thermodynamics 5th edition
solution ( Chapter 1)INTRODUCTION AND BASIC CONCEPTS
Thermodynamics
1–1C What is the difference between the classical and thestatistical approaches to thermodynamics?1–2C Why does a bicyclist pick up speed on a downhill
road even when he is not pedaling? Does this violate the conservation of energy principle?1–3C An office worker claims that a cup of cold coffee on
his table warmed up to 80°C by picking up energy from the
surrounding air, which is at 25°C. Is there any truth to his
claim? Does this process violate any thermodynamic laws?
Thermodynamics1-1C 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
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 What is the difference between pound-mass andpound-force?1–5C What is the difference between kg-mass and kgforce?1–6C What is the net force acting on a car cruising at a
constant velocity of 70 km/h (a) on a level road and (b) on an
uphill road?1–7 A 3-kg plastic tank that has a volume of 0.2 m3 is filled
with liquid water. Assuming the density of water is 1000
kg/m3, determine the weight of the combined system.1–8 Determine the mass and the weight of the air contained in
a room whose dimensions are 6 m 6 m 8 m. Assume the
density of the air is 1.16 kg/m3. Answers: 334.1 kg, 3277 N1–9 At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g a bz,
where a 9.807 m/s2 and b 3.32 10 6 s 2. Determine
the height above sea level where the weight of an object will
decrease by 1 percent. Answer: 29,539 m1–10E A 150-lbm astronaut took his bathroom scale (a
spring scale) and a beam scale (compares masses) to the
moon where the local gravity is g 5.48 ft/s2. Determine
how much he will weigh (a) on the spring scale and (b) on
the beam scale. Answers: (a) 25.5 lbf; (b) 150 lbf1–11 The acceleration of high-speed aircraft is sometimes
expressed in g’s (in multiples of the standard acceleration of
gravity). Determine the upward force, in N, that a 90-kg man
would experience in an aircraft whose acceleration is 6 g’s
Mass, Force, and Units1-4C 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 kgV =0.2 m3mw =ρV =(1000 kg/m3)(0.2 m3) = 200 kg H2Omtotal = mw + mtank = 200 + 3 = 203 kg
Thus,1991 N1 kg m/s
1 N
(203 kg)(9.81 m/s2) 2 =
⋅W = mg =
1-21-8 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.ROOM
AIR
6X6X8 m3Analysis The mass of the air in the room ism = ρV = (1.16 kg/m 3 )(6× 6×8 m 3 ) = 334.1 kgThus, = 3277 N
⋅
= =2
21 kg m/s
1 NW mg (334.1 kg)(9.81 m/s )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.
zAnalysis The weight of a body at the elevation z can be expressed asW m = = g m( . 9 807 - 3.32 ×10-6z)
In our case,W W = = 0. . 99 s s 0 99mg = 0.99(m)(9.807)
Substituting,
0.99(9.81) = (9.81- 3.32×10 -6 z) → z = 29,539 mSea level1-10E 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: = 25.5 lbf
⋅
= =2
232.2 lbm ft/s
1 lbfW mg (150 lbm)(5.48 ft/s )
(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 = 529
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 kgV =0.2 m3mw =ρV =(1000 kg/m3)(0.2 m3) = 200 kg H2Omtotal = mw + mtank = 200 + 3 = 203 kg
Thus,1991 N1 kg m/s
1 N
(203 kg)(9.81 m/s2) 2 =
⋅W = mg =
1-21-8 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.ROOM
AIR
6X6X8 m3Analysis The mass of the air in the room ism = ρV = (1.16 kg/m 3 )(6× 6×8 m 3 ) = 334.1 kgThus, = 3277 N
⋅
= =2
21 kg m/s
1 NW mg (334.1 kg)(9.81 m/s )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.
| 0 |
In our case,W W = = 0. . 99 s s 0 99mg = 0.99(m)(9.807)
Substituting,
0.99(9.81) = (9.81- 3.32×10 -6 z) → z = 29,539 mSea level1-10E 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: = 25.5 lbf
⋅
= =2
232.2 lbm ft/s
1 lbfW mg (150 lbm)(5.48 ft/s )
(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 = 529
