ku thermo notes - slide set 1

8/4/2019 KU Thermo notes - slide set 1

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Thermodynamics 1.01

● Department of Physics and Astronomy

Syllabus

Heat

Energy

Work 1st Law of Thermodynamics



Thermodynamics is fundamental.

Why do cold packs get cold?

If work can be done to make heat,can heat be applied to make work?

Mechanics, E & M, Quantum, relativity,can't answer all the questions in theuniverse.



What does 'hot' mean?

Joule experiments

Heat is energy.

What is temperature?

What is heat?

And temperature? Well, an attempt to define it leads to the 0th Law of Thermodynamics:

How would we define, scientifically, some basic properties?

There is a property called temperature. This property is necessary and sufficient

to determine thermal equilibrium.



∆U = - ∆W

∆U = ∆Q - ∆W

Conservation of energy (simple spring)

1rst Law of Thermodynamics

Heat term Work term

Internal energy

You have a system which behaves as a simple spring with k = 2 N/m and a

heat capacity = 1 J/ºK.

By how much does the internal energy and temperature change when you

a) stretch the system by 10 cm.b) allow the system to snap back to it's original extension quickly.



∆Q = CF

∆T

∆W = - F ∆x

How many variables are necessary to describe a system?

T, F, and x are state variables

Two.

Are state variables path dependent?

No.

Note that: dU = δQ - đW

Two non state variables add up to a state variable!

For this problem we made the following substitutions:



x0

x

F

T1

x1

T2

(a)

(b)

(c)

Thermodynamic Path

Because we know that the force starts and ends at zero, and we know that thetemperature has changed, then the final length must be different from the initial length.

What is the thermodynamic pathway taken in the spring pulling experiment?



The most common work term you will see is

F dx = P A dx = P dV

Equations of state are state variables written asfunctions of the other state variables.

P = P(T,V) T=T(P,V)

The most famous equation of state is the ideal gasequation of state.

pv = RT (Ideal Gas)



Some necessary questions and clarifications:

What is the system?Define the boundary.

What are the boundary conditions?Open or closed?Fixed or floating?

Is the system in equilibrium?Equilibrium or steady state?

Is the process reversible?

What is work and what is heat?Depends on the boundary definition.

What exactly is internal energy, U?As a good start, think Hamiltonian.



Let's say you have a piston which you move from point 1 to point 2 along oneor the other of the paths below:

What is the boundary of the system?Are the boundaries open or closed, fixed or variable?What is the work done on path (a) and path (b)?How does ∆U from path (a) compare to ∆U on path (b)?



Carnot Cycle

Path1,2 - Isothermal

Path4,1

- Adiabatic

Path3,4

- Isothermal

Path2,3

- Adiabatic

Do work, take in heat ∆Q1

Do negative work, lose heat ∆Q2

Do work

Do negative work



Carnot Cycle

∆W = Total word done (area inside the cycle)

Efficiency: (the fraction of heat going in that comes out as work)

∆Q = ∆Q1 - ∆Q2

=W

Q1

=Q1−Q2

Q1



2nd Law of Thermodynamics

Can ∆Q2 = 0?

No. No system can have a cycle such that all of theheat is transferred into work.



Carnot Engines in Parallel

T1

T2

∆Q1

∆Q1

- ∆W1

∆W1

∆W2

∆Q1

- ∆W1

∆Q1

- ∆W1

+ ∆W2

Two engines with different efficiencies. Engine 2 is throttled

to pump as much heat up as engine 1 lets pass through.

Total work done: ∆W = ∆W1

- ∆W2

Total heat change in reservoir 1: ∆Q = ∆Q1

- ( ∆Q1

- ∆W1

- ∆W2)

= ∆W

Thus, the second law is violated. Therefore the two engines cannot have differentefficiencies, and the efficiency is a function only of the two temperatures.



Carnot Efficiency

First Law: ∆U = ∆Q - ∆W

Implying . . . ∆W = ∆Q1

- ∆Q2

Where 1= 'empirical' temperature

The cycle returns to the samestate, therefore: ∆U = 0

=W

Q1

=1−Q2

Q1

Q2

Q1

= f 1 ,2

An efficiency can be defined as:

Considering two engines inparallel reveals that:

REVIEW



Carnot Engines in Series

∆Q1

∆Q2

∆Q2

∆Q3

∆W1

∆W2

Q2

Q1

= f 1 ,2

Q3

Q2= f 2 ,3

Q3

Q1

= f 1 ,3

Considering eachengine separately

Considering bothengines together



Carnot Engines in Series

∆Q1

∆Q2

∆Q2

∆Q3

∆W1

∆W2

Q3

Q1

=Q3

Q2

×Q2

Q1

f 3 ,

1=

f 3 ,

2

f 2 ,1



Thermodynamic Temperature

Only possible if: 2 does not appear in the equation

Let's call f ( ) the 'thermodynamic' temperature, T

f 3 ,

1=

f 3 ,

2

f 2 ,

1

f 3 ,

1=

f 3 f

1

Q3

Q1

= f 3 ,1=T

3

T 1

=1−T low

T highEfficiency is a function of temperature:



Entropy

Q2

Q1

=T 2

T 1

Q2

T 2=Q1

T 1

Q2

T 2−Q1

T 1=0

∑closed , reversable path

Qi

T i =0

S =Q

T We can therefore define a new state variablecalled the entropy , S, such that ∆S is given by:

From the Carnot Cycle we know that:



First Law (again)

S =

Q

T

Remember that: dU = δQ - đW

The work can be written as: đW = P dV

Now that we have a state variable

called entropy . . .

The First Law can be written as: dU = T dS - P dV



Problem #1

Path1,2

- Isothermal

Path4,1

- Adiabatic

Path3,4

- Isothermal

Path2,3

- Adiabatic

Do work, take in heat ∆Q1

Do negative work, lose heat ∆Q2

Do work

Do negative work

T

S

?



Problem #2

P

V

p0

2p0

v0 2v

03v

0

a b

What is the change in internal energy as a system movesadiabatically along the designated path from a to b?



Problem #3

You construct two identical systems for performing a Joule experiment, exceptthat one of the experiments is adiabatic and the other diathermal.

For the adiabatic case you drop the mass M a total height H0

and

measure a temperature change ∆T. To get he same temperaturechange in the diathermal case you need a height H.

What is the heat transferred in the diathermal case?What assumptions have you made measuring H?



Problem # 4

A small explosion takes place inside a balloon with radius Rreleasing an amount of energy . Assume that the balloon skin is

adiabatic and that it exerts a constant additional pressure of Pb

on

the gas inside. The explosion increases the balloon volume by 10%.What is the change in the internal energy?



Problem #5

The surface of the sun is about 6000 ˚K. The averagetemperature of the earth is 15 ˚C. About 120,000 TW hitsthe earth every day from the sun.

Regarding the energy from the sun which hits the earth,

what is the change in the sun's entropy?

What is the change in the earth's entropy?

Why does the sun not 'run out' of entropy?

What is the most efficient engine that can be established

between the earth and the sun?



Problem #6

A 1 kΩ resistor is heating a fluid with a 100 mA current.The resistor is in steady state equilibrium with atemperature 80˚C. After 5 minutes, how much has theentropy of the resistor changed?



Problem 7

Can the efficiency of the Carnot cycle below be improved?

Given that an automobile engine is limited in its temperature range,

how can the engine be made to get more energy out of each cycle?

ku thermo notes - slide set 1

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