Thermodynamic temperature Second law of thermodynamics
for arbitrary heat engine, efficiency is:
η
=
w
n
q
h
=
q
h
−
q
c
q
h
=
1
−
q
c
q
h
(
1
)
{\displaystyle \eta ={\frac {w_{n}}{q_{h}}}={\frac {q_{h}-q_{c}}{q_{h}}}=1-{\frac {q_{c}}{q_{h}}}\qquad (1)}
where wn net work done per cycle. efficiency depends on qc/qh.
carnot s theorem states reversible engines operating between same heat reservoirs equally efficient. thus, reversible heat engine operating between temperatures t1 , t2 must have same efficiency, say, efficiency function of temperatures only:
q
c
q
h
=
f
(
t
h
,
t
c
)
(
2
)
.
{\displaystyle {\frac {q_{c}}{q_{h}}}=f(t_{h},t_{c})\qquad (2).}
in addition, reversible heat engine operating between temperatures t1 , t3 must have same efficiency 1 consisting of 2 cycles, 1 between t1 , (intermediate) temperature t2, , second between t2 andt3. can case if
f
(
t
1
,
t
3
)
=
q
3
q
1
=
q
2
q
3
q
1
q
2
=
f
(
t
1
,
t
2
)
f
(
t
2
,
t
3
)
.
{\displaystyle 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}).}
now consider case
t
1
{\displaystyle t_{1}}
fixed reference temperature: temperature of triple point of water. t2 , t3,
f
(
t
2
,
t
3
)
=
f
(
t
1
,
t
3
)
f
(
t
1
,
t
2
)
=
273.16
⋅
f
(
t
1
,
t
3
)
273.16
⋅
f
(
t
1
,
t
2
)
.
{\displaystyle f(t_{2},t_{3})={\frac {f(t_{1},t_{3})}{f(t_{1},t_{2})}}={\frac {273.16\cdot f(t_{1},t_{3})}{273.16\cdot f(t_{1},t_{2})}}.}
therefore, if thermodynamic temperature defined by
t
=
273.16
⋅
f
(
t
1
,
t
)
{\displaystyle t=273.16\cdot f(t_{1},t)\,}
then function f, viewed function of thermodynamic temperature, simply
f
(
t
2
,
t
3
)
=
t
3
t
2
,
{\displaystyle f(t_{2},t_{3})={\frac {t_{3}}{t_{2}}},}
and reference temperature t1 have value 273.16. (of course reference temperature , positive numerical value used—the choice here corresponds kelvin scale.)
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