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 of energy, 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 a 1-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 the combined 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
H2O
mtotal = mw + mtank = 200 + 3 = 203 kg
Thus,
1N
W = mg = (203 kg)(9.81 m/s 2 ) = 1991 N
2
1 kg ⋅ m/s
, 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 ROOM
3 3 AIR
m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg
Thus, 6X6X8 m3
1N
W = mg = (334.1 kg)(9.81 m/s 2 ) = 3277 N
1 kg ⋅ m/s 2
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
0.99(9.81) = (9.81 − 3.32 × 10 −6 z)
→ z = 29,539 m
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 a body:
1 lbf
W = mg = (150 lbm)(5.48 ft/s 2 ) = 25.5 lbf
32.2 lbm ⋅ ft/s 2
(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 be determined.
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 ) = 5297 N
1 kg ⋅ m/s 2
, 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 ) = 48.95 N
1 kg ⋅ m/s 2
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 Stone
F 101.05 N 1 kg ⋅ m/s 2
= 20.2 m/s 2
a= =
m 5 kg 1N
1-13 EES Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the
numerical results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
W=m*g"[N]"
m=5"[kg]"
g=9.79"[m/s^2]"
"The force balance on the rock yields the net force acting on the rock as"
F_net = F_up - F_down"[N]"
F_up=150"[N]"
F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law."
F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION
a=20.21 [m/s^2]
F_down=48.95 [N]
F_net=101.1 [N]
F_up=150 [N]
g=9.79 [m/s^2]
m=5 [kg]
W=48.95 [N]
, 1-4
1-14 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The
percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.
Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an
altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent reduction in weight is equivalent to the percent reduction in the
gravitational acceleration, which is determined from
∆g 9.807 − 9.767
%Reduction in weight = %Reduction in g = × 100 = × 100 = 0.41%
g 9.807
Therefore, the airplane and the people in it will weight
0.41% less at 13,000 m altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
Systems, Properties, State, and Processes
1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the
system.
1-16C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries
of the system.
1-17C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
1-18C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout
but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing
pressure with depth in a fluid, for example, should be balanced by increasing weight.
1-19C A process during which a system remains almost in equilibrium at all times is called a quasi-
equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The
work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium
processes are used instead of nonquasi-equilibrium processes.
1-20C A process during which the temperature remains constant is called isothermal; a process during
which the pressure remains constant is called isobaric; and a process during which the volume remains
constant is called isochoric.
1-21C The state of a simple compressible system is completely specified by two independent, intensive
properties.
1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated
room is a simple compressible system.
1-23C A process is said to be steady-flow if it involves no changes with time anywhere within the system
or at the system boundaries.
1-24C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to
the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O =
1000 kg/m3). That is, SG = ρ / ρ H2O . When specific gravity is known, density is determined from
ρ = SG × ρ H2O .
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 of energy, 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 a 1-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 the combined 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
H2O
mtotal = mw + mtank = 200 + 3 = 203 kg
Thus,
1N
W = mg = (203 kg)(9.81 m/s 2 ) = 1991 N
2
1 kg ⋅ m/s
, 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 ROOM
3 3 AIR
m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg
Thus, 6X6X8 m3
1N
W = mg = (334.1 kg)(9.81 m/s 2 ) = 3277 N
1 kg ⋅ m/s 2
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
0.99(9.81) = (9.81 − 3.32 × 10 −6 z)
→ z = 29,539 m
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 a body:
1 lbf
W = mg = (150 lbm)(5.48 ft/s 2 ) = 25.5 lbf
32.2 lbm ⋅ ft/s 2
(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 be determined.
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 ) = 5297 N
1 kg ⋅ m/s 2
, 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 ) = 48.95 N
1 kg ⋅ m/s 2
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 Stone
F 101.05 N 1 kg ⋅ m/s 2
= 20.2 m/s 2
a= =
m 5 kg 1N
1-13 EES Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the
numerical results with proper units.
Analysis The problem is solved using EES, and the solution is given below.
W=m*g"[N]"
m=5"[kg]"
g=9.79"[m/s^2]"
"The force balance on the rock yields the net force acting on the rock as"
F_net = F_up - F_down"[N]"
F_up=150"[N]"
F_down=W"[N]"
"The acceleration of the rock is determined from Newton's second law."
F_net=a*m
"To Run the program, press F2 or click on the calculator icon from the Calculate menu"
SOLUTION
a=20.21 [m/s^2]
F_down=48.95 [N]
F_net=101.1 [N]
F_up=150 [N]
g=9.79 [m/s^2]
m=5 [kg]
W=48.95 [N]
, 1-4
1-14 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The
percent reduction in the weight of an airplane cruising at 13,000 m is to be determined.
Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an
altitude of 13,000 m.
Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent reduction in weight is equivalent to the percent reduction in the
gravitational acceleration, which is determined from
∆g 9.807 − 9.767
%Reduction in weight = %Reduction in g = × 100 = × 100 = 0.41%
g 9.807
Therefore, the airplane and the people in it will weight
0.41% less at 13,000 m altitude.
Discussion Note that the weight loss at cruising altitudes is negligible.
Systems, Properties, State, and Processes
1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the
system.
1-16C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries
of the system.
1-17C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
1-18C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout
but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing
pressure with depth in a fluid, for example, should be balanced by increasing weight.
1-19C A process during which a system remains almost in equilibrium at all times is called a quasi-
equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The
work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium
processes are used instead of nonquasi-equilibrium processes.
1-20C A process during which the temperature remains constant is called isothermal; a process during
which the pressure remains constant is called isobaric; and a process during which the volume remains
constant is called isochoric.
1-21C The state of a simple compressible system is completely specified by two independent, intensive
properties.
1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated
room is a simple compressible system.
1-23C A process is said to be steady-flow if it involves no changes with time anywhere within the system
or at the system boundaries.
1-24C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to
the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O =
1000 kg/m3). That is, SG = ρ / ρ H2O . When specific gravity is known, density is determined from
ρ = SG × ρ H2O .
