Phys 2101
Gabriela González
2!
Ideal heat engines use a cycle of reversible thermodynamic processes. A heat
engine transforms energy extracted as heat from thermal reservoirs, into
mechanical work.
Consider a Carnot engine: a cycle with two isothermal processes at a high
temperature TH (a→b) and and a low temperature TL (c →d), and two adiabatic
processes (b→c, d→a).
 
ΔEint = 0 = Q – W → W = Q = |QH|-|QL| > 0
 
Ideally, ΔS = 0 = |QH|/TH - |QL|/TL
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/carnot.htm
3!
We use the heat QH to get work W done, so efficiency is
defined as
 
ε = |W|/|QH|
 For a Carnot engine,
 
W = |QH|-|QL|
 
|QH|/TH = |QL|/TL
 so
εC = (|QH|-|QL|)/ |QH| = 1- |QL|/|QH| = 1- TL/TH < 1
Carnot’s theorem: The most efficient cycle with maximum temperature TH
and minimum temperature TL, is the Carnot cycle.
4!
Three ideal Carnot engines operate between
(a) 400K and 500K,
(b) 500K and 600K, and
(c) 400K and 600K.
Rank them according to their efficiencies, greatest first.
ε=1-TL/TH : εc > εa > εb
a)  20% (1-4/5=0.2)
b)  16% (1-5/6=0.16)
c)  33% (1-4/6=0.33)
5!
6!
A cycle with two isothermal processes, and two constant
volume processes:
Carnot engine:
7!
http://wright.nasa.gov/airplane/otto.html
http://auto.howstuffworks.com/engine.htm
8!
http://www.shermanlab.com/science/physics/thermo/engines/OttoG.php
A refrigerator requires work to extract heat.
Coefficient of performance: K = |QL|/|W|
For a Carnot refrigerator, |W| = |QH|-|QL| and |QH|/TH = |QL|/TL
so KC = QL/(|QH|-|QL| ) = TL/(TH-TL)
http://www.facstaff.bucknell.edu/mvigeant/therm_1/fridge/
9!
You can change a refrigerator’s coefficient of performance by (a)
running the cold chamber at a slightly higher temperature;
(b) running the cold chamber at a lower temperature;
(c) moving the unit to a slightly warmer room;
(d) moving the unit to a slightly cooler room.
If the magnitudes of temperature changes are the same in all
cases, rank the performance of the refrigerators.
K=TL/(TH-TL)
(a) 
(b) 
(c) 
(d) 
TL up, TH-TL slightly bigger: K increases
TL down, TH-TL bigger: K decreases
TH up, TH-TL bigger: K decreases
TH down, TH-TL smaller: K increases
10!
2nd law of thermodynamics : ΔS ≥ 0 , or,
no engine is more efficient than an ideal Carnot engine
operating between the same temperatures.
3rd law of thermodynamics: it is impossible to reach absolute zero temperature,
or, all engines are less than 100% efficient.
11!