Homework 10
Problems: 19.29, 19.63, 20.9, 20.68
Problem 19.29
An automobile tire is inflated with air originally at 10 ºC and normal atmospheric
pressure. During the process, the air is compressed to 28% of its original volume
and the temperature is increased to 40 ºC. (a) What is the tire pressure? (b) After
the car is driven at high speed, the tire’s air temperature rises to 85 ºC and the tire’s
interior volume increases by 2%. What is the new tire pressure (absolute)?
a) With a good approximation, we can assume that air is an ideal
gas. Originally (state 1) and after the gas was compressed (state 2), its
pressure, volume and temperature were related by the equation of state
for an ideal gas.
P2 V2 = nRT2
P1V1 = nRT1
Dividing the corresponding sides and solving for the pressure of the
compressed air yields
V T
1 (273 + 40)K
P2 = 1 ⋅ 2 ⋅ P1 =
⋅
⋅1atm = 3.95 atm
V2 T1
0.28 (273 + 10)K
or
1 (273 + 40 )K
P2 =
⋅
⋅ 1,013 hPa = 4,000 hPa
0.28 (273 + 10 )K
b) In the third state (after the car was driven) the air in the tires still
satisfied the equation of state for an ideal gas
P3 V3 = nRT3
Therefore
P3 =
V2 T3
1
(273 + 85)K ⋅ 3.95 atm = 4.23 atm = 4,485 hPa
⋅ ⋅ P2 =
⋅
(1 + 0.02) (273 + 40)K
V3 T2
Problem 19.63
The relationship Lf = Li(1+αΔT) is an approximation that works when αΔT is
small. If αΔT is large, one must integrate the relationship dL = αLdT to determine
the final length. (a) Assuming the average coefficient of linear expansion of a
material is constant as L varies, determine a general expression for the final length
of a rod made of the material. Given a rod of length 1.0 m and a temperature
change of 100 °C, determine the error caused by the approximation when (b)
α = 2⋅10-5 °C-1 (a typical value for a metal) and when (c) α = 0.02 °C-1 (an
unrealistically large value for comparison).
a)
The definition of the coefficient of linear expansion
dL = αLdT
leads to the approximate relationship only if we assume that in the
process the change in size of the object insignificantly affects the result
of integration. In such a case one can assume that both the coefficient,
as well as the object’s dimension are constants in the integral.
When this assumption is not valid, one must integrate the above
expression with a temperature dependent dimension. Before performing
the integration we have to separate the variables (- we cannot integrate
the right-hand side of the equation without explicit knowledge of
function L(T))
dL
= αdT
L
From which
Lf
dL Tf
= ∫ αdT
∫
L
Li
Ti
From the fundamental theorem of calculus
Lf
L
T
ln f = ln L = αT Tf = α(Tf − Ti )
i
Li
L
i
Solving for the final dimension
Lf = Li eα (Tf − Ti )
b)
is
Using the “exact” expression, the expected final length of the rod
Lf = 1.0m ⋅ e(2⋅10
−5
)
°C −1 ⋅100° C
= 1.002002m
From the approximate expression
( (
)
)
L'f = Li (1 + αΔT ) = 1.00m ⋅ 1 + 2 ⋅ 10− 5°C −1 ⋅ 100°C = 1.002000
The error is about
L'f −Lf = 0.00002m
(constitutes only 0.1% of the change in length)
c) If there were a material with a very high coefficient of linear
expansion the difference in using both formulas could be significant
Lf = 1.0m ⋅ e(0.02°C )⋅100°C = 7.39m
L'f = 1.00m ⋅ 1 + 0.02°C−1 ⋅ 100°C = 3.00m
−1
( (
)
The difference in the results
L'f −Lf = 4.39m
is comparable with the dimensions of the object.
)
Problem 20.9
A 1.5-kg iron horseshoe initially at 600 ºC is dropped into a bucket containing
20 kg of water at 25 ºC. What is the final temperature of the system? Ignore the
heat capacity on the container and assume a negligible amount of water boils away.
mFe, Th
mH2O, Tc
When the system, which includes the water in
the bucket and the iron horseshoe, reaches thermal
equilibrium, a single temperature Tf will be
established. The temperature of each subsystem
changes because of heat transfer between them.
We can express heat absorbed (or given up) by
each subsystem in terms of the information given in
the problem and the final temperature. In the
considered temperatures, both water and
aluminum have a specific heat not dependent on
temperature, which simplifies the integration
Tf
(1)
ΔQ H 2 O = ∫ m H 2 O ⋅ c H 2 O ⋅ dT = m H 2 O ⋅ c H 2 O ⋅ (Tf − Tc )
Tc
Tf
(2)
ΔQ Fe = ∫ m Fe ⋅ c Fe ⋅ dT = m Fe ⋅ c Fe ⋅ (Tf − Th )
Th
Since there was an exchange of heat only between these two
subsystems (approximately), the total heal absorbed by the system is
zero.
(3)
ΔQ H 2 O + ΔQ Fe = 0
The rest is math. There are three unknowns ( ΔQ H 2 O , ΔQ Fe and Tf ) in
this set of equations. Solving for the equilibrium temperature we find
Tf =
=
m H 2 O ⋅ c H 2 O ⋅ Tc + m Fe ⋅ c Fe ⋅ Th
m H 2 O ⋅ c H 2 O + m Fe ⋅ c Fe
=
20 kg ⋅ 4,186J / kg°C ⋅ 25°C + 1.5 kg ⋅ 448J / kg°C ⋅ 600°C
= 29.6°C
20 kg ⋅ 4,186J / kg°C + 1.5 kg ⋅ 448J / kg°C
Note that we did not need to convert all units into the SI system.
We can get away with this because equations (1), (2) and (3) are valid
also for temperatures expressed in the Celsius scale.
Problem 20.68
(a) The inside of a hollow cylinder is maintained at a temperature Ta while the
outside is at a lower temperature Tb. The wall of the cylinder has a thermal
conductivity k. Ignoring the end effects, show that the rate of energy conduction
from the inner toe the outer wall in the radial direction is
⎡ T − Tb ⎤
dQ
= 2πLk ⎢ a
⎥
dt
⎣ ln(b a ) ⎦
For a concentric cylindrical shell of radius r and
thickness dr (drawn in broken line in the figure),
the rate of heat transfer is proportional to the
area of the shell and the temperature gradient
dQ
dT
1)
= − kA
dr
dt
L
Expressing the area in explicitly in terms of its
radius and rearranging the equation we can
relate the (differential) temperature difference
with the thickness of the shell
1 dQ dr
2)
dT = −
r
2πkL dt r
Under a stationary condition, the rate at which
heat flows across the differential shell does not depend on its radius.
Therefore, the integration of equation (2) relates the temperature
difference with the radii of the object
Tb
1 dQ b dr
1 dQ
1 dQ
b
b
Tb − Ta = ∫ dT = −
⋅∫ =−
⋅ ln r a = −
⋅ ln
2πkL dt a r
2πkL dt
2πkL dt
a
Ta
b
a
Solving for the rate
dQ 2πkL(Ta − Tb )
=
b
dt
ln
a