Entropy Change Calculator
Calculate entropy change for an isothermal process as heat divided by temperature, or for an isobaric process from the temperature ratio.
Entropy Change Calculator
Calculator
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Formula: ΔS = Q/T (isothermal) | ΔS = nCₚln(T₂/T₁) (isobaric)
Worked example — 3.33 J/K
Formula
ΔS = Q/T (isothermal) | ΔS = nCₚln(T₂/T₁) (isobaric)
Entropy measures disorder. Second law: total entropy of isolated system always increases.
Worked Examples
Example 1: Isothermal
Problem:Q=1000 J at T=300 K
Solution:ΔS=1000/300=3.33 J/K
Result:3.33 J/K
Frequently Asked Questions
What is entropy change?
Entropy change (ΔS) measures the change in disorder or randomness of a system. For a reversible isothermal process, ΔS = Q/T, where Q is heat transferred in joules (J) and T is the absolute temperature in kelvin (K). The SI unit of entropy is J/K (joules per kelvin).
How do you calculate entropy change?
For an isothermal process, use ΔS = Q/T. If 1000 J of heat flows into a system at 300 K: ΔS = 1000 / 300 ≈ 3.33 J/K. For an isobaric process with n moles of an ideal gas, ΔS = nCₚ ln(T₂/T₁). A positive ΔS means heat flows in; negative means heat flows out.
What units does the entropy change calculator use?
Heat transfer Q is entered in joules (J) — use positive values for heat absorbed and negative for heat released. Temperature T must be in kelvin (K), not Celsius; convert with T(K) = T(°C) + 273.15. The output ΔS is in J/K (joules per kelvin), the standard SI unit for entropy.
Can entropy decrease?
Locally, yes — a refrigerator lowers the entropy of its food compartment by pumping heat out of it, and freezing or crystallisation does the same to a sample. But the compressor’s work leaves the appliance as waste heat delivered to the warmer room, and dividing that larger heat by the higher room temperature adds more entropy than was removed, so the fridge-plus-room total still increases, in full compliance with the Second Law of Thermodynamics.
What is the difference between a reversible and an irreversible process for entropy?
The relation delta-S = Q/T applies to heat exchanged along a reversible path. For a real irreversible change, entropy is still a state function, so you compute delta-S along any convenient reversible path linking the same initial and final states. The Clausius inequality then says the actual heat divided by temperature is less than delta-S, with the shortfall representing entropy generated inside the system by friction, mixing or unrestrained expansion.
Why must temperature be entered in kelvin rather than Celsius?
Entropy is defined through the absolute temperature scale, so T must be measured from absolute zero. Convert with T(K) = T(C) + 273.15. Using Celsius produces badly wrong answers: 1000 J delivered at 27 C is 1000/300.15 = 3.33 J/K, but entering 27 would give 37.0 J/K, more than ten times too large. At 0 C the Celsius value would even divide by zero, and below it the sign would flip.
How do you calculate entropy change when the temperature itself changes?
Integrate dQ/T rather than dividing once. For a solid or liquid of mass m with roughly constant specific heat c, delta-S = m x c x ln(T2/T1). Heating 1 kg of water (c = 4186 J/kg K) from 300 K to 350 K gives delta-S = 4186 x ln(350/300) = 4186 x 0.1542 = 645 J/K. For an ideal gas use n x Cp x ln(T2/T1) at constant pressure or n x Cv x ln(T2/T1) at constant volume.
What is the entropy change during melting or boiling?
A phase transition happens at constant temperature, so the isothermal form applies directly with Q equal to the latent heat: delta-S = L/T. Melting one kilogram of ice at 273.15 K, with a latent heat of fusion of about 334 kJ/kg, gives delta-S = 334000/273.15 = 1223 J/K. The value is positive because a liquid has far more accessible molecular arrangements than an ordered crystal at the same temperature.
What is the entropy change when an ideal gas expands isothermally?
For n moles expanding from V1 to V2 at fixed temperature, delta-S = n x R x ln(V2/V1) with R = 8.314 J/mol K. Doubling the volume of one mole gives 8.314 x ln 2 = 5.76 J/K. Strikingly, a free expansion into vacuum transfers no heat and does no work, yet produces the same entropy change - proof that entropy depends only on the end states, not on the path taken.
What does entropy mean at the microscopic level?
Boltzmann's relation S = k_B x ln(W) connects entropy to W, the number of microscopic arrangements consistent with the observed macroscopic state, with k_B = 1.381 x 10^-23 J/K. Spreading energy over more accessible microstates raises S, which is why heat flows from hot to cold spontaneously. The third law follows naturally: a perfect crystal at absolute zero has a single arrangement, W = 1, so its entropy is zero.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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