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# Second law

#### Heat engines A heat engine converts thermal energy into mechanical work.

Heat is transferred from a thermal energy store at high temperature to a thermal energy sink of lower temperature. This movement enables work to be done on the surroundings. $$Q_\textrm{H} =W_\textrm{out}+Q_\textrm{C}$$ A refrigerator or heat pump is the opposite. Work is done on the gas to transfer heat from a low temperature energy store to a high temperature store.

$$W_\textrm{in}+Q_\textrm{C}=Q_\textrm{H}$$

#### Second law of thermodynamics The second law of thermodynamics can be expressed in three ways.

1. Clausius: Heat cannot pass from a colder to a warmer body without work being done.
2. Kelvin: It is impossible to convert all heat into mechanical work during a cyclic process with 100% efficiency.
3. Entropy: In any isolated system, entropy will either remain the same or increase.

Clausius' and Kelvin's statements are qualitative and equivalent. The statement regarding entropy implies that we can collect measurements and quantitatively determine whether or not a process will take place.

Change in entropy is defined as the ratio of the heat added to a system to the temperature at which it is transferred:

$$\Delta S={\Delta Q\over T}$$

• $$\Delta S$$ is change in entropy (JK-1)
• $$\Delta Q$$ is change in heat (J), where $$\Delta Q>0$$ for heat added to the gas
• $$T$$ is absolute temperature (K)

If entropy is constant, the process is reversible and theoretical only. If entropy increases, the process is irreversible and possible. Let's consider the transfer of heat from a high temperature thermal energy store to a low temperature thermal energy sink: $$\Delta S=-{Q\over T_\textrm{H}}+{Q\over T_\textrm{C}}=Q({1\over T_\textrm{C}}-{1\over T_\textrm{H}})$$

Since $$T_\textrm{C}< T_\textrm{H} \Rightarrow {1\over T_\textrm{C}}-{1\over T_\textrm{H}}>0 \Rightarrow \Delta S>0$$. The process will occur.

The opposite process would involve the transfer of heat from low to high temperature: $$\Delta S={Q\over T_\textrm{H}}-{Q\over T_\textrm{C}}=Q({1\over T_\textrm{H}}-{1\over T_\textrm{C}})$$

Since $$T_\textrm{C}< T_\textrm{H} \Rightarrow {1\over T_\textrm{H}}-{1\over T_\textrm{C}}<0 \Rightarrow \Delta S<0$$. The process will not occur.

#### Heat death Some philosophically minded physicists suggest that ever-increasing entropy will lead to a maximum value. This would bring about the heat death of the universe.

In this state of thermodynamic equilibrium, all regions would have the same temperature. No heat would be transferred and so no work could be done.