SECOND LAW OF THERMODYNAMICS AND ENTROPY 291

dharm

/M-therm/th5-4.pm5

W

Q 1

Q– W 1

System

T = 250 K 1

HE

Reservoir

T = 125 K 2

(ii)Entropy change of source 1 =

−Q

T

1

1

= −^1500

500

= – 3 kJ/K. (Ans.)

Entropy change of sink 2 =

Q

T

2

2

=^1600400 = 4 kJ/K. (Ans.)

Entropy change of source 3 = −TQ^3

3

= −^300
300
= – 1 kJ/K. (Ans.)
(iii) Net change of the entropy = – 3 + 4 – 1 = 0
As the cycle is completed, the net change in entropy must be zero because entropy is a
property.
It may be observed from the new arrangement that the engine takes heat from source 1 and
source 3 and rejects to source 2, only i.e., the equipment does both a heat engine and a heat pump
function.
Example 5.45. The heat capacity of a system at constant volume is given by Cv = ZT^2
where Z = 0.045 J/K^3
A system is originally at 250 K, and a thermal reservoir at 125 K is available. Determine
the maximum amount of work that can be recovered as the system is cooled down to the tempera-
ture of the reservoir.
Solution. Refer Fig. 5.4 8.
Initial temperature of the system, T 1 = 250 K
Temperature of the reservoir, T 2 = 125 K
Heat capacity of the system at constant volume is
Cv = ZT^2 = 0.045 T^2

Theproduct of mass and specific heat ( ) is called the

of the substance. The capital letter ,

or is used for heat capacity

mc

heat capacity C C

C

p

v

L

N

M

M

M

O

Q

P

P

P

Heat removed from the system

QCdTv TdT

T

T

1

2

250

125

1

2

==zz0. 045

= 0.045

T^3

250

125

3

L

NM

O

QP

= 0 045. 3 (125^3 – 250^3 )

= – 205.08 × 10^3 J

(∆ S)system = C dT

T

T dT

250 v T

125

250

(^1252)
zz= 0.^045

L
N
M
M
O
Q
P
z P
0 045 0 045
250 2
125 2
250
125
..TdT T
= 0 045. 2 (125^2 – 250^2 ) = – 1054.7 J/K
(∆ S)reservoir =
QW
T
1 −
reservoir =
205 08 10
125

. ×−^3 W
J/K

Fig. 5.48