# Thermodynamic Possibility Puzzlers

UM-StL Physics and Astronomy, Copyright (2003) P. Fraundorf

Find here examples of how 1st and 2nd Law intution tells us what is likely NOT (and what may be) possible, without delving into details about how to pull it off. The focus is on energy and uncertainty flows into, and out of, ``steady-state engines'' of virtually any type. The equations treat the world outside the engine as a contained universe. This is possible since steady-state engines (by definition) process energy and information (often cyclicly) while their own state of being is maintained (i.e. at the end of each cycle, the state of the engine itself is to first order unchanged). This page helps set up the problems, writing out color-coded terms for 1st and 2nd law equations, and it provides the numeric answer for one example problem in each case as a reality check for your own deductions. However, there is a catch!

We ask that you work out for yourself the final, generally useful, formula for solving each problem type.

Can you do it?

The templates provided may, of course, help solve possibility problems we haven't thought of as well. Suggestions for problem types to add to this list are invited. E-mail your suggestions to pfraundorf[AT]umsl[DOT]edu.

### ''Science of the possible'' equations, for flows into and out of steady state engines, follow from...

1st Law: Heat_OUT plus Work_OUT equals Heat_IN plus Work_IN
2nd Law:
Uncertainty_INCREASE minus Correlation_INCREASE equals

Net_Surprisal_Irreversibly_Lost, which is greater than or equal to Zero.

Note that Uncertainty_INCREASE can often be expressed as...

Heat_OUT/Temperature_OUT minus Heat_IN/Temperature_IN,

and Correlation is information on the relation between subsystems in [J/K] or [bits].

### ...which allow us to put limits (for example) on:

1).  Work available from a heat engine (e.g. operating on fossil fuels or via photosynthesis)...

Qexhaust + Wout = Qhot, and Qexhaust/Texhaust - Qhot/Thot  >= 0,
implies that Carnot Efficiency Wout/Qhot <= (1-Texhaust/Thot).
Case Study:
An automobile engine, with Thot = 429K
and Texhaust = 300K, has an efficiency limit of 30%.

2).  Work needed to keep the frost on a 6-pack: Refrigerator coefficient of performance...

Qroom = Qcold + Win, and Qroom/Troom - Qcold/Tcold >= 0,
implies a Refrigerator C.O.P. of Qcold/Win <= Tcold/(Troom-Tcold).
Case Study:
For each joule of electricity, a freezer in a 295K room
may remove up to 9.9 joules of heat from its 268K air.

3).  Work needed to pump winter heat from the outside in, or heat pump C.O.P....

Qroom = Qcold + Win , and Qroom/Troom - Qcold/Tcold >= 0,
implies a Heat Pump C.O.P. of Qroom/Win <= Troom/(Troom-Tcold).
Case Study:
For each joule of electricity, heat pumps might
bring inside up to 7.4 joules of heat from a 0 F backyard.

4).  Reversible home heating with a flame:  Getting lots more BTU's for the buck...

Qroom = Qflame + Qcold , and Qroom/Troom – (Qflame/Tflame + Qcold/Tcold) >= 0,
implies a Reversibility Gain of Qroom/Qflame <= (Troom/Tflame)(Tflame-Tcold)/(Troom-Tcold)
Case Study:
When Toutside=273K, a furnace irreversibly heating
a room to Troom=298K with Tflame=1000K uses 8.67 times the fuel required.

5).  Zero energy ovens for eskimos, or how to cook food in cold weather for free...

Qoven = Qroom + Qcold , and Qoven/Toven - ( Qroom/Troom + Qcold/Tcold) >= 0,
implies a Heat Transfer Ratio of Qoven/Qroom <= (Toven/Troom)(Troom-Tcold)/(Toven-Tcold).
Case Study:
After "the turkey is done", this transfer may be
spontaneously reversed (see previous example) for no net heat loss!

6).  Reversibility losses from an oven leaking heat...

Qroom = Qoven, and Sirr >= (Qroom/Troom - Qoven/Toven),

implies minimal net_surprisal losses of Sirr = Qoven(1/Troom-1/Toven)
Case Study:
A 473K oven irreversibly raises state uncertainty
by nearly 10^20 nats, per joule of heat leaked to a 295K room.

7).  Reversibility losses from ice melting in the North Sea (or a glass of iced tea)...

Qice = Qliquid, and Sirr >= (Qice/Tmelt - Qliquid/Tmelt),

implies minimal net_surprisal losses of Sirr = Qice(1/Tmelt – 1/Tliquid).
Case Study:
Chipped ice in an ice-cold slurry melts reversibly, resulting in Sirr = 0.

8).  Reversibility losses as your coffee gets cold...

dQroom = dQcoffee, and dSirr >= dQroom/Troom - dQcoffee/Tcoffee, can be integrated
from Tcoffee down to Troom using dQcoffee = HeatCapacity dTcoffee,

to get Net_Surprisal_Loss >= HeatCapacity × ξ[Tcoffee/Troom], where ξ[x] ≡ x - 1 - ln[x].
Case Study:
For water cooled from a boil, with dimensionless
HeatCapacity = 9/molecule, Sirr >= 0.298 nat/molecule.

9).  Ice water invention: Convert hot water to cold reversibly, with ambient exhaust, work-free...

dQroom = dQhot  + dQcold, and dQroom/Troom – (dQhot/Thot+dQcold/Tcold) >= 0 can be integrated
to Troom for dQhot from Thot, and for dQcold from Tcold, using dQwater = HeatCapacity dTwater,

to get HeatCapacity (ξ[Thot/Troom] - ξ[Tcold/Troom]) >= 0, so that Troom <= (Thot-Tcold)/ln[Thot/Tcold].
Case Study:
For water cooled from a boil,

this invention will make ice water as long as Troom <= 100/Ln[373/273] = 320.4[K] or 47.4[C].

10).  Energy required to clear one's mind (or a quantum computer's memory)...

Qambient = Win, and Qambient/Tambient - Iopen >= 0,

implies work to erase old data (and clear space for new) of Win >= Iopen × Tambient.
Case Study:
At room temperature, nature thus requires Win = 1/40 eV per nat of data erased.

11).  Maximum astrophysical (or other) observation rates, per observer per meal...

Qambient = Wfood, and Qambient/Tambient - Irecorded >= 0,

limits mutual information created to Irecorded <= Wfood/Tambient.
Case Study:
Human observers with typical caloric intake must therefore themselves

create less than 10^21 Gigabytes of correlation information per day.
(Aside: Some of us, yours truly included, produce MUCH LESS!)

See also this page on the ice water invention, our page on correlation based complexity and heat capacity in bits, and (oldest of all) our information physics page.

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