michael a boles thermodynamics an engineering approach 7th edition solution manual 2011
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1-1
Solutions Manual for
Thermodynamics: An Engineering Approach
Seventh Edition
Yunus A. Cengel, Michael A. Boles
McGraw-Hill, 2011
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
PROPRIETARY AND CONFIDENTIAL
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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 The “pound” mentioned here must be “lbf” since thrust is a force, and the lbf is the force unit in the English system.
You should get into the habit of never writing the unit “lb”, but always use either “lbm” or “lbf” as appropriate since the
two units have different dimensions.
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 There is no acceleration, thus the net force is zero in both cases.
1-7E The weight of a man on earth is given. His weight on the moon is to be determined.
Analysis Applying Newton's second law to the weight force gives
W 210 lbf ⎛ 32.174 lbm ⋅ ft/s 2 ⎞
W = mg ⎯
⎯→ m = = ⎜ ⎟ = 210.5 lbm
g 32.10 ft/s 2 ⎜ 1 lbf ⎟
⎝ ⎠
Mass is invariant and the man will have the same mass on the moon. Then, his weight on the moon will be
⎛ 1 lbf ⎞
W = mg = (210.5 lbm)(5.47 ft/s 2 )⎜ 2
⎟ = 35.8 lbf
⎝ 32.174 lbm ⋅ ft/s ⎠
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
6X6X8 m3
Thus,
⎛ 1N ⎞
W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎟ = 3277 N
⎜ 1 kg ⋅ m/s 2 ⎟
⎝ ⎠
, 1-4
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 0.5% 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 = 0.995W s = 0.995mg s = 0.995(m)(9.81)
Substituting, 0
−6
0.995(9.81) = (9.81 − 3.32 × 10 z) ⎯
⎯→ z = 14,774 m ≅ 14,770 m Sea level
1-10 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 ) = 1920 N
1-11E The constant-pressure specific heat of air given in a specified unit is to be expressed in various units.
Analysis Applying Newton's second law, the weight 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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