TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 295

dharm

/M-therm/th5-4.pm5

System

(0 + 273

= 273 K)

Q

Reservoir

(90 + 273 = 363 K)

(i)Entropy of water,

(∆ s)water =

mcdT

T T

T

1

2

z = mc loge^

T

T

2

1

(c = Specific heat of water)

= 1 × 4.187 × loge (^363273)
= 1.193 kJ/kg K. (Ans.)
(ii) The temperature of the reservoir remains con-
stant irrespective of the amount of the heat withdrawn.
Amount of heat absorbed by the system from the
reservoir,
Q = 1 × 4.187 × (363 – 273) = 376.8 kJ
∴ Entropy change of the reservoir,
(∆ s)reservoir = – QT = – 376.8
363
= – 1.038 kJ/K. (Ans.)
(–ve sign indicates decrease in entropy).
(iii)Entropy change of the universe,
(∆ s)universe = (∆ s)water (or system) + (∆ s)reservoir
= 1.193 + (– 1.038) = 0.155 kJ/K. (Ans.)
(b) The heating of water is being carried out in two stages, first from 0°C (273 K) to 40°C
(i.e., 313 K) by bringing in contact with a reservoir at 40°C (313 K), and then from 40°C (313 K) to
90°C (363 K) by bringing in contact with a second reservoir at 90°C (363 K).
(∆ s)water = mcdTT mcdTT
273
313
313
363
zz+ = 1 × 4.187 logeelog
313
273
363

• 313
  F
  H
  I
  K
  = 4.187 (0.1367 + 0.1482) = 1.1928 kJ/K.
  (∆ s)res. I = –
  1 313 273
  313
  ××−4.187 ( )
  = – 0.535 kJ/K
  (∆ s)res. (^) II = –^1 363 313
  363
  ××−4.187 ( ) = – 0.576 kJ/K
  ∴ (∆ s)univ. = (∆ s)water + (∆ s)res. (^) I + (∆ s)res. (^) II
  = 1.1928 + (– 0.535) + (– 0.576) = 0.0818 kJ/K.
  i.e., Entropy change of universe = 0.0818 kJ/K. (Ans.)
  (c) The entropy change of universe would be less and less, if the water is heated in more and
  more stages, by bringing the water in contact successively with more and more heat reservoirs,
  each succeeding reservoir being at a higher temperature than the preceding one. When water is
  heated in infinite steps, by bringing in contact with an infinite number of reservoirs in succession,
  so that at any instant the temperature difference between the water and the reservoir in contact is
  infinitesimally small, then the entropy change of the universe would be zero and the water would
  be reversibly heated.
  Example 5.50. 1 kg of ice at – 5°C is exposed to the atmosphere which is at 25°C. The ice
  melts and comes into thermal equilibrium.
  (i)Determine the entropy increase of the universe.
  Fig. 5.50