Solutions Exercises Lecture 2 - AES1030
1. 20 kg water with a temperature of 20oC are in a container. The water is stirred, so that
0.25 kW work is done on the system. How long does it take to increase the
temperature of the water to 30oC. You can assume that there is no heat transition from
the water to the surroundings? Further the heat capacity of the water at constant
pressure is given: C*P = 4.18 kJ kg-1 oC-1.
Solution
The system is isolated (no heat exchange with the surroundings, dq = 0) and consists
of 20 kg water of 200C.
Work is done on the system by stirring so that the temperature in the system
increases. The stirring is done with 0.25 kW or 0.25 kJ s-1. Or other, 0.25 kW = w/ t ,
with t indicating the time. To get an answer to the problem, first it needs to be
determined how much work needs to be done to heat 20 kg of water for 10 K. With
the help of the first law of thermodynamics the work can be determined recalling that
the process is adiabatic (dq=0)
dU = dq + dw = dw and dU = dw ⇒ ∆U = w
But how can the work be computed from this? Therefore, we assume that the volume
of the water and the pressure in the water stay constant for the described process
(assuming the water is incompressible). Typically for the thermodynamic description
of the changes of state functions for an isobaric process is done starting with the
computation of the change of the enthalpy. The relation between the enthalpy and the
internal energy is given via the definition of the enthalpy H = U + pV and dH= dU +
pdV + Vdp.
For an isobaric and isochoric process the total differential for the enthalpy simplifies
to dH = dU ⇒∆H = ∆U. The toal differential can also be defined in terms of the
 ∂H 
 ∂H 
pressure and the temperature with H = H(p,T) so that dH = 
 dp + 
 dT .
 ∂T  p
 ∂p T
The partial derivative of the enthalpy with respect to the pressure at constant pressure
 ∂H 
*
is equal to the heat capacity: 
 = CP = m ⋅ CP . With this the total differential of
 ∂T  p
the enthalpy in terms of T and p simplifies for an isobaric process (dp = 0) to:
T2 = 303 K
T2 = 303 K
T2 = 303 K
 ∂H 
*
∆H = ∫ 
 dT = ∫ C p , H 2O dT = ∫ mH 2O CP , H 2O dT
∂
T
p
T1 = 293 K 
T1 = 293 K
T1 = 293 K
∆H = 20kg ⋅ 4.18
kJ
⋅ ( 303 − 293) K = 836kJ
kg ⋅ K
.From this the time required to stir the system to heat the system for 10 K can be
computed:
∆H = ∆U = w = 836kJ = 0.25kW * t
⇒t =
836kJ
= 3344 s
kJ
0.25
s
This means one needs to stirred for 3344 s to heat 20 kg water for 10 K.
2. Liquid water of a temperature of 180oC and 1002.7 kPa has an internal energy of
762.0 kJ kg-1 and a specific volume of 1.128 cm3 g-1.
(a) Calculate the enthalpy H of the water at the given state.
(b) Then the water is brought into its gaseous state at which the gas has a temperature
of 300 oC and a pressure of 1500 kPa. The internal energy U is
2784.4 kJ kg-1 and the specific volume 169.7 cm3 g-1.
Calculate ∆U en ∆H for this process.
Solution:
The system to be described is liquid water of 180oC and 1002.7 kPa. The internal energy
of the water is given EH2O(L)= 762,0 kJ kg-1 and has a specific volume of
V* = 1.128 cm3 g-1.
Further it is given that 1Pa = 1Nm-2; 103 g = 1 kg and 106 cm3 = 1 m3.
a) To compute the enthalpy of the water at the given state, we make use of the
definition f the enthalpy: H = U + pV so that
H H 2O ( L ) 180o C ,1002.7 Pa = U H 2O ( L ) 180o C ,1002.7 kPa + p ⋅ VH 2O ( v ) 180o C ,1002.7kPa
(
)
(
)
(
J
N
10−6 m3
+ 1002.7 ⋅1000 2 ⋅1.128 −3
kg
m
10 kg
J
kJ
= 763131.046
= 763.131
kg
kg
= 762.0 ⋅1000
b) The water is brought into the gaseous state at 300oC and 1500kPa. The change of
the internal energy of the enthalpy for this process needs to be calculated:
∆U=UH2O(V) (300oC,1500kPa)-UH2O(L)(180oC,1002.7kPa)
∆H=HH2O(V) (300oC,1500kPa)-HH2O(L)(180oC,1002.7kPa)
For the computation of the change of the internal energy all properties are given:
∆U = U H 2O (V ) ( 300o C ,1500kPa ) − U H 2O ( L ) (180o C ,1002.7 kPa )
kJ
kJ
kJ
− 763.131 = 2021.269
kg
kg
kg
For the computation of the change in the enthalpy first HH2O(V) (300oC,1500kPa)
needs tob e computed:
= 2784.4
)
(
)
(
)
(
H H 2O (V ) 300o C ,1500kPa = U H 2O (V ) 300o C ,1500kPa + p ⋅VH 2O (V ) 300o C ,1500kPa
J
N
10−6 m3
+ 1500 ⋅1000 2 ⋅169.7 −3
kg
m
10 kg
J
kJ
= 3038950
= 3038.95
kg
kg
= 2784.4 ⋅1000
So that we finally get:
∆H = H H 2O (V ) 300o C ,1500kPa − H H 2O ( L ) 180o C ,1002.7 kPa
(
= 3038.95
)
(
)
kJ
kJ
kJ
− 763.131 = 2275.819
kg
kg
kg
200 g water is heated isobarically from 10 0C to 20 0C.
a. Compute the change of entropy if the process is done reversibly.
b. What is the change in entropy if the process is done irreversibly? Support
your answer.
It can be assumed that the heat capacity is independent of temperature and that
water is incompressible. The heat capacity of water is given by: CP = 75,3 J
K-1 mol-1 and the molar mass by MH2O=18,02 g mol-1.
Solution:
The change opf the entropy can be computed based on
dq
dS = rev .
T
The entropy is a state function; thus the value of the change of the entropy is independent
on the chosen path. If the real process is irreversible, then the reversible process from the
same initial to the same end state needs to be defined for the computation of the change
of entropy. The answer to exerecise b) is thus already given, the change of entropy for the
irreversible and the reversible process is the same as long as the initial and end state are
the same.
a)
The process is reversible and isobaric. The system consist of water (thus not ideal gas!).
We start with the first law of thermodynamics: dU = dq + dw. It is more practival to
wrok with the enthalpy because the process is isobaric (the equations are simpler):
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 ⋅ dT
 ∂T  p
 ∂p T
3.
 ∂H 
dp = 0 ⇒ dH = 
 ⋅ dT = CP ⋅ dT
 ∂T  p
The relation between the enthalpy and the internal energy is given via the
definition of the enthalpy H ≡ U + pV:
dH = dU + d ( p ⋅ V ) = dU + V ⋅ dp + p ⋅ dV 
 ⇒ dH = dqrev + V ⋅ dp
dU = dqrev + dwrev = dqrev − p ⋅ dV

dp = 0 ⇒ dH = dqrev
)
For an reversible and isobaric process, the change in the enthalpy of the
system is equal to the heat exchanged between the system and the
surroundings the change of the enthalpy of an isobaric process is equal to dH
= CP dT so that the change of the entropy can be computed by
T2
dqrev dH CP ⋅ dT
C ⋅ dT
dS =
=
=
⇒ ∆S = ∫ dS = ∫ P
T
T
T
T
T1
T2
CP = const ⇒ ∆S = CP ⋅ ∫
T1
T 
dT
= CP ⋅ ln  2 
T
 T1 
With
w
200 g
J
J
⋅ CP ⇒ CP =
⋅ 75.3
= 835.74
g
M
K ⋅ mol
K
18.02
mol
Giving (Remember the temperature needs to be converted to Kelvin before
the actual change of the entropy can be computed):
T 
 200 C 
J
J
 293.15 K 
∆S = CP ⋅ ln  2  = 835, 74 ⋅ ln  0  = 835, 74 ⋅ ln 

K
K
 283.15 K 
 10 C 
 T1 
CP = n ⋅ CP =
∆S = 29
J
K