- Thermodynamics Problems And Answers
- Thermodynamics Entropy Problems And Solutions Pdf
- Thermodynamics Questions And Solutions
- Physics Thermodynamics Pdf
- Thermodynamics Problems And Answers Pdf
Introduction to thermodynamics of materials 5th edition pdf Guide book of maths for class 12 pdf, P.D.F Introduction to the Thermodynamics of Materials, Fifth Edition.Full Pages. By David R. Solutions to sample quiz problems and assigned problems Sample Quiz Problems Quiz Problem 1. Prove the expression for the Carnot e ciency for a perfectly reversible Carnot cycle using an ideal gas. Solution: The ideal Carnot cycle consists of four segments as follows (1) An isothermal expansion during which heat Q H is added to the system at.
91
Τ
Ν
M
θ
θ
Choose the x direction along the edge
X
Fx = Max
T ° W sin µ = 0
T = W sin µ = Mg sin µ
where we have used the fact that m ø M so that the mass of the string does not aÆect the tension. Thus the wave speed is
= s | ||||||
MgL sin µ | ||||||
v = | T | = | Mg sin µ | |||
s | s | |||||
m=L | m | |||||
To get the time t for the wave to travel from one end to the other, simply use v = Lt giving
t = v | = LrMgL sin µ | = s | ||
Mg sin µ | ||||
L | m | mL | ||
92 | CHAPTER 14. WAVES - II |
4.A stationary train emits a whistle at a frequency f. The whistle sounds higher or lower in pitch depending on whether the moving train is approaching or receding. Derive a formula for the diÆerence in frequency ¢f, between the approaching and receding train whistle in terms of u, the speed of the train, and v, the speed of sound. [Serway, 5th ed., p. 541, Problem 54]
SOLUTION The Doppler eÆect is summarized by
f0 = f v ß vD v ® vs
where f is the stationary frequency, f0 is the observed frequency, vD is the speed of the detector, vs is the speed of the source and v is the speed of sound.
In this example vD = 0. If the train is approaching the frequency increases, with vs ¥ u, i.e.
f0 | = f | v |
v ° u |
and if the train recedes then the frequency decreases, i.e.
f00 = f | v | ||||||||||||||
v + u | |||||||||||||||
The diÆerence in frequencies is then | |||||||||||||||
¢f = f | 0 ° f00 = f v ° u ° v + u | ||||||||||||||
v | v | ||||||||||||||
= f | v(v + u) ° v(v ° u) | = f | v2 + vu ° v2 + vu | ||||||||||||
(v ° u)(v + u) | v2 ° u2 | ||||||||||||||
= | f | 2vu | |||||||||||||
v2 ° u2 | |||||||||||||||
= | f | 2vu=v2 | = f | 2(u=v) | |||||||||||
v2=v2 ° u2=v2 | ° (u=v)2 | ||||||||||||||
1 | |||||||||||||||
Let Ø ¥ uv . Thus | ¢f = | 2Ø | |||||||||||||
f | |||||||||||||||
1 | ° Ø2 | ||||||||||||||
Chapter 15
TEMPERATURE, HEAT & 1ST LAW OF THERMODYNAMICS
94CHAPTER 15. TEMPERATURE, HEAT & 1ST LAW OF THERMODYNAMICS
1.The coldest that any object can ever get is 0 K (or -273 C). It is rare for physical quantities to have an upper or lower possible limit. Explain why temperature has this lower limit.
SOLUTION
From the kinetic theory of gases, the temperature (or pressure) depends on the speed with which the gas molecules are moving. The slower the molecules move, the lower the temperature. We can easily imagine the situation where the molecules are completely at rest and not moving at all. This corresponds to the coldest possible temperature (0 K), and the molecules obviously cannot get any colder.
95
2.Suppose it takes an amount of heat Q to make a cup of coÆee. If you make 3 cups of coÆee how much heat is required?
Thermodynamics Problems And Answers
SOLUTION
The heat required is
Q = mc¢T
For Øxed c and ¢T we have
Q / m
Thus if m increases by 3, then so will Q. Thus the heat required is 3Q (as one would guess).
96CHAPTER 15. TEMPERATURE, HEAT & 1ST LAW OF THERMODYNAMICS
3.How much heat is required to make a cup of coÆee? Assume the mass of water is 0.1 kg and the water is initially at 0±C. We want the water to reach boiling point.
Give your answer in Joule and calorie and Calorie.
(1 cal = 4.186 J; 1 Calorie = 1000 calorie.
For water: c = 1calgC = 4186kgJC ; Lv = 2:26£106 kgJ ; Lf = 3:33£105 kgJ )
SOLUTION
The amount of heat required to change the temperature of water from 0±C to 100±C is
Q= mc ¢T
=0:1 kg £ 4186kgJ £ 100 C
=41860 J = 4186041:186cal = 10; 000 cal
=10 Calorie
Thermodynamics Entropy Problems And Solutions Pdf
97
4.How much heat is required to change a 1 kg block of ice at °10±C to steam at 110±C ?
Give your answer in Joule and calorie and Calorie.
(1 cal = 4.186 J; 1 Calorie = 1000 calorie.
cwater = 4186kgJC ; cice = 2090kgJC ; csteam = 2010kgJC
For water, Lv = 2:26 £ 106 kgJ ; Lf = 3:33 £ 105 kgJ )
SOLUTION
To change the ice at °10±C to ice at 0±C the heat is
Q = mc¢T = 1 kg £ 2090kgJC £ 10C = 20900J
To change the ice at 0±C to water at 0±C the heat is
Thermodynamics Questions And Solutions
Q = mLf = 1 kg £ 3:33 £ 105 kgJ
To change the water at 0±C to water at 100±C the heat is
Q = mc¢T = 1 kg £ 4186kgJC £ 100 = 418600 J To change the water at 100±C to steam at 100±C the heat is
Q = mLv = 1 kg £ 2:26 £ 106 kgJ
Physics Thermodynamics Pdf
To change the steam at 100±C to steam at 110±C the heat is
Q = mC¢T = 1 kg £ 2010kgJC £ 10 C = 20100 J
The total heat is
(20900 + 3:33 £ 105 + 418600 + 2:26 £ 106 + 20100)J = 3:0526 £ 106 J = 3:0526 £ 106 41:186cal = 7:29 £ 105 cal = 729 Cal
98CHAPTER 15. TEMPERATURE, HEAT & 1ST LAW OF THERMODYNAMICS
99
100 | CHAPTER 16. KINETIC THEORY OF GASES |
Thermodynamics Problems And Answers Pdf
1.
A)If the number of molecules in an ideal gas is doubled, by how much does the pressure change if the volume and temperature are held constant?
B)If the volume of an ideal gas is halved, by how much does the pressure change if the temperature and number of molecules is constant?
C)If the temperature of an ideal gas changes from 200 K to 400 K, by how much does the volume change if the pressure and number of molecules is constant.
D)Repeat part C) if the temperature changes from 200 C to 400 C.
SOLUTION
The ideal gas law is
P V = nRT
where n is the number of moles and T is the temperature in Kelvin. This can also be written as
P V = NkT
where N is the number of molecules, k is Boltzmann's constant and T is still in Kelvin.
A)For V and T constant, then P / N. Thus P is doubled.
B)For T and N constant, then P / V1 . Thus P is doubled.
C)In the idea gas law T is in Kelvin. Thus the Kelvin temperature has doubled. For P and N constant, then V / T . Thus V is doubled.
D)We must Ørst convert the Centigrade temperatures to Kelvin. The conversion is
K = C + 273
where K is the temperature in Kelvin and C is the temperature in Centigrade. Thus
200C = 473K
400C = 673K
Thus the Kelvin temperature changes by 673473 . As in part C, we have V / T . Thus V changes by 673473 = 1:4
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