Entropy Change Calculator

Calculate entropy change for isothermal, heating, phase change, and mixing processes.

Water vaporization: 40,700 J/mol
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Entropy Tracks Energy Dispersal and Thermodynamic Irreversibility

Entropy is a state function whose change can be calculated along a convenient reversible path even when the real process is irreversible. For reversible heat transfer, dS=δQrev/T. Heating an idealized material with constant heat capacity gives ΔS=mc ln(T2/T1), while an isothermal phase change gives ΔS=Qrev/T=ΔH/T.

For an isolated system, total entropy cannot decrease. A reversible process produces no entropy, while real friction, finite-temperature heat transfer, mixing, and other irreversible processes generate entropy. Entropy change of one subsystem can be negative as long as the total change of system plus surroundings satisfies the second law.

dS=δQrev/T,   heating: ΔS=mc ln(T2/T1)
SymbolMeaningWhy it appears / units
SEntropyJ/K.
QrevReversible heat transferJ along a reversible path.
TAbsolute temperatureK.
cSpecific heat capacityJ/(kg·K) for the constant-c heating approximation.

Entropy is not directly “disorder” in a mechanical sense. In thermodynamics it quantifies the number or distribution of accessible energy states and constrains which macroscopic processes can occur spontaneously.

Entropy-change signs should follow the chosen system and process. Heating a body through a positive temperature rise gives a positive entropy change, while cooling gives a negative one. For mixing ideal components, the entropy of mixing should be nonnegative; a negative result usually indicates a sign or mole-fraction error.

Worked Examples

Example 1: Boiling water: ΔH_vap=40700 J/mol, T=373K
ΔS=40700/373
Result: 109 J/mol·K — entropy large increase on vaporization
Disorder increases dramatically: liquid→gas
Example 2: 50:50 ideal gas mixing
ΔS=−R(0.5ln0.5+0.5ln0.5)
Result: R ln2 = 5.76 J/mol·K
Maximum mixing entropy for equal moles
Example 3: Heating a solid
m=1kg, c=500J/kgK, 300K→600K
Result: ΔS=500ln2≈346.6J/K
Entropy depends on the temperature ratio rather than a simple ΔT in this model.
Example 4: Reversible melting
Q=10kJ at T=300K
Result: ΔS=33.3J/K
During an isothermal phase change, divide reversible latent heat by absolute temperature.

Common Mistakes

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Using Celsius inside entropy logarithms

Temperature ratios and denominators in thermodynamics require absolute Kelvin temperature.

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Equating zero heat transfer with zero entropy change in every case

Irreversible adiabatic processes can generate entropy even though Q=0.

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Requiring every subsystem’s entropy to increase

A subsystem can decrease in entropy if surroundings increase by enough to keep total entropy production nonnegative.

Frequently Asked Questions

2nd law and entropy?
ΔS_universe ≥ 0 for any real process (equality for reversible). Entropy always increases in isolated systems. This arrow of time explains why heat flows hot→cold but never spontaneously cold→hot.
Absolute entropy (3rd law)?
At absolute zero (0 K), entropy of a perfect crystal = 0 (3rd law). This allows calculation of absolute (not just relative) entropy values from calorimetric data.
Why can entropy be calculated using a reversible path?
Entropy is a state function, so its change depends only on initial and final equilibrium states. A hypothetical reversible path can be chosen for convenient calculation.
Does entropy always increase?
The entropy of an isolated total system cannot decrease. Individual systems can decrease if they exchange heat or matter with surroundings that gain more entropy.
What is entropy generation?
It is the positive entropy produced by irreversibilities such as friction, mixing, viscous flow, electrical resistance, and heat transfer across a finite temperature difference.
Why must temperature be in Kelvin?
Entropy relations derive from absolute thermodynamic temperature. Celsius zero is arbitrary and would make ratios or 1/T terms physically meaningless.
Why is Q/T used with reversible heat in entropy calculations?
Entropy is a state function, and the differential definition is dS=δQrev/T. For an irreversible real process, the actual heat transfer divided by boundary temperature is not generally equal to the system entropy change. Use a reversible path between the same states when evaluating ΔS from thermodynamic relations.

Formula Explorer connections

Interpretation: This formula tracks heat, temperature, work, entropy or transport in a thermodynamic system. Assumption: Use absolute temperature where required and consistent energy units. Constant properties, equilibrium, ideal gases or negligible losses may be assumed.

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