Thermal & Kinetic Lecture 19

Changes in Entropy; The Carnot cycle

LECTURE 19 OVERVIEW

Calculating changes in entropy

Misinterpretations of the 2nd law

The Carnot cycle

Last time…

Adiabatic work

Calculating changes in entropy

Functions of stateReservoir at 293.1 K

293.1 K

Reservoir at 293.2 K

293.2 K

Reversible and irreversible processes: calculation of entropy

When the water is at a temperature T and it’s heated to T + T, the heat entering (reversibly) is dQ = CPT.

T

dQdS From

the entropy change of the water at each reversible step is: T

dTCdS P

What do we now need to do to evaluate the total change in entropy???

ANS: Integrate. 373

293 T

dTCS P

A block of mass 1 kg, temperature 100°C and heat capacity 100 JK-1 is placed in a lake whose temperature is 10°C. (Consider the lake as a reservoir whose temperature doesn’t change.) The change in entropy of the block is:

0 J

K-1

+27

.61

JK-1

-27.

61 J

K-1

None

of the

se

4% 0%

92%

4%

a) 0 JK-1

b) +27.61 JK-1

c) -27.61 JK-1

d) None of these

The change in entropy of the block, SBlock, is given by: 283

373 T

dQSBlock

BlockS 27.61 JK-1

Calculating changes in entropy: examples

A block of mass 1 kg, temperature 100°C and heat capacity 100 JK-1 is placed in a lake whose temperature is 10°C. (Consider the lake as a reservoir whose temperature doesn’t change.) The change in entropy of the lake is:

-31.

8 JK

-1

+31

.8 J

K-1

0 J

K-1

+63

.6 J

K-1

0% 0%0%

100%a) -31.8 JK-1

b) +31.8 JK-1

c) 0 JK-1

d) +63.6 JK-1

The entropy gain of the lake is:

lakeT

TC

(Lake acts as a thermal reservoir which is so large there’s no change in its temperature).

= 100 x 90/283 = +31.80 JK-1

This means that the change in entropy of the universe is:

Posi

tive

Neg

ativ

e Z

ero

Infin

ite

88%

4%4%4%

a) Positive

b) Negative

c) Zero

d) Infinite

The same block (mass 1 kg and heat capacity 100 JK-1) at a temperature of 10°C is dropped into the lake (temperature 10°C) from a height of 10 metres. What is the change in entropy of the block?

0

+0.

35 J

K-1

-0.3

5 JK

-1

21.

12 J

K-1

15%

2%

44%40%a) 0

b) +0.35 JK-1

c) -0.35 JK-1

d) 21.12 JK-1

The same block (mass 1 kg and heat capacity 100 JK-1) at a temperature of 10°C is dropped into the lake (temperature 10°C) from a height of 10 metres. What is the change in entropy of the lake?

0 J

K-1

+0.

35 J

K-1

-0.3

5 JK

-1

-21.

12 J

K-1

0% 0%

33%

67%a) 0 JK-1

b) +0.35 JK-1

c) -0.35 JK-1

d) -21.12 JK-1

(ii) The block is in the same state (at the same temperature) before and after the process. Although the temperature of the lake remains constant because it is a thermal reservoir, the kinetic energy of the block is transferred as heat energy into the lake. So there’s a positive change of entropy for the lake:

lake

lake T

mghS 1 x 9.81 x 10/283 = +0.35 JK-1

Calculating changes in entropy: examples

This means that the change in entropy of the universe is:

Posi

tive

Neg

ativ

e Z

ero

Infin

ite

100%

0%0%0%

a) Positive

b) Negative

c) Zero

d) Infinite

The same block at 10°C absorbs a photon of light ( = 600 nm). Calculate the change in entropy of the block.

Calculating changes in entropy: examples

??

Some ‘abuses’ of the 2nd law and the concept of entropy

Your sock drawer (or bedroom) does not become disordered due to ‘entropy’ – the change in thermodynamic entropy here is zero (we aren’t changing the number of accessible microstates). (Same thing applies to playing cards!).

“The entropy of a body never decreases – it always increases.”

OK, then how does a fridge work? Heat is ‘taken out’, therefore entropy decreases!

Entropy is a measure of disorder.

Humans and animals are complex, ordered beings.

2nd law states disorder always increases.

Therefore order can’t ‘evolve’ from disorder – theory of evolution can’t be correct…….

Using thermal processes to do work: heat engines

Carnot noted that work is obtained from an engine because there are heat sources at different temperatures.

Furthermore, he realised that heat could also flow from a hot to a cold body with no work being done.

A temperature difference may be used to produce work OR it can be ‘squandered’ as heat.

Engine

How do we convert thermal energy transfer into useful work? (e.g. a steam engine)

How efficient can we make this cycle?

QH

QL

W

The most efficient process: the Carnot cycle

Engine

QH

QL

W

In an ideal engine the temperature difference between the two reservoirs should yield the maximum amount of work possible.

Carnot realised that this meant that all transfers of heat should be between bodies of nearly equal temperature.

The Carnot engine involves reversible processes (these are the most efficient processes in terms of exploiting a temperature difference to do work).

Heat supplied from high temp. reservoir: QH

Heat rejected into lower temp. reservoir: QL

TH

TL

A Carnot engine operates between only two reservoirs and is reversible. All the heat that is absorbed is absorbed at a constant high temperature (QH) and all the heat that is rejected is rejected at a constant lower temp. (QL).

The most efficient process: the Carnot cycle

Carnot engine is an idealisation.

We’ll use an ideal gas as our working substance.

Carnot cycle may be constructed from a combination of adiabatic and isothermal compressions and expansions.

Adiabatic

Isotherm

P

V

A

B

CD

QH

QL

W

Animation

Reversible and irreversible processes

We will show later on why the most efficient heat engine (the Carnot engine) involves reversible processes.

Irreversible processes play a role in any real heat engine.

Friction

• Block sliding across a table slows down due to friction.• Friction converts kinetic energy to heat energy in block total entropy of Universe increases. (dS=dQ/T)• Irreversible process

TH

TC

• Thermal energy transfer between two objects increases the total entropy of the Universe.• Bring two blocks of different temperatures together (see Tutorial Work Set 3).• The smaller the temperature difference between the blocks the closer to a reversible process we get.