Transformadores

1-1

Chapter 1
INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics
is based on the average behavior of large groups of particles.
1-2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and
thus the bicyclist picks up speed. There is no creation ofenergy, and thus no violation of the conservation
of energy principle.
1-3C There is no truth to his claim. It violates the second law of thermodynamics.

Mass, Force, and Units
1-4C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One
pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight
of a1-lbm mass at sea level is 1 lbf.
1-5C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force
required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1
kg-force.
1-6C There is no acceleration, thus the net force is zero in both cases.

1-7 A plastic tank is filled with water. The weight of thecombined system is to be determined.
Assumptions The density of water is constant throughout.
Properties The density of water is given to be ρ = 1000 kg/m3.
Analysis The mass of the water in the tank and the total mass are

mtank = 3 kg
3

V =0.2 m

mw =ρV =(1000 kg/m )(0.2 m ) = 200 kg
3

3

mtotal = mw + mtank = 200 + 3 = 203 kg
Thus,
 1N


W = mg = (203 kg)(9.81 m/s 2 )
2 1 kg ⋅ m/s = 1991 N



H2O

1-2

1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be
determined.
Assumptions The density of air is constant throughout the room.
Properties The density of air is given to be ρ = 1.16 kg/m3.
Analysis The mass of the air in the room is
3

ROOM
AIR

3

m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) =334.1 kg

6X6X8 m3

Thus,

1N
W = mg = (334.1 kg)(9.81 m/s 2 )
 1 kg ⋅ m/s 2



 = 3277 N



1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The
height at which the weight of a body will decrease by 1% is to be determined.
z
Analysis The weight of a body at the elevation z can be expressed as
W = mg = m(9.807 − 3.32 ×10−6 z )

In our case,
W = 0.99Ws = 0.99mgs = 0.99(m)(9.807)

Substituting,
0.99(9.81) = (9.81 − 3.32 × 10 −6 z ) 
→ z = 29,539 m

0
Sea level

1-10E An astronaut took his scales with him to space. It is to be determined how much he will weigh on
the spring and beam scales in space.
Analysis (a) A spring scale measures weight, which is the local gravitational force applied on abody:

1 lbf
W = mg = (150 lbm)(5.48 ft/s 2 )
 32.2 lbm ⋅ ft/s 2



 = 25.5 lbf



(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration.
The beam scale will read what it reads on earth,
W = 150 lbf

1-11 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is
to bedetermined.
Analysis From the Newton's second law, the force applied is

1N
F = ma = m(6 g) = (90 kg)(6 × 9.81 m/s 2 )
 1 kg ⋅ m/s 2



 = 5297 N



1-3

1-12 [Also solved by EES on enclosed CD] A rock is thrown upward with a specified force. The
acceleration of the rock is to be determined.
Analysis The weight of the rock is

1N
W = mg = (5 kg)(9.79 m/s 2 )
 1 kg ⋅ m/s 2

 = 48.95 N



Then the net force that acts on the rock is
Fnet = Fup − Fdown = 150 − 48.95 = 101.05 N

From the Newton's second law, the acceleration of the rock becomes
a=

F 101.05 N  1 kg ⋅ m/s

=
m
5 kg 
1N


2

Stone


 = 20.2 m/s 2



1-13 EES Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the
numerical...