceptions

Entropy is not 'disorder' in the colloquial sense — it is the number of microstates $(\Omega )$ consistent with the macrostate, measured by $S = k_B\cdot \ln (\Omega )$
Entropy of a subsystem CAN decrease (e.g., a refrigerator cools food), but the total entropy of the universe always increases
A reversible process $has \Delta S$ _universe $= 0$ , $not \Delta S$ _system $= 0$ — the system's entropy can change if compensated by the surroundings

Experimental verification

Verified by calorimetric measurements of heat exchange during phase transitions (ice melting at 273 K:

\Delta S = 334000/273 \approx 1224\,\mathrm{J}/(kg\cdot K)

per kg). Residual entropy of glasses and crystals with frozen-in disorder confirms the Third Law

(S \to 0

as

T \to 0

for perfect crystals).

Derivation

Clausius defined entropy via the reversible heat integral:

dS = \delta Q_rev/T

.

For a finite process:

\Delta S = \int (\delta Q_rev/T)

.

For constant temperature:

\Delta S = Q/T

.

For temperature change at constant heat capacity:

\Delta S = nC\cdot \ln (T_f/T_i)

.

Boltzmann later gave the statistical interpretation:

S = k_B\cdot \ln (\Omega )

, where

\Omega is

the number of microstates.

Limiting cases

T \to \infty

⟶

\Delta S \to 0

At very high temperature, adding heat causes negligible entropy change.

T \to 0^{+}

⟶

\Delta S \to \infty

Adding heat near absolute zero produces enormous entropy increase (Third Law prevents reaching

T = 0)

.

Q = 0

(adiabatic reversible)⟶

\Delta S = 0

(isentropic)No heat exchange in a reversible process means entropy is conserved.

What if…

What if the two reservoirs are at the same temperature?

$\Delta S$ _universe $= 0$ . Heat flow between equal-temperature bodies is reversible (and in practice, no net flow occurs). This is thermodynamic equilibrium.

What if entropy could decrease in an isolated system?

This would violate the Second Law. A gas could spontaneously compress into one corner, coffee could unmix from milk, eggs could unscramble. The overwhelming improbability of microstates lining up this way (for macroscopic systems) makes it effectively impossible.

1

Heat flow between reservoirs

Given ·

Q:: 500
T H:: 400
T C:: 300

Find · Delta_S_universe

Steps

Hot reservoir loses heat: $\Delta S_H = -500/400 = -1.25\,\mathrm{J/K}$
Cold reservoir gains heat: $\Delta S_C = +500/300 = +1.667\,\mathrm{J/K}$
$\Delta S$ _universe $= -1.25 + 1.667 = +0.417\,\mathrm{J/K}$ > 0 (irreversible, as expected)

Result ·

\Delta S = -Q/T_H + Q/T_C = -500/400 + 500/300 = -1.25 + 1.667 = 0.417\,\mathrm{J/K}

2

Heating water from 20°C to 80°C

Given ·

m:: 1
c:: 4186
T i:: 293
T f:: 353

Find · Delta_S

Steps

$Use \Delta S = mc\cdot \ln (T_f/T_i)$ for constant-pressure heating
$T_i = 293\,\mathrm{K}$ , $T_f = 353\,\mathrm{K}$
$\ln (353/293) = \ln (1.2048) = 0.1863$
$\Delta S = 1 \times 4186 \times 0.1863 = 779.9\,\mathrm{J/K}$

Result ·

\Delta S = mc\cdot \ln (T_f/T_i) = 1 \times 4186 \times \ln (353/293) = 1 \times 4186 \times 0.1863 = 779.9\,\mathrm{J/K}

First Law of Thermodynamics→Carnot Efficiency→Maxwell-Boltzmann Speed Distribution→