# 2nd Law of Thermodynamics

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 = (T_{h}- T_{c}) / T_{h}= 1 - T_{c}/ T_{h }(1)

where

η= efficiency

T_{h}= temperature high level (K)

T_{c}= temperature low level (K)

As a consequence, to attain maximum efficiency - *T _{c}* has to be as cold as possible. For 100% efficiency

*T*should equals to

_{c}*0 K*. This is in real life impossible so the efficiency will always be less than

*1 (100%)*.

Change in entropy > 0 processirreversible | Change in entropy = 0 processreversible | Change in entropy < 0 processimpossible |

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 = d H / T _{a} (2)*

*where *

*dS = change in entropy (kJ/kg K)*

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

*T _{a} = average temperature (K)*

dT = 0 and dE Since _{1} = 0.dE = H + w, w. For ideal gases, _{1} = - H_{1}dE is dependent on temperature only.State 2 to State 3: Adiabatic ExpansionThe gas is cooled from the high temperature, T, to the low temperature, _{h}T. _{c}dE and _{2} = w_{2}H (adiabatic)._{2} = 0State 3 to State 4: Isothermal CompressionThis is the reverse of the process between states 1 and 2. The gas is compressed at T. _{c}dT = 0 and dE. _{3} = 0w_{3} = - H_{3}State 4 to State 1: Adiabatic CompressionThis is the reverse of the process between states 2 and 3. dE and _{4} = w_{4}H (adiabatic)._{4} = 0The 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 = H_{1} + H_{2} + H_{3} + H_{4}If you decrease T, then the quantity _{c}-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 ( |

### Example - Entropy Heating Water

A process raises *1 kg* of water from *0 to 100 ^{o}C (273 to 373 K)* under atmospheric conditions.

Specific enthalpy at

*0*(from steam tables) (Specific - per unit mass)

^{o}C (h_{f}) = 0 kJ/kgSpecific enthalpy of water at

*100*(from steam tables)

^{o}C (h_{f}) = 419 kJ/kgChange in specific entropy:

dS = dH / T_{a}= [(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 *100 ^{o}C (373 K)* to saturated steam at

*100*under atmospheric conditions.

^{o}C (373 K)Specific enthalpy of steam at

*100*

^{o}C (373 K)**before**evaporating =

*418 kJ/kg*(from steam tables)

Specific enthalpy of steam at *100 ^{o}C (373 K)*

**after**evaporating =

*2675 kJ/kg*(from steam tables)

Change in specific entropy:

dS = dH / T_{a}

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

= 6.054 kJ/kgK

The total change in specific entropy from water at *100 ^{o}C* to saturated steam at

*100*is the sum of the change in specific entropy for the water, plus the change of specific entropy for the steam.

^{o}C### Example - Entropy Superheated Steam

A process superheats *1 kg* of saturated steam at atmospheric pressure to *150 ^{o}C (423 K)*.

Specific total enthalpy of steam at *100 ^{o}C (373 K) *=

*2675 kJ/kg*(from steam tables)

Specific total enthalpy of superheated steam at

*150*(from steam tables)

^{o}C (423 K) = 2777 kJ/kgChange in specific entropy:

dS = dH / T_{a}

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

= 0.256 kJ/kgK

## Related Topics

### • Thermodynamics

Work, heat and energy systems.

## Related Documents

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### Equilibrium

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