294 ENGINEERING THERMODYNAMICS
dharm
/M-therm/th5-4.pm5
Solution. Mean temperature of the mixture = GHFTT^12 + 2 IKJ.
Thus change in entropy is given by
∆S = mc dT
T
mc dT
T T
TT
TT
T
1
12
12
(^22)
2
()/
()/
- zz+
−
= mc loge TT^12 T
(^21)
F +
HG
I
KJ
- mc loge 22
12
T
TT+
F
HG
I
KJ
= mc loge TT^12 T
(^21)
F +
HG
I
KJ + mc loge^
TT
T
12
(^22)
F +
HG
I
KJ
= mc loge
()TT
TT
122
(^412)
- = mc loge TT
TT
12
12
2
2
L +
N
M
M
O
Q
P
P
= 2 mc loge TT
TT
12
(^212)
L +
N
M
M
O
Q
P
P
= 2 mc loge ()/TT
TT
12
12
L + 2
N
M
M
O
Q
P
P
i.e., Resultant change of entropy of universe
= 2 mc loge (T T ) / 2
TT
12
12
L +
N
M
M
O
Q
P
P
...Proved
The arithmetic mean TT^12 + 2 is greater than geometric mean TT 12.
Therefore, loge ()/TT
TT
12
12 - (^2) is + ve.
Thus, the entropy of the universe increases.
+Example 5.49. (a) One kg of water at 0°C is brought into contact with a heat reservoir
at 90°C. When the water has reached 90°C, find :
(i)Entropy change of water ;
(ii)Entropy change of the heat reservoir ;
(iii)Entropy change of the universe.
(b) If water is heated from 0°C to 90°C by first bringing it in contact with a reservoir at
40 °C and then with a reservoir at 90°C, what will the entropy change of the universe be?
(c) Explain how water might be heated from 0°C to 90°C with almost no change in the
entropy of the universe.
Solution. Mass of water, m = 1 kg
Temperature of water, T 1 = 0 + 273 = 273 K
Temperature of the heat reservoir, T 2 = 90 + 273 = 363 K.
(a) Refer Fig. 5.50. Water is being heated through a finite temperature difference. The
entropy of water would increase and that of the reservoir would decrease so that net entropy
change of the water (system) and the reservoir together would be positive definite. Water is being
heated, irreversibly, and to find the entropy change of water, we have to assume a reversible path
between the end states, which are at equilibrium.