Solutions Manual for 1-1
Thermodynamics: An
Engineering Approach 9th
Edition
Yunus A. Çengel, Michael A .
Boles, Mehmet Kanoğlu
All Chapters Solutions Manual Supplement files
download link at the end of this file.
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-2
Thermodynamics
1-1C 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.
1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of
the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between
two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.
1-3C There is no truth to his claim. It violates the second law of thermodynamics.
1-4C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the
average behavior of large groups of particles.
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-3
Mass, Force, and Units
1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time.
Hence, this product forms a distance dimension and unit.
1-6C 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.
1-7C There is no acceleration, thus the net force is zero in both cases.
1-8 The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
W mg (200 kg)(9.6 m/s2 ) 1920 N
1-9E The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
1 lbf
W mg (10 lbm)(32.0 ft/s2 ) 9.95 lbf
32.174 lbm ft/s2
1-10 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
1 N
F ma m(6 g) (90 kg)(6 9.81 m/s2 ) 5297 N
1 kg m/s2
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-4
1-11 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
%Reduction in weight %Reduction in g
g 9.807 9.767
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.
1-12 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 r = 1000 kg/m3.
Analysis The mass of the water in the tank and the total mass are mtank = 3 kg
mw = V = (1000 kg/m3)(0.2 m3) = 200 kg V = 0.2 m3
mtotal = mw + mtank = 200 + 3 = 203 kg H2O
Thus,
1 N
W mg (203 kg)(9.81 m/s2 ) 1991 N
1 kg m/s2
1-13 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
1 N
W mg (2 kg)(9.79 m/s 2 ) 2
19.58 N
1 kg m/s
Then the net force that acts on the rock is
Fnet Fup Fdown 200 19.58 180.4 N
Stone
From the Newton's second law, the acceleration of the rock becomes
F 180.4 N 1 kg m/s 2
a 90.2 m / s2
m 2 kg 1 N
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
Thermodynamics: An
Engineering Approach 9th
Edition
Yunus A. Çengel, Michael A .
Boles, Mehmet Kanoğlu
All Chapters Solutions Manual Supplement files
download link at the end of this file.
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-2
Thermodynamics
1-1C 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.
1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of
the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between
two marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.
1-3C There is no truth to his claim. It violates the second law of thermodynamics.
1-4C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the
average behavior of large groups of particles.
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-3
Mass, Force, and Units
1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time.
Hence, this product forms a distance dimension and unit.
1-6C 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.
1-7C There is no acceleration, thus the net force is zero in both cases.
1-8 The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
W mg (200 kg)(9.6 m/s2 ) 1920 N
1-9E The mass of an object is given. Its weight is to be determined.
Analysis Applying Newton's second law, the weight is determined to be
1 lbf
W mg (10 lbm)(32.0 ft/s2 ) 9.95 lbf
32.174 lbm ft/s2
1-10 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
1 N
F ma m(6 g) (90 kg)(6 9.81 m/s2 ) 5297 N
1 kg m/s2
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
, 1-4
1-11 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
%Reduction in weight %Reduction in g
g 9.807 9.767
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.
1-12 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 r = 1000 kg/m3.
Analysis The mass of the water in the tank and the total mass are mtank = 3 kg
mw = V = (1000 kg/m3)(0.2 m3) = 200 kg V = 0.2 m3
mtotal = mw + mtank = 200 + 3 = 203 kg H2O
Thus,
1 N
W mg (203 kg)(9.81 m/s2 ) 1991 N
1 kg m/s2
1-13 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
1 N
W mg (2 kg)(9.79 m/s 2 ) 2
19.58 N
1 kg m/s
Then the net force that acts on the rock is
Fnet Fup Fdown 200 19.58 180.4 N
Stone
From the Newton's second law, the acceleration of the rock becomes
F 180.4 N 1 kg m/s 2
a 90.2 m / s2
m 2 kg 1 N
PROPRIETARY MATERIAL ã 2019 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.
If you are a student using this Manual, you are using it without permission.
