سش

Thermodynamics I
Spring 1432/1433H (2011/2012H)
Saturday, Wednesday 8:00am 10:00am & Monday 8:00am - 9:00am
MEP 261 Class ZA
Dr. Walid A. Aissa
Associate Professor, Mech. Engg. Dept.
Faculty of Engineering at Rabigh, KAU, KSA
Chapter #4
October XX, 2011
Announcements:
Dr. Walid’s e-mail and Office Hours
[email protected]
Office hours for Thermo 01 will be every
Sunday and Tuesday from 9:00 – 12:00 am
in Dr. Walid’s office (Room 5-213).
Text book:
Thermodynamics An Engineering Approach
Yunus A. Cengel & Michael A. Boles
7th Edition, McGraw-Hill Companies,
ISBN-978-0-07-352932-5, 2008
Chapter 4
ENERGY ANALYSIS OF CLOSED
SYSTEMS
Objectives of CH4: To
• Examine the moving boundary work or P dV
work.
• Identify the first law of thermodynamics as
simply a statement of the conservation of
energy principle for closed (fixed mass)
systems.
• Develop the general energy balance applied to
closed systems.
• Define the specific heat at constant volume and
the specific heat at constant pressure.
* Relate the specific heats to the calculation of
the changes in internal energy and enthalpy of
ideal gases.
*Describe incompressible substances and
determine the changes in their internal energy
and enthalpy.
*Solve energy balance problems for closed
(fixed mass) systems that involve heat and work
interactions for general pure substances, ideal
gases, and incompressible substances
Chapter 4
ENERGY ANALYSIS OF CLOSED SYSTEMS
4–1 ■ MOVING BOUNDARY WORK
One form of mechanical work frequently
encountered in practice is associated with the
expansion or compression of a gas in a
piston–cylinder device.
the expansion and
compression work is
often called moving
boundary work, or
simply boundary
work.
The boundary work is
+ve during expansion &
-ve during compression
The total boundary work done during the
entire process as the piston moves is
So, p = f (V) should be available.
The boundary work done during a process
depends on the path followed as well as the
end states
Net work done during a cycle is the difference
between the work done by the system and
the work done on the system.
a) Constant Volume (V = C) process
EXAMPLE 4–1 Boundary Work for a
Constant-Volume Process.
A rigid tank contains air at 500 kPa
and 150°C. As a result of heat transfer
to the surroundings, the temperature
and pressure inside the tank drop to
65°C and 400 kPa, respectively.
Determine the boundary work done
during this process
Solution:
V = C , Hence dV = 0
b) For isobaric process (p = C).
EXAMPLE 4–2 Boundary Work for a
Constant-pressure Process.
A frictionless piston–cylinder device
contains 4.53 kg of steam at 413.57 kPa
and 160°C.
° Heat is now transferred to the
steam until the temperature reaches 204.4
°C. If the piston is not attached to a shaft
and its mass is constant, determine the
work done by the steam during this
process.
Solution:
p , kPa
p0 = 413.57 kPa.
m = 4.53 kg
p = 413.57 kPa.
v, m3/kg
T
T 2 = 204.4 °C
2
T1 =
160 °C
1
v1
v2
v
The total boundary work done during the
entire process as the piston moves is
But as p = C
Hence, Wb= p *( V2 - V1 ) = p *m* ( v2 - v1 ) kJ
Where, Wb is the total boundary work done
during the entire process as the piston moves.
Evaluation of Wb implies evaluation of v1 & v2 .
From Table A-5, psat corresponding to T =
160°C = 618.23 kPa. As psat > p (= 413.57
kPa). Hence, state 1 is superheated steam.
1) Evaluation of v1
From Table A-6, For p = 0.4 MPa
T, °C
150
160
200
v, m3/kg
0.47088
v
0.53434
u, kJ/kg
2564.4
u
2647.2
h, kJ/kg
2752.8
h
2860.9
Hence, vT= 160°C,
p = 0.4 MPa
= 0.4836 m3/kg
From Table A-6, For p = 0.5 MPa
T, °C
Tsat =
151.83°C
160
200
v, m3/kg
0.37483
u, kJ/kg h, kJ/kg
2560.7 2748.1
v
0.42503
u
2643.3
h
2855.8
Hence, vT= 160°C,
p = 0.5 MPa
= 0.3833443 m3/kg
Hence by interpolation, For T = 160°C
p, kPa
0.4
0.41357
0.5
v, m3/kg
0.4836
v
0.3833443
Hence, v1= vT= 160°C,
p = 0.41357 MPa
= 0.47 m3/kg
Similarly,
2) Evaluation of v2 For p =0.4 MPa
T, °C v, m3/kg u, kJ/kg h, kJ/kg
200
0.53434 2647.2 2860.9
204.4 v
u
h
250
0.5952 2726.4 2964.5
Hence, vT= 204.4°C,
3/kg
=
0.5397
m
p = 0.4 MPa
Similarly, From Table A-6, For p = 0.5 MPa
T, °C
200
204.4
250
v, m3/kg
0.42503
v
0.47443
Hence, vT= 204.4°C,
u, kJ/kg
2643.3
u
2723.8
h, kJ/kg
2855.8
h
2961.0
3/kg
=
0.42938
m
p = 0.5 MPa
Hence by interpolation, For T = 204.4°C
p, kPa
0.4
0.41357
0.5
v, m3/kg
0.5397
v
0.42938
Hence, v2= vT= 204.4°C,
3/kg
=
0.52473
m
p = 0.41357 MPa
Hence, V1= m * v1 = 4.53 kg * 0.47 m3/kg
= 2.129 m3 .
& V2= m * v2 = 4.53 kg * 0.52473 m3/kg=
2.377 m3 .
Hence, Wb= p *( V2 - V1 ) = 0.41357 MPa *
(2.377 m3 - 2.129 m3)= 0.41357×
× 106 Pa *
(2.377 - 2.129) m3 = 1.026 × 105 J = 102.6 kJ.
c) For isothermal process T= Constant).
pV = mRT
Hence, for isothermal process :
Hence, pV = C
Hence, p = C V-1
(#-1)
Hence,
(#-2)
Where, C = p1V 1= p2V 2
(#-3)
EXAMPLE 4–3 Isothermal
Compression of an Ideal Gas.
A piston–cylinder device initially contains
0.4 m3 of air at 100 kPa and 80°C.
The air is now compressed to 0.1 m3 in
such a way that the temperature
inside the cylinder remains constant.
Determine the work done during this
process.
Solution:
V 1 = 0.4 m3 ,p1 = 100 kPa and T = 80°C.
V 2 = 0.1 m3.
Hence, From Eq. (#-3)
Where, C = p1V 1= 100 kPa * 0.4 m3
= 40 kJ
Hence, From Eq. (#-2)
Wb= 40 kJ * ln (0.1 m3 /0.4 m3) = -55.45kJ
The negative sign indicates that the
work is done on the system.
d) For polytropic process ;
n
pV = Constant).
n
As, pV = C
Hence, p = C V-n
(#-4)
Hence,
n
But, C = p2V2
Hence,
=
n
p1V1
(4-9)
But, From Equation of State
pV = mRT Hence,
p1V1 = mRT1& p2V2 = mRT2
(#-5)
By substituting by p2V2 & p1V1 from
Eq. (#-4) in Eq. (4-9) to get
(4-10)
EXAMPLE 4–4 Expansion of a Gas
against a Spring.
A piston–cylinder device contains 0.05
m 3 of a gas initially at 200 kPa. At
this state, a linear spring that has a
spring constant of 150 kN/m is
touching the piston but exerting no
force on it. Now heat is transferred to
the gas, causing the piston to rise and
to compress the spring until the
volume inside the cylinder doubles. If
the cross-sectional area of the piston
is 0.25 m 2, determine (a) the final
pressure inside the cylinder, (b) the
total work done by the gas, and (c)
the fraction of this work done against
the spring to compress it.
Fig. (EXAMPLE 4–4)
V1 = 0.05 m 3 , p1 = 200 kPa, k = 150
kN/m , V2 = 2*V1 , A = 0.25 m 2
Required?
p2 = ?, Wb = ?, Fraction of work done
against the spring to compress it?
Solution:
V2 = 2 *V1 = 2 * 0.05 m3 = 0.1 m3
Displacement of the spring; x
i.e.;
Force applied by the linear spring at
the final state; F is
F = k x = (150 kN/m)*(0.2 m)= 30 kN
Additional pressure applied by the
spring on the gas at this state; p is
Without the spring, the pressure of the
gas would remain constant at
200 kPa while the piston is rising.
But under the effect of the spring, the
pressure rises linearly from 200 kPa to
200 + 120 = 320 kPa at the final state.
i.e.
p2 = 320 kPa
(b) One way for finding the work done
is to plot the process on a P-V diagram
and find the area under the process
curve. From Fig. (Example 4–4) the area
under the process curve (a trapezoid) is
Hence; W = 13 kJ
+ve sign indicates that the work is done
by the system.
( c) Total work consists of two
portions:
I) Work represented by the
rectangular area (region I) is done
against the piston and the
atmosphere, and
II) the work represented by the
triangular area (region II) is done
against the spring.
Thus,
- Work done against the piston and
the atmosphere (region I),
- Fraction of work done against the
spring to compress it; (region II),
4–2 ■ ENERGY BALANCE FOR
CLOSED SYSTEMS
Energy balance for any system undergoing any
kind of process (as shown in Chap. 2) was
expressed as:
(4-11)
or, in the rate form, as
(4-12)
For constant rates, the total quantities
during a time interval ∆t are related to
the quantities per unit time as
(4-13)
The energy balance can be expressed
on a per unit mass basis as
(4-14)
For a closed system undergoing a cycle,
the initial and final states are identical,
and thus,
For a cycle
∆E =0, thus
Q = W.
For a closed system undergoing a cycle,
the initial and final states are identical,
Hence, Eq. (4-11) can be re-written as
∆Esystem = E2 – E1 = 0
(#-6)
Ein – Eout = 0 i.e. Eout = Ein
&
(#-7)
Energy balance for a cycle can be
expressed in terms of heat and work
interactions as
Wnet, out = Qnet, in i.e.
(4-16)
Energy balance (first-law) relation for
a closed system becomes
Q- W =∆E = ∆U+∆PE+∆KE (4-17)
where,
Qnet, in= Qin - Qout is the net heat input
&
Wnet, out= Wout - Win is the net work
output
Various forms of the first-law
relation for closed systems
EXAMPLE 4–5 Electric Heating of a
Gas at Constant Pressure.
A piston–cylinder device contains 25 g
of saturated water vapor that is
maintained at a constant pressure of 300
kPa. A resistance heater within the
cylinder is turned on and passes a
current of 0.2 A for 5 min from a 120-V
source. At the same time, a heat loss of
3.7 kJ occurs.
(a) Show that for a closed system
the boundary work Wb and the
change in internal energy ∆U
in the first-law relation can be
combined into one term, ∆H, for a
constant pressure process.
(b) Determine the final temperature
of the steam.
Solution:
Qin
Schematic and P-v diagram for
Example 4–5.
m = 25 g, p2= p1= 300 kPa
I = 0.2 A , V = 120 V, t= 5 min
Qout= 3.7 kJ
From Eq. (4-17)
Qnet, in- Wnet, out=∆E= ∆U+∆PE+∆KE
As system is stationary, hence ∆PE=0,∆KE=0
Hence, Eq. (4-17) can be re-written as
0
0
Qnet, in- Wnet, out =∆E = ∆U+∆PE+∆KE
i.e., Qnet, in- Wnet, out =∆U (#-8)
P = I2R = I V= 0.2 A* 120 V= 24 W
Qin= P * t = 24 W* (5*60) s= 7200 J
Qin= 7.2 kJ
i.e.
Wnet, out = Wb
(#-9)
But, from Eq. (4-2)
But, p2 =p1= p0 . Hence, Eq. (4-2) can be
re-written as:
Wb = p0 ( V2 – V1)
(#-10)
By substituting by Wb from Eq. (#-10) in
Eq. (#-9) to get:
Wnet, out = Wb = p0 ( V2 – V1) (#-11)
By substituting from Eq. (#-11) in
Eq. (#-8) to get:
Qnet, in= Wnet, out +∆U=
= (p2 V2 – p1V1) + (U2 –U1)
(#-12)
But, By definition;
pV + U = H
Hence, Eq. (#-12) can be re-written as:
Qnet, in= H2 –H1
(#-13)
Hence,
Qin= m(h2 –h1)
(#-13′′)
From Table (A-5);
Pressure table
Saturated
water-
h1 = hg( p = 300 kPa) = 2724.9 kJ/kg
Qnet, in= Qin – Qout = 7.2 kJ- 3.7 kJ =3.5 kJ
Hence, 3.5 kJ = (25/1000) kg* (h2 2724.9 kJ/kg)
Hence, h2 = 2864.9 kJ/kg
It is clear that h2 (= 2864.9 kJ/kg) > hg
(p2 = 300 kPa)
Hence, state 2 is superheated stream.
Hence, from Table (A-6); superheated
stream
150 2761.2
T2 2864.9
200 2865.9
Hence, T2 = 199.5224 °C
EXAMPLE 4–6 Unrestrained Expansion
of Water.
A rigid tank is divided into two
equal parts by a partition. Initially,
one side of the tank contains 5 kg
of water at 200 kPa and 25°C, and
the other side is evacuated. The
partition is then removed, and the
water expands into the entire tank.
The water is allowed to exchange
heat with its surroundings until the
temperature in the tank returns to
the initial value of 25°C. Determine
(a) the volume of the tank,
(b) the final pressure, and
(c) the heat transfer for this process.
Solution:
Schematic and p-v diagram for
Example 4–6
m = 5 kg , p1 = 200 kPa , T1 = 25°C,
T2 = T1 = 25°C. V = ?, p2 = ?,
Qheat transfer with soroundings= ?,
(a) Evaluation of the volume of the tank, Vtank
From Table (A-5) for Saturated waterPressure table
Tsat (p =200 kPa ) =120.21 °C
It is clear that
T1 < [Tsat (p =200 kPa ) =120.21 °C].
Hence, state 1 is compressed liquid.
It is clear that no data for
compressed liquid at p =200 kPa.
Hence, v1 ≈ vf [T= 25°C]= 0.001003
m3/kg, u1 ≈ uf [T= 25°C]= 104.83 kJ/kg,
h1 ≈ hf [T= 25°C]= 104.83 kJ/kg.
Initial volume of the water (V1) is
V1 = m* v1 =5 kg * (0.001003
0.005 m3.
3
m /kg)
=
Total volume of the tank (Vtank )is twice
the initial volume of the water (V1)
Vtank = 2*V1 = 2* 0.005 m3= 0.01 m3
(b) Evaluation of final pressure, p2
v2 = Vtank / m = 0.01
m3/kg
3
m
/5 kg = 0.002
Specifying conditions at state 2:
At state 2, T2 = 25°C, v2 = 0.002 m3/kg
From Table (A-4);
Temperature table
Saturated
water-
At T2 = 25°C, vf = 0.001003 m3/kg , vg =
43.34 m3/kg
Hence, state 2 is saturated liquid-vapor
mixture.
i.e., x = 2.30047
From Table (A-4);
Temperature table
-5
Saturated
water-
p2 = psat (T = 25°C) = 3.1698 kPa
From Eq.(#-8)
0
i.e., Qnet, in- Wnet, out =∆U =m(u2 – u1 )
From Table (A-4);
Temperature table
(#-14)
Saturated water-
uf (T = 25°C) = 104.83 kJ/kg
ufg (T = 25°C) = 2304.3 kJ/kg
u2 = uf + (x* ufg) =104.83 +(2.30047 10 -5
* 2304.3) kJ/kg = 104.883kJ/kg
Hence, from Eq. (#-14)
i.e., Qnet, in =m(u2 – u1 )
Hence,
Qnet, in = 5 kg (104.883 –104.83) kJ/kg
Qnet, in = 5 kg (104.883 –104.83) kJ/kg
= 0.265 kJ
The +ve sign indicates that heat is
transferred to the water
4–3 ■ SPECIFIC HEATS
The specific heat is defined as the energy
required to raise the temperature of a unit
mass of a substance by one degree.
specific heat at
specific heat at
constant volume constant volume
; cv.
; cp.
specific heat at constant volume
; cv.: the energy required to raise the
temperature of the unit mass of a
substance by one degree as the
volume is maintained constant
specific heat at constant pressure
; cp.: the energy required to raise the
temperature of the unit mass of a
substance by one degree as the
pressure is maintained constant
(4-19)
Formal definitions of cv.
Cv dT = du
at constant volume
(4-20)
Formal definitions of cp.
Cp dT = dh
at constant pressure
Eqs. (4-20) & (4-20) are properties
relations and as such are independent of
the type of processes.
A common unit for specific heats is
kJ/kg ·°C or kJ/kg·K. Notice that
these two units are identical since T(°C)
T(K), and 1°C change in temperature is
equivalent to a change of 1 K.
4–4 ■ INTERNAL ENERGY, ENTHALPY,
AND SPECIFIC HEATS OF IDEAL GASES
Ideal gas is a gas whose temperature,
pressure, and specific volume are related
by
p v = RT
(3-10)
u = u(T) only
(4-21)
But by definition; enthalpy and internal
energy of an ideal gas r related by:
h= u + p v
(#-15)
By substitution by p v from Eq. (3-10)
in Eq. (#-15) to get:
h= u + RT
(#-16)
Since R is constant and u = u(T). Hence,
enthalpy of an ideal gas is also a
function of temperature only:
i.e.
(4-22)
h = h(T) only
As u = u(T) only; Eq. (4-21) & h = h(T)
only; Eq. (4-22) . Hence, the partial
derivatives; (4-19) and (4-20) turn to
ordinary derivatives; i.e.
(4-23)
du = Cv (T) dT
(4-24)
dh = Cp (T) dT
The change in internal energy or enthalpy
for an ideal gas during a process from
state 1 to state 2 is determined by
integrating these equations:
(4-25)
and
(4-26)
To carry out these integrations, we need
to have relations for Cv and Cp as
functions of temperature.
The variation of specific heats with T over
Small temperature intervals (a few hundred
degrees or less)may be approximated as linear.
Therefore the specific heat functions in
Eqs. 4–25 and 4–26 can be replaced by the
constant average specific heat values. Then
the integrations in these equations can be
performed, yielding:
(4-27)
(4-28)
The specific heat values for some
common gases are listed as a function of
temperature in Table A–2b. The average
specific heats Cp,avg, and Cv,avg are
evaluated from this table at the average
temperature (T1 + T2)/2, as shown in Fig.
4–26.
FIGURE 4–26 For small T intervals, the
specific heats may be assumed to vary linearly
with T.
If the final temperature T2 is not known,
the specific heats may be evaluated at T1
or at the anticipated average temperature.
Then T2 can be determined by using these
specific heat values. The value of T2 can
be refined, if necessary, by evaluating
the specific heats at the new average
temperature.
Specific Heat Relations of Ideal Gases
Differentiating the relation (#-16)
h = u + RT
to get
dh = du + RdT
(#-16)
(#-17)
By substituting by du & dh from Eqs.
(4-23)&(4-24) in Eq. (#-17) to get
Cp dT= Cv dT + R dT
Hence,
Cp = Cv + R
(kJ/kg.K) (4–29)
In addition, Cp & Cv are related by idealgas property called specific heat ratio k,
defined as;
k = Cp / Cv
(4–31)
In molar basis;
(4–31)
where;
,
&
(4–30)
To summarize
Three ways of calculating u
EXAMPLE 4–7 Evaluation of the ∆u of
an Ideal Gas.
Air at 300 K and 200 kPa is heated at
constant pressure to 600 K.
Determine the change in internal
energy of air per unit mass, using (a)
data from the air table (Table A–17),
(b) the functional form of the specific
heat (Table A–2c), and (c) the
average specific heat value (Table
A–2b).
Solution:
(a) Evaluation ∆u using Table A–17
u1 = u (T=300 K) =214.07 kJ/kg
u2 = u (T=600 K) =434.787 kJ/kg
Hence, ∆u = u2 –u1 = 434.79 kJ/kg 214.07 kJ/kg = 220.72 kJ/kg
(b) Evaluation of ∆u using functional
form of the specific heat (Table A–2c)
kJ/kmol.K
-2
a = 28.11, b = 0.1967
-5
c = 0.4802
d = -1.966
Ru = 8.31447 kJ/kmol.K
-9
Cv (T) = Cp (T) - Ru
Cv (T) = (a- Ru) + bT+ cT2+dT3
But, from Eq. (4–30)
,
From Table (A–1); M (Air) =
28.97 kg/kmol
Hence,
Cv (T) = [(a- Ru) + bT+ cT2+dT3]/M
∆u = (1/M) {[(a- Ru) T+ b(T2/2) +
c(T3/3) +d(T4/4)]T=600K - [(a- Ru) T+
b(T2/2) + c(T3/3) +d(T4/4)]T=300K }
∆u = (1/ 28.97) {[(28.11 - 8.31447)
-2) [(6002(600-300)]+ (0.1967
-5
3002) /2] + (0.4802
[(6003 -9
3003)/3] +(-1.966
(60043004)/4] } = 223 kJ/kg
(c) Evaluation of ∆u using the
average specific heat value (Table
A–2b).
Cv (T=600 K) = 0.764 kJ/kg.K
Cv (T=300 K) = 0.718 kJ/kg.K
Cv, average = [Cv(T=600 K) + Cv (T=300
K)]/2= (0.764+0.718)/2 kJ/kg.K=
0.741 kJ/kg.K
Hence, ∆u = Cv, average * [600 – 300] =
0.741 kJ/kg.K * 300 K = 222.3 kJ/kg
EXAMPLE 4–8 Heating of a Gas in a
Tank by Stirring.
An insulated rigid tank initially
contains 0.5 kg of helium at 30°C and
3 kPa. A paddle wheel with a power
rating of 15 W is operated within the
tank for 30 min. Determine (a) the final
temperature and (b) the final pressure
of the helium gas.
Solution:
p , kPa
m= 0.5 kg
T1= 30°C
p1= 3 kPa
m = 0.5 kg, T1= 30°C , p1= 3 kPa, P
=15 W , t = 30 min. (a) T2= ? and
(b) p2= ? .
From Eq. (4-17) (4-17)
0
0
0
Q- W =∆E = ∆U+∆PE+∆KE
0
Wnet,out = Wout - Win
Hence,
(#-18)
(#-19)
Win = P *∆t
By substituting
Eq. (#-18) to get
from
Eq.
P *∆t =∆U=m (u2 – u1)
Hence,
∆t =30*60 = 1800 s
(#-19)
in
(#-20)
From Table (A-2) ,C
v=
3.1156 kJ/kg.K
T1= 30+273 K= 303 K
Hence, Eq. (#-20) can be re-written as:
P *∆t =∆U=m (u2 – u1)
Hence,
(15/1000) kW* 1800 s = 0.5 kg *
3.1156 kJ/kg.K *(T2 - 303 K)
T2 =320.32 K= 47.3 K
Treating Helium as an ideal gas. Hence,
pv = RT
As v2= v1
Hence, p2/T 2= p1/T 1
p2= p1* (T2/T1 )= 3 kPa * (320.32 K / 303 K)
Hence,
p2= 3 kPa * (320.32 K / 303 K)= 3.17 kPa
EXAMPLE 4–9 Heating of a Gas by a
Resistance Heater
A piston–cylinder device initially contains
0.5 m 3 of Nitrogen gas at 400 kPa and
27°C. An electric heater within the device
is turned on and is allowed to pass a
current of 2 A for 5 min from a 120-V
source. Nitrogen expands at constant
pressure, and a heat loss of 2800 J occurs
during the process. Determine the final
temperature of nitrogen..
Solution:
Schematic and P-V diagram for Example 4–9.
From Eq. (4-17)
0
0
Qnet, in- Wb,out=∆E= ∆U+∆PE+∆KE
Qnet, in= Qin –Qout=(IV.∆t)- Qout
i.e.
Qnet, in= [2*120 *(5*60)]/1000 kJ2.8 kJ = 69.2 kJ
Qnet, in=∆H
V1= 0.5 m 3 , p1= 400 kPa , T1= 27°C.
Treating Nitrogen as ideal gas. Hence,
pv = RT
From Table A-2, RN2 = 0.2968 kJ/Kg.K
, Cp,N2 =1.039kJ/Kg.K
p1v1 = RT1 Hence,
400 kPa *v1 = 0.2968 kJ/Kg.K *(27+273 K)
v1 = 0.55725 m 3 /Kg
v1 = V1 /m
0.2226 m 3 /Kg = 0.5 m 3 /m
Hence,
m = 2.212 kg
Qnet, in=∆H= m*∆h= m*(h2-h1)
i.e., Qnet, in= m*Cp*(T2-T1)
69.2 kJ = 2.212 kg * 1.039 kJ/Kg.K
*(T2-300 K)
T2= 330.1 K i.e., t2= 57.1 °C.
Homework
4–4C, 4–5, 4–6, 4–7, 4–8, 4–9, 4–12,
4–13, 4–14, 4–18, 4–21, 4–28, 4–56,
4–61.