Entropy Change Calculator

Physics Thermodynamics • Heat Engines and Second Law

Written by STEM Calculators Team Published January 4, 2026 Updated February 24, 2026

4. Entropy Change Calculator

Compute entropy changes using $\Delta S = \dfrac{Q_{\rm rev}}{T}$ (reservoir / reversible steps) and $\Delta S = \int \dfrac{\delta Q_{\rm rev}}{T}$ . Includes system + surroundings and the total entropy change $\Delta S_{\rm tot} \ge 0$ .

Inputs support: pi, e, sqrt(), sin, cos, exp, log. Use * for multiplication.

Mode + Inputs

Process / scenario: Energy unit for Q: Heat $Q$ (positive = to system): Temperature $T$ (K) (for reversible heat / isothermal): Hot reservoir $T_h$ (K) (heat transfer mode): Cold reservoir $T_c$ (K) (heat transfer mode): Amount $n$ (mol) (ideal gas modes): $V_1$ (any units) (ideal gas modes): $V_2$ (same units) (ideal gas modes): Multi-step builder: Show ΔS_total check:

Sign convention: $Q>0$ means heat enters the system. Then the reservoir/surroundings entropy change is typically negative: $ \Delta S_{\rm surr} = -Q/T_{\rm res} $ (for a reservoir at constant temperature).

Multi-step builder (optional)

Add steps and the calculator will sum $ \Delta S_{\rm sys} $, $ \Delta S_{\rm surr} $, and $ \Delta S_{\rm tot} $. Use this to compare reversibility across a sequence.

Step type: Step $Q$: Step $T$ (K): Step $T_h$ (K): Step $T_c$ (K): Step $n$ (mol): Step $V_1$: Step $V_2$:

0 step(s)

#	Type	Inputs	$\Delta S_{\rm sys}$ (J/K)	$\Delta S_{\rm surr}$ (J/K)	$\Delta S_{\rm tot}$ (J/K)
No steps added.

Animation + diagram controls

Progress $\tau\in[0,1]$: Play speed: In-diagram labels: Show ΔS bars:

Entropy flow diagram illustrates the sign convention and the second-law check: reversible transfers can give $\Delta S_{\rm tot}=0$, while irreversible ones yield $\Delta S_{\rm tot}>0$.

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Steps

Enter values and click Solve.

Entropy flow diagram (animated)

System + surroundings entropy changes

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4. Entropy Change Calculator — Theory

Entropy $S$ is a thermodynamic state function. For any process, the system entropy change $\Delta S_{\rm sys}$ depends only on the initial and final states, even when the process itself is irreversible.

Core definition

\[ dS \;=\; \frac{\delta Q_{\rm rev}}{T} \] \[ \Delta S \;=\; \int_{1}^{2}\frac{\delta Q_{\rm rev}}{T} \]

Important: $\delta Q_{\rm rev}$ means heat along a reversible path between the same end states. Even if the real process is irreversible, you can compute $\Delta S$ by choosing a reversible “imaginary” path.

Reservoir (constant-temperature source/sink)

A reservoir stays at constant temperature $T_{\rm res}$. If it loses heat $Q$, its entropy decreases:

\[ \Delta S_{\rm res} = -\frac{Q}{T_{\rm res}} \]

System + surroundings and the second law

The “universe” entropy change is the sum:

\[ \Delta S_{\rm tot}=\Delta S_{\rm sys}+\Delta S_{\rm surr} \] \[ \Delta S_{\rm tot}\ge 0 \]

Reversible idealization: $\Delta S_{\rm tot}=0$.
Irreversible process: $\Delta S_{\rm tot}>0$.

Common processes included in the calculator

1) Reversible heat at constant temperature

If the system undergoes a reversible heat transfer at constant $T$:

\[ \Delta S_{\rm sys}=\frac{Q_{\rm rev}}{T} \] \[ \Delta S_{\rm surr}=-\frac{Q_{\rm rev}}{T} \quad\Rightarrow\quad \Delta S_{\rm tot}=0 \]

2) Heat transfer from hot to cold reservoirs (irreversible)

Consider heat $Q>0$ flowing from a hot reservoir $T_h$ to a cold reservoir $T_c$ with $T_h>T_c$.

\[ \Delta S_h=-\frac{Q}{T_h},\qquad \Delta S_c=+\frac{Q}{T_c} \] \[ \Delta S_{\rm tot} =Q\left(\frac{1}{T_c}-\frac{1}{T_h}\right) > 0 \]

3) Ideal-gas reversible isothermal

For an ideal gas expanding/compressing isothermally and reversibly:

\[ \Delta S_{\rm sys}=nR\ln\!\left(\frac{V_2}{V_1}\right) =nR\ln\!\left(\frac{P_1}{P_2}\right) \]

In a reversible isothermal process, the surroundings/reservoir entropy change is $-\Delta S_{\rm sys}$, so $\Delta S_{\rm tot}=0$.

4) Reversible adiabatic (isentropic)

If a process is reversible and adiabatic, then $\delta Q_{\rm rev}=0$ everywhere:

\[ \Delta S_{\rm sys}=0 \]

5) Free expansion of an ideal gas (irreversible)

Free expansion into a vacuum has $Q=0$ and $W=0$, but entropy increases because the gas occupies a larger volume. Using a reversible isothermal path between the same end states gives:

\[ \Delta S_{\rm sys}=nR\ln\!\left(\frac{V_2}{V_1}\right) > 0 \] \[ \Delta S_{\rm surr}=0 \quad\Rightarrow\quad \Delta S_{\rm tot}=\Delta S_{\rm sys}>0 \]

Sign conventions (used in the calculator)

$Q>0$: heat enters the system.
Reservoir/surroundings typically have $\Delta S_{\rm surr}=-Q/T_{\rm res}$ for a constant-$T$ reservoir.

Website tip

The entropy-flow arrows visualize where entropy “goes” and the total. Reversible idealizations make $\Delta S_{\rm tot}=0$, while irreversible cases (like hot→cold heat transfer or free expansion) produce $\Delta S_{\rm tot}>0$.

Steps

Entropy flow diagram (animated)

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