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:
[work available] [work required] [heating costs] [irreversibility losses] [information engines]
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
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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