Playground
Interactive entropy visualization: adjust heat and temperature to see disorder level with particle simulation and entropy gauge.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Entropy changeoutput Change in entropy of the system | J/K | M·L²·T⁻²·Θ⁻¹ | -10 – 10 | |
| Heat transferred Heat energy transferred reversibly | J | M·L²·T⁻² | 0 – 10000 | |
| Temperature Absolute temperature at which heat transfer occurs | K | Θ | 50 – 2000 |
Deep dive
Derivation
Clausius defined entropy via the reversible heat integral: dS = δQ_rev/T. For a finite process: ΔS = ∫(δQ_rev/T). For constant temperature: ΔS = Q/T. For temperature change at constant heat capacity: ΔS = nC·ln(T_f/T_i). Boltzmann later gave the statistical interpretation: S = k_B·ln(Ω), where Ω is the number of microstates.
Experimental verification
Verified by calorimetric measurements of heat exchange during phase transitions (ice melting at 273 K: ΔS = 334000/273 ≈ 1224 J/(kg·K) per kg). Residual entropy of glasses and crystals with frozen-in disorder confirms the Third Law (S → 0 as T → 0 for perfect crystals).
Common misconceptions
- Entropy is not 'disorder' in the colloquial sense — it is the number of microstates (Ω) consistent with the macrostate, measured by S = k_B·ln(Ω)
- Entropy of a subsystem CAN decrease (e.g., a refrigerator cools food), but the total entropy of the universe always increases
- A reversible process has ΔS_universe = 0, not ΔS_system = 0 — the system's entropy can change if compensated by the surroundings
Real-world applications
- Second law efficiency analysis of engines and refrigerators
- Chemical equilibrium prediction (Gibbs free energy G = H − TS)
- Information theory — Shannon entropy is mathematically identical to Boltzmann entropy
- Black hole thermodynamics (Bekenstein-Hawking entropy)
Worked examples
Heat flow between reservoirs
Given:
- Q:
- 500
- T_H:
- 400
- T_C:
- 300
Find: Delta_S_universe
Solution
ΔS = −Q/T_H + Q/T_C = −500/400 + 500/300 = −1.25 + 1.667 = 0.417 J/K
Heating water from 20°C to 80°C
Given:
- m:
- 1
- c:
- 4186
- T_i:
- 293
- T_f:
- 353
Find: Delta_S
Solution
ΔS = mc·ln(T_f/T_i) = 1 × 4186 × ln(353/293) = 1 × 4186 × 0.1863 = 779.9 J/K
Scenarios
What if…
- scenario:
- What if the two reservoirs are at the same temperature?
- answer:
- ΔS_universe = 0. Heat flow between equal-temperature bodies is reversible (and in practice, no net flow occurs). This is thermodynamic equilibrium.
- scenario:
- What if entropy could decrease in an isolated system?
- answer:
- This would violate the Second Law. A gas could spontaneously compress into one corner, coffee could unmix from milk, eggs could unscramble. The overwhelming improbability of microstates lining up this way (for macroscopic systems) makes it effectively impossible.
Limiting cases
- condition:
- T → ∞
- result:
- ΔS → 0
- explanation:
- At very high temperature, adding heat causes negligible entropy change.
- condition:
- T → 0⁺
- result:
- ΔS → ∞
- explanation:
- Adding heat near absolute zero produces enormous entropy increase (Third Law prevents reaching T = 0).
- condition:
- Q = 0 (adiabatic reversible)
- result:
- ΔS = 0 (isentropic)
- explanation:
- No heat exchange in a reversible process means entropy is conserved.
Context
Rudolf Clausius · 1865
Clausius coined the word 'entropy' from the Greek τροπή (transformation). He stated: 'The entropy of the universe tends to a maximum' — one of the most profound sentences in physics.
Hook
Why can't you unscramble an egg — even though no energy is lost?
500 J of heat flows reversibly from a reservoir at 400 K to one at 300 K. Find the total entropy change of the universe.
Dimensions: [ΔS] = [Q]/[T] → M·L²·T⁻²·Θ⁻¹ = (M·L²·T⁻²)/(Θ) = M·L²·T⁻²·Θ⁻¹ ✓
Validity: The integral form applies to reversible processes. For irreversible processes, ΔS > Q/T (Clausius inequality). Statistical mechanics generalizes via S = k_B ln Ω.