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 ofenergy, 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. 1-4C 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 twomarks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.
Mass, Force, and Units 1-5C 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. 1-6C In thisunit, 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.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If youare a student using this Manual, you are using it without permission.
1-8E 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 = mg ⎯ ⎯→ m =
W 180 lbf = g 32.10 ft/s 2
⎛ 32.174 lbm ⋅ ft/s 2 ⎜ ⎜ 1 lbf ⎝
⎞ ⎟ = 180.4 lbm ⎟ ⎠
Mass is invariant and the man will have the same masson the moon. Then, his weight on the moon will be
1 lbf ⎞ ⎛ W = mg = (180.4 lbm)(5.47 ft/s 2 )⎜ ⎟ = 30.7 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠
1-9 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 theair in the room is
m = ρV = (1.16 kg/m 3 )(6 × 6 × 8 m 3 ) = 334.1 kg
ROOM AIR 6X6X8 m3
⎛ 1N W = mg = (334.1 kg)(9.81 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝
⎞ ⎟ = 3277 N ⎟ ⎠
1-10 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 abody 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)
0.99(9.81) = (9.81 − 3.32 × 10 −6 z) ⎯ ⎯→ z = 29,539 m
0 Sea level
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student usingthis Manual, you are using it without permission.
1-11E 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/s 2 )⎜ ⎟ = 9.95 lbf 2 ⎝ 32.174 lbm ⋅ ft/s ⎠
1-12 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to bedetermined. 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 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ ⎞ ⎟ = 5297 N ⎟ ⎠
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
⎛ 1N W = mg = (5 kg)(9.79 m/s 2 )⎜ ⎜ 1 kg ⋅ m/s 2 ⎝
Then the net force that acts on the rock is...
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