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Calculate Entropy Change During Phase Transitions

Q: What does ΔS = Q/T actually mean physically?

ΔS measures how much the microscopic 'disorder' or number of accessible microstates increases when a substance undergoes a phase change. Dividing heat Q (the energy driving the transition) by absolute temperature T normalizes it: at lower temperatures, the same heat causes a proportionally larger entropy jump because the system starts from a more ordered state. For example, melting 1 kg of ice at 273.15 K gives ΔS ≈ 1,223 J/K, reflecting the collapse of the ice crystal lattice.

Q: Why must temperature be in kelvin and not Celsius?

The formula ΔS = Q/T derives from the second law of thermodynamics where T is thermodynamic (absolute) temperature, defined in kelvin. The Celsius scale places 0 at water's freezing point — an arbitrary reference — so 0 °C does not mean 'zero thermal energy.' Using °C would give nonsensical or infinite results near 0 °C. Always add 273.15 to convert: 0 °C = 273.15 K, 100 °C = 373.15 K.

Q: What is Trouton's Rule, and how does it relate to this calculator?

Trouton's Rule (1884) states that the molar entropy of vaporization at the normal boiling point is approximately **85–88 J/(mol·K)** for most non-polar liquids. This means ΔS_vap = ΔH_vap / T_b ≈ 87 J/(mol·K). You can verify this with the calculator: for benzene, ΔH_vap = 30,720 J/mol at 353.2 K → ΔS = 87.0 J/(mol·K). Exceptions include water (~109 J/mol·K) due to hydrogen bonding, and ethanol (~110 J/mol·K).

Q: What is the difference between entropy change during melting vs. vaporization for water?

For 1 kg of water: **melting** at 273.15 K gives ΔS ≈ 1,223 J/K, while **vaporization** at 373.15 K gives ΔS ≈ 6,057 J/K — nearly 5 times larger. This reflects the dramatic increase in molecular freedom when liquid water becomes steam: gas-phase molecules have vastly more translational, rotational, and configurational microstates than liquid molecules, requiring much more energy input per degree of disorder increase.

Q: Can ΔS be negative, and does that violate the second law?

Yes — ΔS of the **system** can be negative during exothermic transitions like freezing or condensation (Q < 0). This does NOT violate the second law, which states that the total entropy of the **universe** (system + surroundings) must be ≥ 0. When water freezes, ΔS_system ≈ −1,223 J/K, but the heat released to the surroundings increases their entropy by at least 1,223 J/K, keeping ΔS_universe ≥ 0.

Q: How do I calculate ΔS for a quantity other than 1 kg?

Simply scale Q proportionally. The formula is Q = m × L, where m is mass and L is specific latent heat. For example, melting 500 g of ice: Q = 0.5 kg × 334,000 J/kg = 167,000 J; ΔS = 167,000 / 273.15 = **611.7 J/K**. For molar quantities, use Q = n × ΔH_fus (in J/mol) and the same T. The calculator accepts any Q value in joules.

Q: Is this formula valid for alloys, solutions, or mixtures?

Not directly. The formula ΔS = Q/T applies strictly to **pure substances** undergoing reversible phase transitions at a fixed temperature. Alloys and solutions typically melt over a temperature *range* (liquidus–solidus interval), so T is not constant and you must integrate ΔS = ∫(dQ/T). Additionally, mixing entropy (ΔS_mix = −R Σ xᵢ ln xᵢ) must be accounted for separately in multicomponent systems.

Q: What are typical units for entropy change, and how do J/K relate to J/(mol·K)?

Entropy change is reported in **J/K** (for a given mass) or **J/(mol·K)** (molar entropy change, also written J·mol⁻¹·K⁻¹). To convert: ΔS [J/(mol·K)] = ΔS [J/K] × M / m, where M is molar mass (g/mol) and m is mass in grams. For water (M = 18.015 g/mol): 1,223.4 J/K for 1,000 g → 1,223.4 × 18.015 / 1,000 = **22.04 J/(mol·K)** — the molar entropy of fusion, matching the accepted value of 22.0 J/(mol·K).

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Was this calculator helpful?

The Entropy Change Calculator computes ΔS — the change in thermodynamic entropy during a phase transition — using the fundamental formula ΔS = Q / T, where Q is the heat absorbed or released (in joules) and T is the absolute temperature at which the transition occurs (in kelvin). Entropy quantifies the degree of disorder in a system: when a solid melts, liquid evaporates, or any reversible phase change occurs at constant temperature and pressure, this ratio precisely captures how much thermodynamic "spreading" takes place per kelvin. Use this calculator whenever you need to determine the entropy change during melting, freezing, vaporization, condensation, or sublimation of a pure substance at its transition temperature.

Last reviewed: April 17, 2026 Verified by Martín Rodríguez Source: NIST WebBook — Thermophysical Properties of Fluid Systems, NIST Chemistry WebBook — Condensed Phase Thermochemistry Data, Wikipedia — Entropy (thermodynamics), Wikipedia — Latent Heat, Wikipedia — Trouton's Rule 100% private

When to use this calculator

Calculating the entropy change when 1 kg of ice melts at 0 °C (273.15 K), absorbing 334,000 J of latent heat — a classic thermodynamics benchmark.
Determining ΔS for the vaporization of water at 100 °C (373.15 K): Q = 2,260,000 J/kg → ΔS ≈ 6,057 J/K per kilogram, used in steam-cycle engineering.
Verifying Trouton's Rule in physical chemistry: most liquids have ΔS_vap ≈ 85–88 J/(mol·K) at their normal boiling point, helping predict boiling-point behavior.
Evaluating entropy production in industrial refrigeration cycles, where the refrigerant undergoes condensation and vaporization at known temperatures and heat loads.
Checking phase-transition data in materials science — e.g., sulfur's monoclinic-to-rhombic transition at 368.5 K with ΔH_trs ≈ 401 J/mol → ΔS ≈ 1.09 J/(mol·K).

Example Calculation

Ice melting: 334 kJ at 273K
ΔS = 334000/273 = 1223.8 J/K

Result: ~1224 J/K (ice → water, 1kg)

How it works

3 min read

How It Is Calculated

The entropy change for a reversible phase transition at constant temperature and pressure is given by:

ΔS = Q / T

Where:
  ΔS = entropy change (J/K)
  Q  = heat transferred during the phase change (J)
       Q > 0 → endothermic (melting, vaporization, sublimation)
       Q < 0 → exothermic (freezing, condensation)
  T  = absolute temperature of the transition (K)
       T(K) = T(°C) + 273.15

This formula is derived from the second law of thermodynamics for a reversible isothermal process: dS = δQ_rev / T. Because phase transitions occur at a fixed temperature (e.g., water always melts at 273.15 K at 1 atm), the integral reduces directly to Q/T. The heat Q equals the product of mass and specific latent heat: Q = m × L.

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Reference Table

Standard latent heats and resulting entropy changes for common substances at 1 atm:

Substance	Transition	T (K)	L (J/g)	Q for 1 kg (J)	ΔS for 1 kg (J/K)
Water (H₂O)	Melting (fusion)	273.15	334	334,000	1,223.4
Water (H₂O)	Vaporization	373.15	2,260	2,260,000	6,057.1
Ethanol (C₂H₅OH)	Vaporization	351.4	841	841,000	2,393.3
Nitrogen (N₂)	Vaporization	77.36	199	199,000	2,572.4
Iron (Fe)	Melting (fusion)	1,811	247	247,000	136.4
Sulfur (S)	Monoclinic→Rhombic	368.5	—	401 J/mol	1.09 J/(mol·K)
Carbon dioxide (CO₂)	Sublimation	194.7	571	571,000	2,932.7

Sources: NIST WebBook, CRC Handbook of Chemistry and Physics.

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Typical Cases

Case 1 — Melting Ice (the canonical example)

Substance: Water, 1 kg

Transition: Solid → Liquid at 0 °C

T = 273.15 K; Q = 1,000 g × 334 J/g = 334,000 J

ΔS = 334,000 / 273.15 = 1,223.4 J/K

Interpretation: The system gains ~1,223 J of entropy per kelvin — the crystalline hydrogen-bond network collapses into a disordered liquid.

Case 2 — Vaporizing Water (steam engineering)

Substance: Water, 1 kg

Transition: Liquid → Gas at 100 °C

T = 373.15 K; Q = 1,000 g × 2,260 J/g = 2,260,000 J

ΔS = 2,260,000 / 373.15 = 6,057 J/K

Interpretation: Nearly 5× more entropy is generated than during melting — gas molecules have vastly more translational freedom than liquid molecules.

Case 3 — Condensing Steam (refrigeration / HVAC)

Substance: Water vapor condensing to liquid at 100 °C

Q = −2,260,000 J (heat released to environment)

ΔS = −2,260,000 / 373.15 = −6,057 J/K

Interpretation: Negative ΔS confirms entropy decreases in the system (disorder decreases as gas → liquid), consistent with the second law only because the surroundings gain at least this much entropy.

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Common Errors

1. Using Celsius instead of Kelvin — The formula requires absolute temperature. Using T = 0 °C instead of 273.15 K would produce a division-by-zero or wildly incorrect result. Always convert: T(K) = T(°C) + 273.15.

2. Forgetting the sign of Q — Heat absorbed by the system is positive (endothermic); heat released is negative (exothermic). Vaporization has ΔS > 0; condensation has ΔS < 0. Mixing up the sign leads to wrong physical interpretation.

3. Applying ΔS = Q/T to non-isothermal processes — This formula is valid only at the phase-transition temperature where T is constant. For heating a substance through a temperature range, you must integrate: ΔS = ∫(Cp/T)dT, not use Q/T directly.

4. Confusing specific latent heat (J/g or J/kg) with molar latent heat (J/mol) — Always match the unit of L to the mass/mole quantity used. The molar enthalpy of fusion of water is 6,010 J/mol, while the specific value is 334 J/g; mixing these gives answers off by a factor of ~18 (the molar mass of water).

5. Ignoring pressure dependence — Standard latent heat values assume 1 atm. At different pressures (e.g., water boiling at 70 °C at altitude), both Q and T change, and the tabulated value of L no longer applies.

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Frequently asked questions

What does ΔS = Q/T actually mean physically?

ΔS measures how much the microscopic 'disorder' or number of accessible microstates increases when a substance undergoes a phase change. Dividing heat Q (the energy driving the transition) by absolute temperature T normalizes it: at lower temperatures, the same heat causes a proportionally larger entropy jump because the system starts from a more ordered state. For example, melting 1 kg of ice at 273.15 K gives ΔS ≈ 1,223 J/K, reflecting the collapse of the ice crystal lattice.

Why must temperature be in kelvin and not Celsius?

The formula ΔS = Q/T derives from the second law of thermodynamics where T is thermodynamic (absolute) temperature, defined in kelvin. The Celsius scale places 0 at water's freezing point — an arbitrary reference — so 0 °C does not mean 'zero thermal energy.' Using °C would give nonsensical or infinite results near 0 °C. Always add 273.15 to convert: 0 °C = 273.15 K, 100 °C = 373.15 K.

What is Trouton's Rule, and how does it relate to this calculator?

Trouton's Rule (1884) states that the molar entropy of vaporization at the normal boiling point is approximately 85–88 J/(mol·K) for most non-polar liquids. This means ΔS_vap = ΔH_vap / T_b ≈ 87 J/(mol·K). You can verify this with the calculator: for benzene, ΔH_vap = 30,720 J/mol at 353.2 K → ΔS = 87.0 J/(mol·K). Exceptions include water (~109 J/mol·K) due to hydrogen bonding, and ethanol (~110 J/mol·K).

What is the difference between entropy change during melting vs. vaporization for water?

For 1 kg of water: melting at 273.15 K gives ΔS ≈ 1,223 J/K, while vaporization at 373.15 K gives ΔS ≈ 6,057 J/K — nearly 5 times larger. This reflects the dramatic increase in molecular freedom when liquid water becomes steam: gas-phase molecules have vastly more translational, rotational, and configurational microstates than liquid molecules, requiring much more energy input per degree of disorder increase.

Can ΔS be negative, and does that violate the second law?

Yes — ΔS of the system can be negative during exothermic transitions like freezing or condensation (Q < 0). This does NOT violate the second law, which states that the total entropy of the universe (system + surroundings) must be ≥ 0. When water freezes, ΔS_system ≈ −1,223 J/K, but the heat released to the surroundings increases their entropy by at least 1,223 J/K, keeping ΔS_universe ≥ 0.

How do I calculate ΔS for a quantity other than 1 kg?

Simply scale Q proportionally. The formula is Q = m × L, where m is mass and L is specific latent heat. For example, melting 500 g of ice: Q = 0.5 kg × 334,000 J/kg = 167,000 J; ΔS = 167,000 / 273.15 = 611.7 J/K. For molar quantities, use Q = n × ΔH_fus (in J/mol) and the same T. The calculator accepts any Q value in joules.

Is this formula valid for alloys, solutions, or mixtures?

Not directly. The formula ΔS = Q/T applies strictly to pure substances undergoing reversible phase transitions at a fixed temperature. Alloys and solutions typically melt over a temperature range (liquidus–solidus interval), so T is not constant and you must integrate ΔS = ∫(dQ/T). Additionally, mixing entropy (ΔS_mix = −R Σ xᵢ ln xᵢ) must be accounted for separately in multicomponent systems.

What are typical units for entropy change, and how do J/K relate to J/(mol·K)?

Entropy change is reported in J/K (for a given mass) or J/(mol·K) (molar entropy change, also written J·mol⁻¹·K⁻¹). To convert: ΔS [J/(mol·K)] = ΔS [J/K] × M / m, where M is molar mass (g/mol) and m is mass in grams. For water (M = 18.015 g/mol): 1,223.4 J/K for 1,000 g → 1,223.4 × 18.015 / 1,000 = 22.04 J/(mol·K) — the molar entropy of fusion, matching the accepted value of 22.0 J/(mol·K).