zeroth law and introductory concepts

8/4/2019 Zeroth Law and Introductory Concepts

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THE ZEROTH LAW OF THERMODYNAMICSAND INTRODUCTORY CONCEPTS

The Zeroth Law

If two bodies are in thermal equilibrium with a third body,

they are also in thermal equilibrium with each other.

The equality of temperature is the only requirement for thermal equilibrium.

Temperature

The absolute zero temperature is defined as the temperaturewhere all molecular activity ceases.



Dimensions and Units

Primary (fundamental) dimensionsMass (m)Length (L)Time (t)Temperature (T)

Secondary dimensions

Volume (V = L3)

Velocity (u = L t-1)Force (F = m L t-2)Energy (E = m L2s-2)

ForceFrom Newton’s 2nd law

Force = (mass) x (acceleration)

Example

An astronaut weighs 730 N in Houston, Texas, wherethe local acceleration of gravity is g = 9.792 m s-2.What are the astronaut’s mass and weight on themoon, where g = 1.67 m s-2 ?



With a = g , Newton’s law is : F = mg . Whence,

)(55.74

)(55.74)(792.9

)(730 21

2

kgm

sm N sm

N gF m

=

=== −

−

This mass of the astronaut is independent of location, butweight depends on the local acceleration of gravity. Thuson the moon the astronaut’s weight is :

)(5.124)(5.124)(67.1)(55.74

2

2

N smkgF smkgmgF

moon

moonmoon

==

×==−

−

PressurePressure is defined as a normal force exerted by a

fluid at rest per unit area

( )

areaunit

ForceressureP normal

=

Example

A dead-weight gauge with a 1 cm diameter piston is used tomeasure pressures very accurately. In a particular instance amass of 6.14 kg (including piston and pan) brings it into balan-ce. If the local acceleration of gravity is 9.82 ms-2 , what is thegauge pressure being measured? If the barometric pressure is748 (torr), what is the absolute pressure?



The force exerted by gravity on the piston, pan and weight is ;

F = mg = (6.14)(9.82) = 60.925 N

( ) ( ))(77.76

1

295.60 2

2

41

−=== cm N

A

F pressureGauge

π

The absolute pressure is therefore:

P = 76.77 + (748)(0.013332)

= 86.74 N cm-2 = 867.4 kPa



Example

At 27 oC the reading on a manometer filled with mercury is

60.5 cm. The local acceleration of gravity is 9.784 m s-2 .To what pressure does this height of mercury correspond?

( )gh

A

g Ah

A

mg

A

F P ρ

ρ ====

P = 60.5 (cm ) x 13.53 (g cm -3 ) x 9 784.(m s -2 )

= 8009 (g m s -2 cm -2 )P = 8.009 (kg m s -2 cm -2 ) =8.009 (N cm -2 )

= 80.09 (kPa) = 0.8009 (bar )

WorkWork W is performed whenever a force acts througha distance

∫ =

=

2

1

l

l

FdlW

FdlW δ



Expansion or compression work

The force exerted by the piston on the fluid( ) ( )

∫ −=

=

×=

2

1

l

l

PAdlW

PAF

area piston pressure force

Because A = constant,

∫ −=

2

1

V

V PdV W

The minus signs are made necessary by the sign convention adopted for work



Example

A gas in a piston-cylinder assembly undergoes an expansionprocess for which the relationship between pressure and volu-me is given by PV n = constant

The initial pressure is 3 bar , the initial volume is 0.1 m-3 , andthe final volume is 0.2 m-3. Determine the work for the process,in kJ, if (a) n = 1.5 , (b) n = 1.0 , and (c) n = 0 .

Substituting the relationshipP = const / V n

( )( ) ( )

n

V const V const W

dV V

const PdV W

nn

V

V

n

V

V

−

−−=

−=−=

−−

∫ ∫

1

1

1

1

2

2

1

2

1



The constant can be evaluated at either end state:

nn V PV Pconst 2211 ==

( )( ) ( )

( )n

V PV PW

n

V V PV V PW

nnnn

−

−−=

−

−−=

−−

1

1

1122

1

111

1

222

This expression is valid for all values of n except for n = 1.0

To evaluate W , the pressure state at P2 is required. Use

nnV PV P

2211 =

( ) 06.12.0

1.03

5.1

2

1

12=

=

=

n

V

V PP

( )

( )( ) ( ) ( ) ( )

( )

( )( )

( )

( )

( )kJ W

m N

kJ

bar

mbar mbar

n

V PV PW

m

N

6.17

.10

1

1

10

5.11

1.032.006.1....

1

3

533

1122

2

−=

−

×−×−=

−

−−=



Energy

By Newton’s 2nd law maF =

The acceleration is defined as

dt

dua =

where u is the velocity of the body

=

dt dumF

Using the chain rule, this can be written as :

udl

dum

dt

dl

dl

dum

dt

dumF

=

=

=

where

dt

dlu =

Substituting F into

∫

∫ ∫

=

==

2

1

2

1

2

1

u

u

l

l

l

l

muduW

dl

dl

dumuFdlW



Kinetic energy

For a constant mass m, this equation may now be integrated fora finite change in velocity from u1 to u2 .

22

22

2

1

2

2

2

1

2

2

2

1

mumuW

uumudumW

u

u

−=

−== ∫

By definition:2

21 mu E kin =

∆=−=

2

2

1,2,

mu E E W kinkin

Potential energy

The weight of a body of mass m is the force of gravity on it

mgmaF ==

where g is the local acceleration of gravity

The work required to raise the body upward from z1 to z2

W = F ( z2 – z1) = mg( z2 – z1)= mgz2 – mgz1

By definition: E pot = mgz

W = ( E pot )2 – ( E pot )1 = ∆( mgz)



Energy conservation

The work of accelerating a body produces a change in

its kinetic energy

∆=∆=

2

2

mu E W kin

The work done on elevating a body produces a change inits potential energy ( )mgz E W pot ∆=∆=

For a freely falling body, ∆ E kin + ∆ E pot = 0

( )

const mgzmu

mgzmu

mgzmgzmumu

=+=+

=−+

−

1

2

1

2

2

2

12

2

1

2

2

22

0

22

ExampleAn elevator with a mass of 2,500 kg rests at a level 10 m above the base ofan elevator shaft. It is raised to 100 m above the base of the shaft, where

the cable holding it breaks. The elevator falls freely to the base of the shaftand strikes a strong spring. The spring is designed to bring the elevator torest and by means of a catch arrangement, to hold the elevator at the posi-

tion of maximum spring compression. Assuming the entire process to befrictionless, and taking g = 9.8 ms-2 , calculate:

(a) The potential energy of the elevator in its initial position relative

to the base of the shaft

(b) The work done in raising the elevator.(c) The potential energy of the elevator in its highest position relative

to the base of the shaft.(d) The velocity and kinetic energy of the elevator just before it

strikes the spring.(e) The potential energy of the compressed spring.



Let subscript-1 designate the initial conditions; subscript-2 , conditions when

the elevator is at its highest position; and subscript-3 , conditions just beforethe elevator strikes the spring.

(a) E pot,1 = mgz1 = (2,500)(9.8)(10 – 0) = 245,000 J

(b) ( )12

2

1

2

1

z zmgmgdlFdlW

z

z

z

z

−=== ∫ ∫

W = (2,500)(9.8)(100 – 10) = 2,205,000 J

(c) E pot,2 = mgz2 = (2,500)(9.8)(100 – 0)

= 2,450,000 J

Note that W = E pot,2 – E pot,1

∆ E kin,2→3 + ∆ E pot ,2→3 = 0 E kin,3 – E kin,2 + E pot ,3 – E pot ,2 = 0

However, E kin,2 = E pot ,3 = 0therefore E kin,3 = E pot ,2 = 2,450,000 J

With

( )

( )1

3

3,2

3

2

32

1

3,

.27.44

500,2

000,450,222

−=

==

=

smu

m

E u

mu E

kin

kin

From the principles of conservation of mechanical energy,

the sum of the kinetic- and potential-energy changes duringthe process from condition-2 to condition-3 is zero,

(d)



Because the changes in the potential energy of the springand the kinetic energy of the elevator must sum to zero,

(e)

∆ E kin,elevator + ∆ E pot ,spring = 0

The initial potential energy of the spring and the final kinetic en-ergy of the elevator are zero; therefore, the final potential ener-gy of the spring must equal the kinetic energy of the elevator

just before it strikes the spring.

E kin,elevator,3 = E pot ,spring,3 = 2,450,000 J

HeatHeat is the energy which is transferred across the boundaries of a system interacting with the surroundings by virtue of a temperature difference

Conductiondx

dT AQ xcond κ −=,

& Fourier’s law

Convection f sconv T T hAQ −=&

Newton’s law of cooling

Radiation ( )4

2

4

1T T AQemit −= εσ &

Stefan-Boltzmann law

zeroth law and introductory concepts

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