= Coefficient of performance
{title2=$\mathrm{COP}=Q_{\rm useful}/W$}
{wiki}
= COP
{c}
{synonym}
For the cyclic two-side <energy> balance in <refrigerator>, define the dimensionless cooling and heating coefficients by
$$
\mathrm{COP}_{\rm r}=\frac{Q_c}{W},\qquad
\mathrm{COP}_{\rm heat}=\frac{Q_h}{W}
=\frac{Q_c+W}{W}=\mathrm{COP}_{\rm r}+1.
$$
This identity applies even to irreversible machines, provided both coefficients refer to the same operating point, <work> input and boundary with no additional <energy> transfer or storage. It does not compare two separately operated appliances. Unlike <thermal efficiency> of a <heat engine>, a <COP> can exceed one: its numerator counts <heat> moved, including <energy> taken from another region. https://labs.phys.utk.edu/mbreinig/phys136core/modules/m3/refrigerators.html[University of Tennessee, Refrigerators, air conditioners, and heat pumps: the two performance ratios]{ref}
To obtain temperature-only bounds, impose stronger assumptions: a cyclic machine exchanges <heat> only with two ideal <thermal reservoirs> maintained at uniform, fixed absolute <thermodynamic temperatures> $0<T_c<T_h$, and receives <work> through an ideal input carrying no <entropy>. Include the machine and the <heat transfer> interfaces in the <entropy production> accounting. Because its own <entropy> returns to its initial value each cycle, the <Second law of thermodynamics> gives
$$
S_{\rm gen}=\frac{Q_h}{T_h}-\frac{Q_c}{T_c}\geq0.
$$
Combining this inequality with $Q_h=Q_c+W$ gives
$$
\boxed{\mathrm{COP}_{\rm r}\leq\frac{T_c}{T_h-T_c}},\qquad
\boxed{\mathrm{COP}_{\rm heat}\leq\frac{T_h}{T_h-T_c}}.
$$
Equality requires a <reversible thermodynamic process> for the entire cycle and its contacts, as in a reversed <Carnot cycle>. These are <Carnot's theorem> bounds independent of the working substance. Use <kelvin> values, not <degree Celsius> readings, in the numerators; the <temperature> difference can be expressed in either unit. https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node25.html[MIT, Refrigerators and heat pumps: reversed Carnot cycle and absolute-temperature cooling bound]{ref} https://openstax.org/books/university-physics-volume-2/pages/4-5-the-carnot-cycle[Ling, Moebs and Sanny, University Physics Volume 2, §4.5: reversible reservoir entropy balance]{ref}
The diagram fixes $T_c=280\,\mathrm K$ and sets $T_h=320\,\mathrm K$, so the reversible bounds are $280/40=7$ and $320/40=8$. The stipulated machine with $\mathrm{COP}_{\rm r}=3$ and $\mathrm{COP}_{\rm heat}=4$ satisfies both. Its reservoir <entropy production> per cycle is $4000/320-3000/280=25/14\,\mathrm{J\,K^{-1}}>0$. Increasing the <temperature> lift $\Delta T=T_h-T_c$ at fixed $T_c$ lowers both bounds. Equivalently, delivering fixed $Q_h$ needs at least
$$
W_{\min}=Q_h\left(1-\frac{T_c}{T_h}\right).
$$
These expressions reveal why a small <temperature> lift is attractive. Their divergence as $T_h-T_c\to0^+$ is an ideal reversible limit, not a promise of infinite heating or cooling <power>. For finite <heat transfer> rates, the working <fluid> must be colder than the cold bath and hotter than the hot bath; these contact differences produce <entropy> and increase the internal <temperature> lift. Dissipative compression and the pressure-reducing restriction of <vapor-compression refrigeration> supply further departures from reversibility. Thus bath <temperatures> set an outer bound, not a prediction of real <COP>. https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node25.html[MIT, Refrigerator hardware: working-fluid temperatures on either side of the baths]{ref}
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