Entropy Change Calculator
Compute entropy change from heat and temperature, or from heating a mass.
What it does
Entropy counts how many ways a system can arrange its energy. When heat moves or a body changes temperature, entropy moves with it, and thermodynamics measures the shift in joules per kelvin.
Two expressions cover the common cases. Heat crossing a boundary at a steady, reversible temperature adds Q/T. A mass warming or cooling adds m times c times the logarithm of the temperature ratio, because each extra joule buys less disorder when the body is already hot.
The defaults show both faces. Ten joules entering a 300 K reservoir raises entropy by 0.0333 J/K, a tiny trickle. Four kilograms of steel cooling from 400 K to 350 K sheds 240.4 J/K, a much larger swing from a modest temperature change.
The sign describes the system you chose, not the universe. A cooling steel block loses entropy locally while the surrounding air gains more than it lost, which is how the second law stays satisfied.
The formulas assume reversible conditions and constant specific heat. Melting, boiling, and chemical reactions need latent heat or reaction terms on top. This page computes the two core cases with a full worked example and practice checks.
How to use
- Pick heat mode for flow at a steady temperature.
- Pick the heating mode to move a mass between two temperatures.
- Enter heat in joules, negative if it leaves.
- Or enter mass, specific heat, and both temperatures in kelvin.
- Read the entropy change and its direction.
- Check the sign against what the system physically did.
Formula
- Q
- Heat entering the system in joules; negative when heat leaves.
- T
- Absolute temperature in kelvin for the constant temperature case.
- m
- Mass in kilograms.
- c
- Specific heat in J/(kg·K): 450 for steel, 4186 for water.
- T1, T2
- Start and end temperatures in kelvin; T2 above T1 means heating.
Temperatures must be kelvin, because entropy is defined from absolute temperature and the logarithm of a Celsius ratio is meaningless. The Q/T form applies strictly to reversible heat flow at a constant T; irreversible processes generate at least this much entropy, never less. Positive ΔS is the usual result, but a negative number is fine for a system that exports heat, as long as the surroundings gain more.
Worked example
Example: 10 J at 300 K, and steel cooling
Heat mode: 10 J added at 300 K (defaults). Temperature mode mirror: 4 kg of steel, c = 450, from 400 K to 350 K.
- Heat mode: divide 10 J by 300 K.
- ΔS = 10 / 300 = 0.03333 J/K, shown as 0.0333 J/K.
- The system absorbs heat, so entropy increases.
- Temperature mode: ratio 350 / 400 = 0.875.
- ln(0.875) = -0.1335; 4 x 450 x -0.1335 = -240.4 J/K.
- Cooling lowers the system's entropy; the air around it gains more.
Reference table
| Process | Entropy change |
|---|---|
| 10 J into 300 K | +0.0333 J/K |
| 450 J into 300 K | +1.5 J/K |
| 2 kg water, 293 to 373 K | +2021 J/K |
| 1 kg water, 273 to 373 K | +1306.5 J/K |
| 4 kg steel, 400 to 350 K | -240.4 J/K |
Limits
- Assumes reversible heat flow for the Q/T case.
- Constant specific heat only; real c drifts with temperature.
- No phase changes, which need latent heat terms instead.
- System level answer: the surroundings need their own calculation.
Common mistakes
- Using Celsius temperatures in the logarithm.
- Reporting a negative ΔS as a second law violation.
- Applying Q/T to a violent irreversible process as if it were reversible.
- Skipping the mass when the whole body changes temperature.
- Ignoring melting or boiling, where latent heat dominates the budget.
Try it yourself
Quick checks. Answers sit tucked under each question.
450 J added at 300 K?
+1.5 J/K, entropy increases.
600 J removed at 200 K?
-3 J/K, entropy decreases.
2 kg of water (c = 4186) heated from 293 K to 373 K?
+2021 J/K.
Questions
What exactly is entropy?
A measure of how spread out energy and microstates are. The second law says the total for an isolated system never falls.
Why does the heating case use a logarithm?
Each joule added to a hot body changes the state count less than the same joule added cold. Integrating dQ/T over the temperature change produces the natural log of the ratio.
Can ΔS be negative?
Yes, for the system you picked, like cooling steel. The surroundings must gain at least as much, so the universe total still rises.
Why kelvin and not Celsius?
Entropy divides heat by absolute temperature. Kelvin starts at absolute zero, where the ratio is defined; Celsius can cross zero and break the math.
What if T2 is below T1?
The ratio drops below one, the log turns negative, and the formula reports cooling. The tool labels the direction for you.
Do phase changes fit?
Not with these two forms alone. Ice melting holds temperature constant but still consumes latent heat, which needs an extra term like mass times latent heat over temperature.
Sources
- NASA Glenn Research Center: second law and entropy
- OpenStax University Physics volume 2, section 4.6: entropy