Chapter 2
Properties of Pure Substances
Pure Substance
The pure substance has the fixed chemical composition throughout and
homogenous. Water, Nitrogen, Helium, and Carbon Dioxide, for example, are all pure
substances. A mixture of various chemical elements or compounds also qualified as a
pure substance as long as the mixture is homogenous. Air, for example, is a mixture of
several gases, but it is often considered to be a pure substance because it has a uniform
chemical composition. However, a mixture of oil and water is not a pure substance.
Since oil is not soluble in water, it will collect on top of the water forming two
chemically dissimilar regions.
Phases of a Pure Substance
There are three principal phases, solid, liquid and gas. Intermolecular bond are strongest
in solids and weakest in gases. One reason is that molecules in a solid are closely
packed together, whereas in gases they are separated by relatively large distances. The
molecules in a solid phase are arranged in a three-dimensional pattern and very small
distances between each other. The attractive forces of molecules are large and keep the
molecules at fixed positions and preventing it from piling up on top and cannot move
relative to each other.
Otherwise, the distances between molecules in the liquid phase is not much different
from that of the solid phase, except the molecules are no longer at fixed positions
relative to each other and they can rotate and translate freely. In a liquid phase, the
intermolecular forces are weaker relative to solids. In the gas phase, the intermolecular
bonds are very weak and the molecules are far apart from each other, and moves at
20
random, continually colliding with each other as shown in Fig. 2-1. Molecules in the gas
phase are at a considerably higher energy level than liquid or solid phase.
Fig. 2-1 the arrangement of molecules in different phases
Phase-change Process of Pure Substance
Our attention will be focused on the liquid and vapor phases and its mixtures. As well
known substance, water will be used to demonstrate the basic principles of phase
change process. Remember that all pure substance exhibit the same general behavior.
To demonstrate the phase change process, put some of water in a free piston-cylinder
device at atmospheric pressure and assumed temperature of 25 oC as shown in Fig. 2-2.
The water exists in liquid phase and is called a compressed liquid or sub-cooled
liquid. With heating the water, the temperature will be increases until start boiling.
When water just boils it is called saturated water and the temperature is called
saturating temperature or boiling temperature and is about 100 oC at atmospheric
pressure.
Once boiling started, the temperature will stop rising until liquid is completely
vaporized. The water temperature will remain constant during the boiling process until
all liquid disappeared. Since the water start boil until the liquid disappeared, the water
and vapor phases in mixture is called saturated liquid-vapor mixture or wet vapor.
After liquid completely vaporized, the temperature still constant and the vapor is called
saturated vapor. With adding heat to the vapor and the pressure kept constant, the
temperature of the vapor increases and the vapor is called superheated vapor. From
experimental data recorded on the temperature-volume diagram Fig. 2-3, there are three
21
regions. The water phase change process and the three distinguish regions of sub-cooled
liquid, saturated liquid-vapor mixture, and superheated vapor are illustrated in T-v
diagram as shown in Fig. 2-3.
Fig. 2-2 Water phase change process at atmospheric pressure
Fig. 2-3 Phase change process at various pressures
22
Fig. 2-4 P-v and T-v diagrams of pure substance
From Fig. 2-3, we can observe three distinguish regions, from state 1 to state 2, the
water temperature is rising up but the water in liquid phase (compressed liquid or subcooled liquid). At state 2, the water boiling just starts (saturated water). From state 2
to state 4, the mixture is called saturated liquid vapor mixture (wet vapor or saturated
mixture) and the heat required in this process is called heat of vaporization or latent
heat. At state 4, all liquid disappeared and all vapor (saturated vapor). From state 4 to
state 5, the vapor temperature is rising up and is called superheated vapor. If this
experiment conducted at various pressures, we can see that the heat required for
vaporization decreased and we can get the T-v diagram also P-v diagram. The general
shape of the P-v diagram of a pure substance is very much like the T-v diagram, but the
T = constant lines on this diagram have a downward trend, as shown in Fig. 2-4.
The two equilibrium P-v and T-v diagrams developed so far represent the equilibrium
stats involving the liquid and vapor phase only. However, these two diagrams can easily
be extended to include the solid phase as well as the solid-liquid and solid-vapor
saturation regions. Most substances contract during a solidification process. But there
are other substances like water expands during solidification. The P-v diagrams for both
groups of substances are given in Fig. 2-5 and Fig. 2-6.
23
Fig. 2-5 P-v diagram of a substance that contracts during solidification
Fig. 2-6 P-v diagram of a substance that expands during solidification
The triple line appears as a point on the P-T diagram and, therefore, is often called the
triple point as shown in Fig. 2-7. The substance can pass from the solid to vapor phase
directly without melts first into a liquid and is called sublimation. For water the
sublimation happened at temperature of 0.01 oC and pressure of 0.6113 kP.
24
Fig. 2-7 P-T diagram of pure substance shown the triple point
Enthalpy
Enthalpy, h, is a combined property and equal to the sum of the internal energy, u, and
flow work, Pv.
H = U + PV
(kJ)
h = u + pv
(kJ/kg)
Property Tables
Table 2-1 properties of saturated liquid and saturated vapor
T
P
(oC) (kPa)
vf
(kJ/kg)
vg
(kJ/kg)
uf
(kJ/kg)
ug
hf
hg
(kJ/kg)
(kJ/kg)
(kJ/kg)
sf
sg
(kJ/kg.K) (kJ/kg.K)
10
50
90
For most substances, the relationships among thermodynamic properties are very
complex to be expressed by simple equation. Therefore, the properties are frequently
presented in tables. For each substance, the thermodynamic properties are listed in more
than one table. In fact, a separated table is prepared for each region of interest such as
25
compressed or sub-cooled liquid, saturated liquid-vapor mixture and superheated vapor
as shown in table 2-1 and table 2-2.
Table 2-2 properties of superheated vapor
P (MPa), Tsat (oC)
T ( oC)
v (m3/kg)
u (kJ/kg)
h (kJ/kg)
s (kJ/kg.K)
150
200
250
The properties that listed in these tables are specific volume, internal energy, enthalpy
and entropy. In saturated tables, the properties take symbol f like vf, uf, hf, and sf for
saturated liquid. Also, the properties take symbol g like vg, ug, hg, and sg for saturated
vapor.
Sub-cooled Liquid (compressed liquid)
For sub-cooled liquid or compressed liquid, the properties to be read from the table at a
given temperature only such as, vf, uf, hf, and sf.
Saturated Liquid
For saturated liquid, the properties should be read from the tables at a given temperature
or pressure such as, vf, uf, hf, and sf, because the temperature and pressure are related
each other at boiling point.
Saturated Vapor
For saturated vapor, the properties should be read from the tables at a given temperature
or pressure such as, vg, ug, hg, and sg, because the temperature and pressure are related
each other at boiling point.
Superheated Vapor
For superheated vapor, the properties should be read from the tables at a given two
parameters such as, temperature and pressure like, v, u, h, and s.
26
Saturated liquid vapor mixture
For saturated liquid vapor mixture, the properties should be calculated as a function of
the properties at saturated liquid and saturated vapor with dryness fraction or quality
x. The quality x is the ratio between the mass of vapor to the total mass of the mixture
as shown in Fig 2-8. Consider a tank contains a saturated liquid vapor mixture which
has a part of total mass as a liquid and other is a vapor at saturation temperature and
pressure. The quality x is defined as,
Fig. 2-8 quality x of saturated liquid vapor mixture
mtotal = mliquid + mvapor = m f + m g
x=
mg
mt
=
mg
m f + mg
or 1 − x =
mf
mt
=
mf
m f + mg
The volume of the mixture can be calculated as,
Total volume = Voume of liquid + Volume of vapor
V =V f +V g
mt v = m f v f + m g v g
v=
mf
mt
vf +
mg
mt
v g = (1 − x)v f + xv g
v = v f + x (v g − v f )
The analysis given above can be repeated for internal energy, enthalpy and entropy with
the following results,
u = u f + x(u g − u f )
h = h f + x(hg − h f )
s = s f + x( s g − s f )
27
It is important to know that the value of x = 0 ~ 1 or x = 0 % to 100 % and the value of
x = 0 for saturated liquid and x = 1 for saturated vapor condition.
Ideal Gas Equation of State
The equation that relates the pressure, temperature, and volume of a substance is called
an equation of state as follows,
PV = mRT
for unit of mass the relation becomes, Pv = RT
Where R is the gas constant and is defined as the ratio between the universal gas
constant Ru and the molecular weight, M, of the substance.
Ru
, where
M
Ru = 8.314 kJ /( kmol.K )
R=
or
Ru = 1.986
Btu /( Ibmol.R)
For a certain gas and fixed mass, the relation between P, T, and V for two cases as the
follows:
P1V1 = mRT1
mR =
and
P2V2 = mRT2
P1V1 P2V2
Pv
Pv
, and for unit mass 1 1 = 2 2
=
T1
T2
T1
T2
Internal Energy, Enthalpy, and Specific Heat of Ideal Gas
The specific heat is defined as the energy required to raising the temperature of a
unit mass of substance by one degree. In general, this energy will depend on how the
process is executed. In Thermodynamics, we are interested in two kinds of specific
heats, specific heat at constant volume, Cv, and specific heat at constant pressure, Cp.
Both the specific heat of constant volume and constant pressure are function of
temperature.
From the definition of Cv , for constant volume process, the heat added to the process is
equal to C v dT . Thus,
28
u = u (T )
du = C v dT
⎛ ∂u ⎞
Cv = ⎜ ⎟
⎝ ∂t ⎠ v
or
Similarly, for constant pressure process, the heat added to the process is equal to C p dT .
Thus, an expression for specific heat at constant pressure can be obtained as,
h = h(T )
dh = C p dT
⎛ ∂u ⎞
Cp = ⎜ ⎟
⎝ ∂t ⎠ p
or
For unit mass, the relation between Pv, u, h as follows,
Pv = RT
and h = u + Pv
or in differential form,
dh = du + RdT
C p dT = C v dT + RdT
C p = Cv + R
or
C p − Cv = R
At this point, we introduce another ideal gas property called the specific heat ratio, k,
defined as,
k=
Cp
Cv
The change in internal energy or enthalpy for ideal gas during a process from state 1 to
state 2 is determined by integrating these equations,
2
Δu = u 2 − u1 = ∫ C v (T )dT
1
2
Δh = h2 − h1 = ∫ C p (T )dT
1
By replacing the values of Cv(T) and Cp(T) with constant values, we obtain,
Δu = u 2 − u1 = C v (T2 − T1 )
Δh = h2 − h1 = C p (T2 − T1 )
29
Internal Energy, Enthalpy, and Specific Heat of Solids and Liquids
It can be mathematically shown that the constant volume and constant pressure specific
heats are identical for incompressible substances. Therefore, for solids and liquids, the
specific heats can be dropped, and both can be represented by a single symbol C.
C p = Cv = C
Internal Energy Change
Like ideal gases, the specific heats of incompressible substances depend on temperature
only. Thus, the partial differential form of u can be obtained as,
du = C v dT = C (T )dT or from state 1 to state 2
2
Δu = u 2 − u1 = ∫ C (T )dT
1
For small temperature interval, a C value at the average temperature can be used and
treated as constant,
Δu = u 2 − u1 ≅ C av (T2 − T1 )
Enthalpy Change
Using the definition of enthalpy for unite mass h = u + Pv and noting that the
volume v = constant, for incompressible substances, we can get,
dh = du + vdp + Pdv
For incompressible substances, dv = 0 , yields,
dh = du + vdP
Δh = Δu + vΔP
For solids, the term of, vΔP ≅ o , and thus Δu ≅ C av ΔT ≅ C av (T2 − T1 )
For liquids, two special cases are commonly encountered:
1- Constant pressure process, as in heaters (ΔP = 0) : Δh = Δu ≅ C av ΔT ≅ C av (T 2−T1 )
h2 − h1 = u 2 − u1 ≅ C av (T 2−T1 )
2- Constant temperature process, as in pumps (ΔT = 0) : Δh = vΔP = v( P 2 − P1 )
h2 − h1 = v( P2 − P1 )
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Examples of Pure Substances Properties
1. Complete the following tables for H2O:
T (oC)
P (kPa)
50
V (m3/kg)
4.16
200
250
400
110
600
T (oC)
P (kPa)
Saturated vapor
h (kJ/kg)
200
140
X
Phase description
0.7
1800
950
80
Phase description
0.0
500
800
3161.7
Solution
T (oC)
P (kPa)
V (m3/kg)
Phase description
50
12.349
4.16
Wet vapor
12.23
200
0.8857
Saturated vapor
250
400
0.5951
Superheated vapor
110
600
0.001052
Sub-cooled liquid
T (oC)
P (kPa)
h (kJ/kg)
X
Phase description
120.23
200
2046.03
0.7
Wet vapor
140
361.3
1800
0.564
Wet vapor
177.69
950
753.02
0.0
Saturated liquid
80
500
334.91
-
Sub-cooled liquid
350
800
3161.7
-
Superheated vapor
2. Complete the following tables for Refrigerant-134a:
T (oC)
P (kPa)
V (m3/kg)
Phase description
31
-8
500
30
0.022
320
100
600
T (oC)
P (kPa)
20
Saturated vapor
u (kJ/kg)
95
-12
Saturated vapor
400
8
Phase description
300
600
Solution
T (oC)
P (kPa)
V (m3/kg)
Phase description
-8
500
0.0007569
Sub-cooled liquid
30
770.06
0.022
Wet vapor
2.48
320
0.0632
Saturated vapor
100
600
0.0479
Superheated vapor
T (oC)
P (kPa)
u (kJ/kg)
Phase description
20
571.6
95
Wet vapor
-12
185.4
220.36
Saturated vapor
88
400
300
Superheated vapor
8
600
60.43
Sub-cooled liquid
3. Complete the following tables for H2O
T (oC)
P (kPa)
140
V (m3/kg)
0.48
800
25
Saturated liquid
750
500
T (oC)
Phase description
0.14
P (kPa)
u (kJ/kg)
400
1825
Phase description
32
220
190
Saturated vapor
2000
4000
3040
Solution
T (oC)
P (kPa)
V (m3/kg)
Phase description
140
361.3
0.48
Wet vapor
170.43
800
0.001115
Saturated liquid
25
750
0.001003
Sub-cooled liquid
500
2500
0.14
Superheated vapor
T (oC)
P (kPa)
u (kJ/kg)
Phase description
143.63
400
1825
Wet vapor
220
2318
2602.4
Saturated vapor
190
2000
806.19
Sub-cooled liquid
466.85
4000
3040
Superheated vapor
4. Water is to be boiled at sea level in stainless steel pan placed on top of a 3 kW
electric heater. If 60 % of the heat generated by the heater is transferred to the water
during boiling. Determined the rate of evaporation of water.
Solution
At sea Level, Patm = 101 kPa,
∴ h fg = 2257 kJ / kg from water saturated table
Q& t = m& w (hg − h f ) = m& w × h fg ,
Q 0.6 × 3 = m& w × 2257
∴ m& w = 7.9752 × 10 − 4
kg / s
5. In problem 4, if it is observed that the water level in the pan drops by 8 cm in 30
min, determine the rate of heat transfer to the pan.
Solution
33
Vw =
π
4
d 2 × h,
∴ Vw =
m w = Vw × ρ w = Vw ×
π
4
× 0.55 2 × 0.08 = 0.01901
m3
1
1
= 0.01901 ×
= 18.209
vf
0.001044
mw
18.209
=
= 0.010116
kg / s
time 30 × 60
Q Q& t = 0.010116 × 2257 = 22.832
Q& t = m& w × h fg ,
kg
m& w =
kW
6. Water is being heated in a vertical piston-cylinder device. The piston has a mass of
20 kg and a cross-sectional area of 100 cm 2. If the local atmospheric pressure is 100
kPa, determine the temperature at which the water will start boiling.
Solution
The water at saturated temperature,
P = Patm +
mp × g
T = 104.62
Ap
o
= 100 +
20 × 9.81
× 10− 3 = 119.62
−4
100 × 10
kPa
C
7. A rigid tank with a volume of 3.5 m3 contains 7.5 kg of saturated liquid-vapor
mixture of water at 65 oC. Now the water is slowly heated. Determine the
temperature at which the liquid in the tank is completely vaporized and the change of
internal energy. Also, show the process on a T-v diagram with respect the saturation
lines.
Solution
Constant volume process, V = C
34
V 3.5
=
= 0.4667 kg / m 3
m 7.5
At complet vaporization, saturated vapor, v g = v 2 = 0.4667 m 3 , ∴ T2 = 143.37
v1 =
o
C
State 1 is a wet vapor (saturated liquid vapor mixture), from the saturated table we can
calculate the quality x from the specific volume at state 1 of 65 oC as follows,
v1 = v f + x(v g − v f )
0.4667 = 0.001020 + x(6.197 − 0.001020)
x=
0.46568
= 0.075162 = 7.52 %
6.19598
The internal energy u1 at state 1 as,
u1 = u f + x(u g − u f ) = u f + xu fg
u1 = 272.02 + 0.0752 × 2191.1 = 436.791
kJ / kg
State 2 is a saturated vapor and the internal energy equal ug at vg = v2 = 0.4667 m3/kg
u 2 = u g at v g of 0.4667 m 3 /kg = 2553.303
kJ / kg
Δu = u 2 − u1 = 2553.303 − 436.791 = 2116.512
Total internal energy change is,
ΔU = mΔu = 7.5 × 2116.512 = 15873.84
kJ / kg
kJ
8. A rigid vessel contains 2 kg of refrigerant-134a at 900 kPa and 80 oC. Determine the
volume of the vessel and the total internal energy.
Solution
The saturated temperature of R-134a that corresponding to 900 kPa is equal to 35.53
o
C, Its mean that the refrigerant is superheated vapor, at 900 kPa, 80 oC. From
superheated vapor table of R-134a, we find,
35
v = 0.02861
m 3 / kg , u = 288.87
kJ / kg
V = m × v = 2 × 0.02861 = 0.05722 m
U = u × m = 288.87 × 2 = 577.74 kJ
3
9. A 0.5 m3 vessel contains 10 kg of R-134a at -20 oC. Determine (a) the pressure, (b)
the total internal energy, and (c) the volume occupied by the liquid phase.
Solution
v = V/m = 0.5/10 = 0.05 m3/kg, at -20 oC, vf < v < vg
The R-134a is in wet vapor region (saturated liquid vapor mixture), the pressure
corresponding to the saturated temperature of -20 oC is,
P = 132.99
kPa
v = v f + x (v g − v f ) ,
∴x =
v −v f
vg − v f
=
0.05 − 0.0007361
= 0.3382
0.1464 − 0.0007361
u = u f + x(u g − u f ) = 24.17 + 0.3382(215.84 − 24.17) = 88.992
kJ / kg
U = u × m = 88.992 × 10 = 889.92 kJ
mg
mf
x=
or 1 − x =
, ∴ m f = mt × (1 − x) = 10 × (1 − 0.3382) = 6.618
mt
mt
V f = v f × m f = 0.0007361× 6.618 = 4.8715 × 10 −3
kg
m3
10. A piston-cylinder device contains 1 kg of water vapor in equilibrium at 800 kPa and
quality of 55 %. Heat is transferred at constant pressure until the temperature reaches
350 oC. (a) what is the initial temperature of the water?, (b) determine the total
volume of mixture at state 1, (c) calculate the final volume, (d) show the process on a
P-v diagram with respect to saturation line.
Solution
Constant pressure process, P = C
36
State 1 is wet vapor (saturated liquid vapor mixture) at pressure of 800 kPa, from
saturated table of water vapor, the saturated temperature is
T1 = Tsat = 170.43
o
C , mt = 1
kg
v1 = v f + x(v g − v f ) = 0.001115 + 0.55 × (0.2402 − 0.001115) = 0.132612 m 3 / kg
V1 = mt × v1 = 1× 0.132612 = 0.132612 m 3 / kg
State 2 is superheated vapor at P2 = 800 kPa, and T 2= 350
v2 = 0.3544
m 3 / kg , V2 = mt × v2 = 1× 0.3544 = 0.3544
o
C
m3
11. A piston-cylinder device initially contains 50 lit of liquid water at 25 oC and 300
kPa. Heat is added to the water at constant pressure until the entire liquid is
vaporized. (a) What is the mass of the water, (b) What is the final temperature, (c)
determine the total enthalpy change, (d) show the process on P-v and T-v diagrams
with respect to saturation line.
Solution
The liquid water initially is sub-cooled liquid and the process of heating is a constant
pressure, P = C,
37
State 1 is compressed liquid at temperature of T1 = 25
v1 = v f = 0.001003 m 3 / kg ,
o
C
∴ mw = Vw / v f = 50 × 10 −3 / 0.001003 = 49.84 kg
h1 = h f = 104.89 kJ / kg
Process 1 → 2 is constant pressure, state 2 is saturated vapor at pressure of P2 = 300 kPa
T2 = 133.55
o
C , h2 = 2725.3 kJ / kg
Δh = h2 − h1 = 2725.3 − 104.89 = 2620.41 kJ / kg
ΔH = m × Δh = 49.84 × 2620.41 = 130601.234 kJ
12. A 0.5 m3 rigid vessel initially contains saturated liquid-vapor mixture of water at
100 oC. The water is now heated until it reached the critical state. Determine the
mass of the liquid water and the volume occupied by the liquid at the initial state.
Solution
For water, Pcr = 22.09 MPa and Tcr = 374.14
o
C and v f = v g = 0.003155 m 3 / kg
The heating process constant volume, v = v1 = v2 = 0.003155 m 3 / kg
mt = V / v = 0.5 / 0.003155 = 158.479
x1 =
v −vf
v g − vv
=
kg
0.003155 − 0.001044
= 0.0013
1.6729 − 0.001044
m f = mt (1 − x ) = 158.479 × (1 − .0013) = 158.273
kg
V f = m f × v f = 158.273 × 0.003155 = 0.49935 m 3
13. Determine the specific volume, internal energy, and enthalpy of compressed liquid
water at 100 oC and 15 MPa.
Solution
For water,at T = 100 o C , P = 15 MPa , from compressed table of water
v = 0.0010361, m 3 / kg ,
u = 414.74
kJ / kg ,
38
h = 430.28 , kJ / kg
14. A rigid tank contains water vapor at 300 oC and an unknown pressure. When the tank
is cooled to 180 oC, the vapor starts condensing. Estimate the initial pressure in the
tank.
Solution
The process is constant volume process, v = C
At T2 = 180 o C , x 2 = 1, ∴ v1 = v 2 = v g = 0.19405
m 3 / kg
at 300 o C , v f < v1 > v g , ∴ the state is super heated vapor , P = 1.33
MPa
15. The pressure gage on a 2 m3 oxygen tank reads 500 kPa. Determine the amount of
oxygen in the tank if the temperature is 25 oC and the atmospheric pressure is 100
kPa.
Solution
Pressure inside the tank , P = Patm + Pg = 100 + 500 = 600
From ideal gas equation, PV = mRT
∴ m = PV / RT = 600 × 2 /(0.2589 × 298) = 15.554
kPa
kg
16. A 1 m3 tank containing air at 25 oC and 500 kPa is connected through a valve to
another tank containing 5 kg of air at 35 oC and 200 kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium with the surroundings,
which at 20 oC. Determine the volume of the second tank and the final equilibrium
pressure of air.
Solution
39
Case 1 , Tank A , mA = PV / RT = 500 × 1/(0.287 × 298) = 5.846
Tank B , VB = mRT / P = 5 × 0.287 × 308 / 200 = 2.2099
Case 2 , m = mA + mB = 5.846 + 5 = 10.846
kg
V = VA + VB = 1 + 2.2099 = 3.2099
m3
kg
m3
Final Pressure , P = mRT / V = 10.846 × 0.287 × 293 / 3.2099 = 284.137
kPa
Problems
1- Water is being heated in a vertical piston-cylinder device. The piston has a mass of
20 kg and a cross-sectional area of 100 cm 2. If the local atmospheric pressure is 100
kPa, determine the temperature at which the water will start boiling. Determine the
specific volume, internal energy, and enthalpy of compressed liquid water at 100 ºC
and 15 MPa.
2- A rigid tank with a volume of 3.5 m3 contains 7.5 kg of saturated liquid-vapor
mixture of water at 65 ºC. Now the water is slowly heated. Determine the
temperature at which the liquid in the tank is completely vaporized. Also, show the
process on a T-v diagram with respect the saturation lines.
3- A rigid vessel contains 2 kg of refrigerant-134a at 900 kPa and 80 ºC. Determine the
volume of the vessel and the total internal energy.
4- A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m3 of water vapor in
equilibrium at 800 kPa. Heat is transferred at constant pressure until the temperature
reaches 350 ºC. (a) what is the initial temperature of the water?, (b) determine the
total mass of the water, (c) calculate the final volume, (d) show the process on a P-v
diagram with respect to saturation line.
40