Hong Life CalculatorsUnderstanding the entropy change calculator — concept and principle
Entropy measures how energy is spread out among the possible microscopic states of a system. This calculator gives the entropy change when a substance of known mass and specific heat is heated or cooled from one temperature to another without changing phase. It is used in thermodynamics courses, in checking the second law for heat-exchange problems, and in estimating the irreversibility of mixing hot and cold water.
How it is calculated
ΔS = m c ln(T₂ / T₁)
- m: mass in kg.
- c: specific heat capacity in J/(kg·K); water is about 4186.
- T₁, T₂: start and end temperatures, entered in °C and converted to kelvin.
- ΔS: entropy change in J/K. It is positive for heating and negative for cooling.
For heat Q absorbed at a single constant temperature, such as during melting or boiling, the simpler rule ΔS = Q / T applies.
The principle
The thermodynamic definition is dS = δQ_rev / T: a small amount of heat added reversibly, divided by the absolute temperature at which it is added. When heating a substance with constant specific heat, δQ = m c dT. Integrating m c dT / T from T₁ to T₂ gives m c ln(T₂/T₁). The division by T means that the same joule of heat raises entropy more at low temperature than at high temperature. That asymmetry is what makes heat flow from hot to cold: the cold body gains more entropy than the hot body loses, so total entropy increases.
Worked example
- Inputs: m = 1 kg, c = 4186 J/(kg·K) (water), T₁ = 20 °C, T₂ = 80 °C.
- In kelvin: T₁ = 293.15 K, T₂ = 353.15 K.
- Ratio: 353.15 / 293.15 = 1.204673; ln(1.204673) = 0.186208.
- ΔS = 1 × 4186 × 0.186208 = 779.4686657 J/K.
- Check: heat added Q = 4186 × 60 = 251,160 J; divided by the mean temperature 323.15 K this gives about 777 J/K, close to the exact result.
| T₂ (°C) | ΔS heating (J/K) |
|---|---|
| 20 | 0 |
| 40 | 276.267149 |
| 60 | 535.4246277 |
| 80 | 779.4686657 |
| 100 | 1,010.065308 |
| 120 | 1,228.619444 |
| 160 | 1,634.212997 |
How to read the result
The sign tells you direction: positive when the substance gains heat. On its own, the entropy change of one body says nothing about whether a process is possible; the second law applies to the total change of the system plus surroundings, which must be zero or positive. For example, if the heat came from a reservoir at a constant 100 °C, the reservoir lost 251,160 / 373.15 ≈ 673 J/K, so the total entropy rose by about 106 J/K.
Common mistakes
- Using °C in the logarithm; ln(80/20) gives a value far too large.
- Applying the formula across a phase change; add Q / T for melting or boiling separately.
- Forgetting that specific heat varies with temperature over very wide ranges.
Frequently asked questions
What is the entropy change of boiling 1 kg of water at 100 °C?
Using a latent heat of about 2257 kJ/kg, ΔS = 2,257,000 / 373.15 ≈ 6050 J/K, far larger than the heating step.
Can entropy decrease?
The entropy of one object can decrease when it is cooled, but the total entropy of an isolated system never decreases.
Why is entropy measured in J/K?
It is heat divided by absolute temperature.
Sources: Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, McGraw-Hill · P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press