Section 2: Second Law Flashcards

(63 cards)

1
Q

second law of thermodynamics

A

ΔS> or =0

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2
Q

concept of entropy and second law derive historically from

A

heat engines

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3
Q

the carnot engine represents an

A

idealised, reversible heat engine of maximal efficiency operating on a working substance between two thermal reservoirs to convert Q into W

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4
Q

the greatest efficiency comes from

A

the greatest temperature differences, but materials melting form a practical limit

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5
Q

the carnot cycle comprises of

A

two isotherms and two (reversible) adiabatics

1-2: isothermal heat absorption
2-3: adiabatic expansion
3-4: isothermal heat loss
4-1: adiabatic compression

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6
Q

carnot cycle is unique in that it is the only

A

thermodynamic cycle operating reversibly between only two resevoris

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7
Q

cycle

A

a series of processes that eventually return a system to its initial state

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8
Q

heat engine

A

device operating on a working substance between two thermal reservoirs to convert heat Q into mechanical energy W

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9
Q

heat engine steps

A
  1. extract heat Q1 from hot reservoir (isothermal heating)
  2. perform work W under cyclic process (isentropic work out)
  3. lose some heat Q2 to cold reservoir (isothermal cooling)
  4. compress fluid again and return to start (isentropic work in)
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10
Q

reversing all the arrows and the sense of rotation of an engine turns

A

heat engine into a refrigerator

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11
Q

efficiency

A

ratio between energy out and heat in

=W/Q = Q1-Q2/Q1 = 1-Q2/Q1

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12
Q

can also define an efficiency for a refrigerator but in that case, we would like to

A

extract heat by putting work in which will give rise to a different reciprocal ratio

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13
Q

it is not possible to build a

A

perfect carnot engine

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14
Q

Kelvin-Planck statement

A

it is impossible to construct a device that, operating in a cycle, will produce no other effect than the conversion of heat into work

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15
Q

Clausius statement

A

it is impossible to construct a device that, operating in a cycle, produces no other effect than the transfer of heat from a colder body to a hotter body

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16
Q

equivalence of Kelvin-Planck and Clausius statements can be seen by

A

considering an engine which violates one formulation and then show that this engine would also violate the other

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17
Q

equivalence

consider a heat engine that violates Clausius, transferring heat Q2 from a cold body at T2 to warm body T1

A

let it run parallel with a normal heat engine that extracts Q1 from warm body, transfers Q2 to cold body and does work Q=Q1-Q2

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18
Q

equivalence - the net result is

A

extraction of heat Q1-Q2 from the warm body

no change to cold body

work done W=Q1-Q2

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19
Q

If Clausius’s statement violates it is possible to

A

construct a machine which extracts heat from the warm body and converts this into work, without any other effect (in particular heat loss to the colder body)

this violates KP

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20
Q

equivalence

now consider a heat engine that violates KP - extracting Q1 from hot body at T1 to perform work W=Q1. Let this

A

run in parallel with normal heat engine (operating in reverse) that extracts Q2 from cold body and transfers Q3 to hot body

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21
Q

equivalence - energy conservation gives Q3=Q2+W = Q2+Q1 so

A

net result is a transfer of heat from bold body to warm body

violates Clausius

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22
Q

a heat engine works by a

A

heat injection from warm reservoir and a heat rejection to cold reservoir

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23
Q

carnot’s theorem can be used to derive

A

a stronger restriction on the efficiency based on the formulations of the second law

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24
Q

carnot’s theorem

A

no engine operating between two thermal reservoirs can be more efficient than a reversible engine operating between those reservoirs

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25
implications of the KP statement
W
26
implication of Claus statement
Q1does not = Q2 and must have some W
27
Carnot's theorem proof (1)
1. one engine run in reverse so Q2A=Q1A-W, Q2B=Q1B-W 2. take reversed engine to be optimal carnot but suppose other is more efficeint 3. show this is transferring heat from cold to hot without work so violates clausius
28
carnot proof (2)
1. 2 engines, both taking Q1 from hot reservoir 2. take A as more efficient irreversible engine 3. run B as reversible heat pump using work from A 4. composite system violates KP 5. A cannot be more efficient than reversible B
29
corollary of carnot's theorem
all Carnot engines operating between the same two temperature reservoirs have the same efficiency
30
proof of carnot corollary
1. already proved for A and B, if B is carnot then nA< or=nB 2. butif A is carnot then nB< or = NA 3. can only be resolved if nA=nB
31
carnot corollary suggests that the efficiency is independent of
the working substance and internal workings of the engine and solely dependent on the reservoir temperatures leads to definition of thermodynamic temp scale
32
thermodynamic temp - for the composite 2-stage process we have
f(T1,T2) = Q2/Q1 f(T2,T3) = Q3/Q2 f(T1,T3) = Q3/Q1 multiplication gives f(T1,T3)=f(T1,T2).f(T2,T3)
33
f(T1,T3)=f(T1,T2).f(T2,T3) suggests the definition
T2/T1 = Q2/Q1 *useful* as independent of the working fluid and not tied to non-linearities, material impurities etc
34
have T2/T1 = Q2/Q1 so can choose
a single reference point - triple point of water so T=273.16 Q/Qtp
35
the efficiency of a refrigerator is typically defined as
the ratio of extracted heat to the work done Q2/W = Q2/Q1-Q2
36
for a carnot refrigerator, n=
T2/T1-T2
37
efficiency of heat pump
Q1/Q1-Q2 = T1/T1-T2
38
commercial ground source heat pump works with
3 loops connected by 2 heat exchangers external loop - water internal loop - water refrigerant loop
39
the maximum efficiency of a heat pump increases as
T2 and T1 converge commercial systems only reach efficiency or around 4
40
the purpose of the Clausius inequality is to
introduce entropy as a thermodynamic function of state
41
the clausius inequality - consider
a system undergoing a cyclic process by N incremental steps of variable temp an external principle thermal reservoir system interacts with environment via N carnot engines
42
clausius inequality - each carnot engine works between
auxiliary reservoirs, the lower reservoir temperature depending on the stage of the process
43
claus ineq - set heat flow so
auxiliary reservoirs are unchanged
44
claus ineq - the ith engine
works between T0 and Ti supplies heat δQi performs work δWi extracts heat T0/Ti δQi
45
claus ineq - after the cycle total work
W is sum of δWi
46
claus ineq - after the cycle total heat extracted
Q= sim of T0/Ti δQi =T0sum of δQi/Ti
47
claus ineq - delta U=
0 for each engine and the composite cycle
48
claus ineq - 1st law gives
delta U=Q-W so 0=T0 sum of δQi/Ti - W W=T0 sum of δQi/Ti
49
W=T0 sum of δQi/Ti but KP would be violated unless
W=Q< or = 0 so sum of δQi/Ti < or =0
50
Clausius inequality
integral δQ/T < or =0
51
for a reversible process, it must be possible to exchange -Q for Q and get a valid result which implies
integral δQ/T =0
52
non-isolated processes lose
heat to their environs
53
there is only a single reservoir Q<0 according to KP so
cyclic processes tend to dump heat leads to the 'real' claus ineq
54
the 'real' Clausius inequality
integral δQ/T0 < or =0
55
the entropy change of the surroundings depends on
the heat transferred and the surrounding temperature
56
for a closed system, we always need to consider
the entropy of the surroundings, which can be expanded to the whole universe as the ultimate closed system
57
heat flows from hot to cold so auxiliary needs to be
very slightly hotter or colder than T0
58
for heat to flow into system
δQ>0 and T0>Tsys
59
for heat to flow out of the system
δQ<0 and T0
60
in both cases of heat flowing in and out, δQ/Tsys > δQ/T0 and for the whole cycle
0> or = integral δQ/Tsys > or = integral δQ/T0
61
an arbitrary cycle can be approximated by
a multitude of carnot cycles
62
if the arbitrary cycle is given by the black curve in a PV diagram, then interested in
the area enclosed by the curve, which can be divided into carnot cycles
63