Library / Thermodynamics / THE-21
Thermodynamics
Entropy change from reversible heat transfer (Clausius)
Entropy gained by a system that takes in heat Q at a constant absolute temperature T by a reversible path.
- Ref
- THE-21
- Rev
- v1
- Field
- Thermodynamics
- Result
- ΔS [J/K]
Clausius defined entropy change by dS = δQ_rev/T. For a reservoir or a phase change that stays at constant temperature, ΔS = Q/T. Heat added gives a positive ΔS; heat removed gives a negative one. T must be absolute temperature in kelvin.
Entropy gained by a system that takes in heat Q at a constant absolute temperature T by a reversible path. Heat added at a low temperature raises entropy more than the same heat at a high temperature.
Calculator
- Entropy change per kilojoule of heat3.333 J/K
- Same heat at 600 K8.333 J/K
- Same heat at 150 K33.33 J/K
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Taking in 5,000 J at 300 K raises the entropy by 16.67 J/K.
- The same heat adds less entropy at a higher temperature. A heat engine gets its work from this difference between hot and cold reservoirs.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J/K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Qrev Heat added reversibly (negative if removed) | −1,000,000 to 1,000,000 J | −3,333 | 3,333 | 6,667 | +40000 J | +800 % |
| T Absolute temperature | 1 to 5,000 K | 1 | 5,000 | 4,999 | +10% | −9.09 % |
Sweep of Absolute temperature
| Absolute temperature (K) | Entropy change (J/K) |
|---|---|
| 1 | 5,000 |
| 2.9 | 1,724 |
| 8.409 | 594.6 |
| 24.38 | 205 |
| 70.71 | 70.71 |
| 205 | 24.38 |
| 594.6 | 8.409 |
| 1,724 | 2.9 |
| 5,000 | 1 |
Sweep one variable
Variables and how each one contributes
Entropy change
What the equation gives you. Everything else feeds it.
Heat added reversibly (negative if removed)
The heat added reversibly, in J (negative if removed). Raises the entropy change in direct proportion. Raising it by 10 % (to 5,500) moves the result by +10 %, from 16.67 to 18.33 J/K.
Absolute temperature
The absolute temperature at which the heat flows, in K. Lowers the entropy change in inverse proportion to T. Raising it by 10 % (to 330) moves the result by −9.09 %, from 16.67 to 15.15 J/K.
Worked example with the values above
- Heat and temperature: Q = 5,000 J at T = 300 K.
- ΔS = Q/T = 5,000 / 300.
- ΔS = 16.6667 J/K.
Assumptions and limits
- Heat transferred reversibly at a constant absolute temperature.
- For irreversible processes the actual heat gives only a bound: ΔS ≥ Q/T.
Common mistakes
- Using degrees Celsius for T.
- Using ΔS = Q/T when the temperature changes during heating (integrate, or use cₚ ln(T₂/T₁)).
- Losing the sign of Q for heat removed.
Statistics and uncertainty
The relative error in ΔS is the combined relative error in Q and T. Because T is absolute, an error of 1 K in a 300 K reservoir is only 0.3 %, which seldom matters.
Where it is used
- Entropy of a heat reservoir, or of melting or boiling at a fixed temperature.
- Checking the second law for a heat exchanger.
Maths you need
- Absolute temperature: kelvin
- Heat: joules