Editing Second law of thermodynamics (section)

=== Thermodynamic temperature ===
{{main|Thermodynamic temperature}}
For an arbitrary heat engine, the efficiency is:
{{NumBlk|: |<math>\eta = \frac {|W_n|}{q_\text{H}} = \frac{q_H+q_\text{C}}{q_\text{H}} = 1 - \frac{|q_\text{C}|}{|q_\text{H}|}</math>|{{EquationRef|1}}}}
where ''W''<sub>''n''</sub> is the net work done by the engine per cycle, ''q''<sub>H</sub> > 0 is the heat added to the engine from a hot reservoir, and ''q''<sub>C</sub> = −{{abs|''q''<sub>''C''</sub>}} < 0<ref name="PlanckBook">{{cite book |last=Planck |first=M. |title=Treatise on Thermodynamics |page=§90 |quote=eq.(39) & (40) |publisher=Dover Publications |year=1945}}.</ref> is waste [[Heat|heat given off]] to a cold reservoir from the engine. Thus the efficiency depends only on the ratio {{abs|''q''<sub>C</sub>}} / {{abs|''q''<sub>H</sub>}}.

[[Carnot theorem (thermodynamics)|Carnot's theorem]] states that all reversible engines operating between the same heat reservoirs are equally efficient. Thus, any reversible heat engine operating between temperatures ''T''<sub>H</sub> and ''T''<sub>C</sub> must have the same efficiency, that is to say, the efficiency is a function of temperatures only: 
{{NumBlk|:|<math>\frac{|q_\text{C}|}{|q_\text{H}|} = f(T_\text{H},T_\text{C}).</math>|{{EquationRef|2}}}}

In addition, a reversible heat engine operating between temperatures ''T''<sub>1</sub> and ''T''<sub>3</sub> must have the same efficiency as one consisting of two cycles, one between ''T''<sub>1</sub> and another (intermediate) temperature ''T''<sub>2</sub>, and the second between ''T''<sub>2</sub> and ''T''<sub>3</sub>, where ''T''<sub>1</sub> > ''T''<sub>2</sub> > ''T''<sub>3</sub>. This is because, if a part of the two cycle engine is hidden such that it is recognized as an engine between the reservoirs at the temperatures ''T''<sub>1</sub> and ''T''<sub>3</sub>, then the efficiency of this engine must be same to the other engine at the same reservoirs. If we choose engines such that work done by the one cycle engine and the two cycle engine are same, then the efficiency of each heat engine is written as the below.
: <math>\eta _1 = 1 - \frac{|q_3|}{|q_1|} = 1 - f(T_1, T_3)</math>,
: <math>\eta _2 = 1 - \frac{|q_2|}{|q_1|} = 1 - f(T_1, T_2)</math>,
: <math>\eta _3 = 1 - \frac{|q_3|}{|q_2|} = 1 - f(T_2, T_3)</math>.

Here, the engine 1 is the one cycle engine, and the engines 2 and 3 make the two cycle engine where there is the intermediate reservoir at ''T''<sub>2</sub>. We also have used the fact that the heat <math>q_2</math> passes through the intermediate thermal reservoir at <math>T_2</math> without losing its energy. (I.e., <math>q_2</math> is not lost during its passage through the reservoir at <math>T_2</math>.) This fact can be proved by the following.
: <math>\begin{align}
  & {{\eta }_{2}}=1-\frac{|{{q}_{2}}|}{|{{q}_{1}}|}\to |{{w}_{2}}|=|{{q}_{1}}|-|{{q}_{2}}|,\\ 
 & {{\eta }_{3}}=1-\frac{|{{q}_{3}}|}{|{{q}_{2}}^{*}|}\to |{{w}_{3}}|=|{{q}_{2}}^{*}|-|{{q}_{3}}|,\\ 
 & |{{w}_{2}}|+|{{w}_{3}}|=(|{{q}_{1}}|-|{{q}_{2}}|)+(|{{q}_{2}}^{*}|-|{{q}_{3}}|),\\ 
 & {{\eta}_{1}}=1-\frac{|{{q}_{3}}|}{|{{q}_{1}}|}=\frac{(|{{w}_{2}}|+|{{w}_{3}}|)}{|{{q}_{1}}|}=\frac{(|{{q}_{1}}|-|{{q}_{2}}|)+(|{{q}_{2}}^{*}|-|{{q}_{3}}|)}{|{{q}_{1}}|}.\\ 
\end{align}</math>

In order to have the consistency in the last equation, the heat <math>q_2</math> flown from the engine 2 to the intermediate reservoir must be equal to the heat <math>q_2^*</math> flown out from the reservoir to the engine 3.

Then
: <math>f(T_1,T_3) = \frac{|q_3|}{|q_1|} = \frac{|q_2| |q_3|} {|q_1| |q_2|} = f(T_1,T_2)f(T_2,T_3).</math>

Now consider the case where <math>T_1</math> is a fixed reference temperature: the temperature of the [[triple point]] of water as 273.16&nbsp;K; <math>T_1 = \mathrm{273.16~K}</math>. Then for any ''T''<sub>2</sub> and ''T''<sub>3</sub>,
: <math>f(T_2,T_3) = \frac{f(T_1,T_3)}{f(T_1,T_2)} = \frac{273.16 \text{ K} \cdot f(T_1,T_3)}{273.16 \text{ K} \cdot f(T_1,T_2)}.</math>

Therefore, if thermodynamic temperature ''T''* is defined by
: <math>T^* = 273.16 \text{ K} \cdot f(T_1,T)</math>
then the function ''f'', viewed as a function of thermodynamic temperatures, is simply
: <math>f(T_2,T_3) = f(T_2^*,T_3^*) = \frac{T_3^*}{T_2^*},</math>
and the reference temperature ''T''<sub>1</sub>* = 273.16 K × ''f''(''T''<sub>1</sub>,''T''<sub>1</sub>) = 273.16 K. (Any reference temperature and any positive numerical value could be used{{snd}}the choice here corresponds to the [[Kelvin]] scale.)