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# 2nd Law of Thermodynamics

## Entropy and disorder

Two classical statements of the Second Law of Thermodynamics:

 Kelvin & Planck "No (heat) engine whose working fluid undergoes a cycle can absorb heat from a single reservoir, deliver an equivalent amount of work, and deliver no other effect" Clausius "No machine whose working fluid undergoes a cycle can absorb heat from one system, reject heat to another system and produce no other effect"

Both statements of the Second Law constrains the First Law of Thermodynamics by identifying that energy goes downhill. The Second Law is concerned with Entropy (S) which is produced by all processes and associated with the loss of ability to do work. The Second Law states that the entropy of the universe increases.

For energy to be available there must be a region with high energy level and a region with low energy level. Useful work must be derived from the energy that flows from the high level to the low level.

• 100% energy can not be transformed to work
• entropy can be produced but never destroyed

### Efficiency of a heat machine

The thermodynamic efficiency of a heat machine working between two energy levels is defined in terms of absolute temperature and can be expressed as

η = (Th - Tc) / Th = 1 - Tc / Th                                                  (1)

where

η = efficiency

Th = temperature high level (K)

Tc = temperature low level (K)

As a consequence, to attain maximum efficiency - Tc has to be as cold as possible. For 100% efficiency Tc should equals to 0 K. This is in real life impossible so the efficiency will always be less than 1 (100%).

 Change in entropy > 0 irreversible process Change in entropy = 0 reversible process Change in entropy < 0 impossible process

Entropy is used to define the unavailable energy in a system and entropy defines the relative ability of one system to act on an other. As energy moves toward lower energy levels where one is less able to act upon the surroundings, the entropy is said to increase.

• For the universe as a whole - the entropy is increasing
• Entropy is not conserved like energy

Go to Thermodyamics key values internationally agreed, Standard state and enthalpy of formation, Gibbs free energy of formation, entropy and heat capacity and Standard enthalpy of formation,  Gibbs energy of formation, entropy and molar heat capacity of organic substances for listing of values for a lot of inorganic and organic substances.

### Thermodynamic Entropy

Change of entropy in a thermodynamic system can be expressed as

dS = dH / Ta                             (2)

where

dS = change in entropy (kJ/kg K)

dH = change in enthalpy or internal energy (kJ/kg K)

Ta = average temperature (K)

 Carnot Heat Cycle In a heat engine, a gas is reversibly heated and then cooled. A model of the cycle is as follows: State 1 --(isothermal expansion) --> State 2 --(adiabatic expansion) --> State 3 --(isothermal compression) --> State 4 --(adiabatic compression) --> State 1 State 1 to State 2: Isothermal Expansion Isothermal expansion occurs at a high temperature Th, dT = 0 and dE1 = 0. Since dE = H + w, w1 = - H1. For ideal gases, dE is dependent on temperature only. State 2 to State 3: Adiabatic Expansion The gas is cooled from the high temperature, Th, to the low temperature, Tc. dE2 = w2 and H2 = 0 (adiabatic). State 3 to State 4: Isothermal Compression This is the reverse of the process between states 1 and 2. The gas is compressed at Tc. dT = 0 and dE3 = 0. w3 = - H3 State 4 to State 1: Adiabatic Compression This is the reverse of the process between states 2 and 3. dE4 = w4 and H4 = 0 (adiabatic). The processes in the Carnot cycle can be graphed as the pressure vs. the volume. The area enclosed in the curve is then the work for the Carnot cycle because w = - integral (P dV). Since this is a cycle, dE overall equals 0. Therefore, -w = H = H1 + H2 + H3 + H4 If you decrease Tc, then the quantity -w gets larger in magnitude. if -w > 0 then H > 0 and the system, the heat engine, does work on the surroundings. The sum of (dH / T) values for each step in the Carnot cycle equals 0. This only happens because for every positive H there is a countering negative H, overall.

### Example - Entropy Heating Water

A process raises 1 kg of water from 0 to 100oC (273 to 373 K) under atmospheric conditions.

Specific enthalpy at 0oC (hf) = 0 kJ/kg (from steam tables) (Specific - per unit mass)

Specific enthalpy of water at 100oC (hf) = 419 kJ/kg (from steam tables)

Change in specific entropy:

dS = dH / Ta

= [(419 kJ/kg) - (0 kJ/kg)] / [((273 K) + (373 K)) / 2]

= 1.297 kJ/kgK

### Example - Entropy Evaporation Water to Steam

A process changes 1 kg of water at 100oC (373 K) to saturated steam at 100oC (373 K) under atmospheric conditions.

Specific enthalpy of steam at 100oC (373 K) before evaporating = 418 kJ/kg (from steam tables)

Specific enthalpy of steam at 100oC (373 K) after evaporating = 2675 kJ/kg (from steam tables)

Change in specific entropy:

dS = dH / Ta

= [(2675 kJ/kg) - (418 kJ/kg)] / [((373 K) + (373 K)) / 2]

= 6.054 kJ/kgK

The total change in specific entropy from water at 100oC to saturated steam at 100oC is the sum of the change in specific entropy for the water, plus the change of specific entropy for the steam.

### Example - Entropy Superheated Steam

A process superheats 1 kg of saturated steam at atmospheric pressure to 150oC (423 K).

Specific total enthalpy of steam at 100oC (373 K) = 2675 kJ/kg (from steam tables)

Specific total enthalpy of superheated steam at 150oC (423 K) = 2777 kJ/kg (from steam tables)

Change in specific entropy:

dS = dH / Ta

= [(2777 kJ/kg) - (2675 kJ/kg)] / [((423 K) + (373 K)) / 2]

= 0.256 kJ/kgK

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## Citation

• Engineering ToolBox, (2003). 2nd Law of Thermodynamics. [online] Available at: https://www.engineeringtoolbox.com/law-thermodynamics-d_94.html [Accessed Day Mo. Year].

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