1 THE ZEROTH LAW OF THERMODYNAMICS AND 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 temperatureis the only requirement for thermal equilibrium. Temperature The absolute zero temperature is defined as the temperature where all molecular activity ceases.
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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 ?
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?
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 .
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.
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 tem- perature difference