Complete Solution Manual for Thermodynamics: An Engineering Approach, 8th Edition by Yunus A. Çengel & Michael A. Boles | Detailed Step-by-Step Solutions and Notes – All Chapters

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Solutions Manual for
Thermodynamics: An Engineering Approach
8th Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2015




Chapter 1
INTRODUCTION AND BASIC CONCEPTS




PROPRIETARY AND CONFIDENTIAL


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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 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.




1-4C There is no truth to his claim. It violates the second law of thermodynamics.




Mass, Force, and Units


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 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.




1-7C There is no acceleration, thus the net force is zero in both cases.




1-8 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 0.3% is to be determined.
z
Analysis The weight of a body at the elevation z can be expressed as
W  mg  m(9.807  332
.  106 z)

In our case,
W  (1  0.)Ws  0.997Ws  0.997mg s  0.997(m)(9.807)

Substituting, 0
6
0.997(9.807)  (9.807  3.32  10 z) 
 z  8862 m Sea level



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1-9 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/s 2 )  1920N




1-10 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  = 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 =V =(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/s2 )   1991 N
2
 1 kg  m/s 




1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units.
Analysis Using proper unit conversions, the constant-pressure specific heat is determined in various units to be

 1 kJ/kg  K 
c p  (1.005 kJ/kg  C)   1.005 kJ/kg K
 1 kJ/kg  C 
 1000 J  1 kg 
c p  (1.005 kJ/kg  C)    1.005 J/g  C
 1 kJ  1000 g 
 1 kcal 
c p  (1.005 kJ/kg  C)   0.240 kcal/kg C
 4.1868 kJ 
 1 Btu/lbm  F 
c p  (1.005 kJ/kg  C)   0.240 Btu/lbm  F
 4.1868 kJ/kg  C 




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If you are a student using this Manual, you are using it without permission.