TITLE.PM5

296 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

(ii)What is the minimum amount of work necessary to convert the water back into ice at

• 5°C?
  Take : cp of ice = 2.093 kJ/kg°C
  Latent heat of fusion of ice = 333.33 kJ/kg.
  Solution. Mass of ice, m = 1 kg
  Temperature of ice = – 5°C (= – 5 + 273 = 268 K)
  Temperature of atmosphere = 25°C (= 25 + 273 = 298 K)
  Heat absorbed by ice from the atmosphere (Fig. 5.51)
  = Heat absorbed in solid phase + latent heat
  
  
  • heat absorbed in liquid phase
    = 1 × 2.093 × [0 – (– 5)] + 1 × 333.33 + 1 × 4.187 × (25 – 0)
    = 10.46 + 333.33 + 104.67 = 448.46 kJ.
    (i)Entropy increase of the universe, (∆ s)universe :
    Entropy change of the atmosphere,
  
  
  
  

(∆ s)atm. = – QT = –
448.46
298 = – 1.5049 kJ/K
Entropy change of system (ice) as it gets heated from – 5°C
to 0°C,

(∆ sI)system = mc

dT

268 p T

273

z = 1 × 2.093 loge^

273

268 = 0.0386 kJ/K

Entropy change of the system as ice melts at 0°C to become water at 0°C.

(∆ sII)system = 333.33

273

= 1.2209 kJ/K

Entropy change of water as it gets heated from 0°C to 25°C

(∆ sIII)system = mcpdTT

273

298

z = 1 × 4.187 loge^

298

273

F

H

I

K = 0.3668 kJ/K

Total entropy change of ice as it melts into water

(∆ s)total = ∆ sI + ∆ sII + ∆ sIII

= 0.0386 + 1.2209 + 0.3668 = 1.6263 kJ/K

Then temperature-entropy diagram for the system as ice at – 5°C converts to water at 25°C

is shown in Fig. 5.52.

∴ Entropy increase of the universe,

(∆ s)univ. = (∆s)system + (∆s)atm.

= 1.6263 + (– 1.5049) = 0.1214 kJ/K. (Ans.)

(ii)Minimum amount of work necessary to convert the water back into ice at

• 5°C, Wmin. :

To convert 1 kg of water at 25°C to ice at – 5°C, 448.46 kJ of heat have to be removed from
it, and the system has to be brought from state 4 to state 1 (Fig. 5.52). A refrigerator cycle, as
shown in Fig. 5.53, is assumed to accomplish this. The entropy change of the system would be the
same, i.e., s 4 – s 1 , with the only difference that its sign will be negative, because heat is removed
from the system (Fig. 5.52).
(∆s)system = s 1 – s 4 (negative)

Fig. 5.51