TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 293

dharm

/M-therm/th5-4.pm5

Solution. Fig. 5.49 shows the isolated system before mixing. When barrier is removed, the
water from two compartments mix each other. Let tm is the final equilibrium temperature after
mixing.

Isolated

system

Barrier

3 kg

at 80°C

(Water)

4 kg

at 15°C

(Water)

Fig. 5.49
Applying first law of thermodynamics to the isolated system :
Total energy before mixing = Total energy after mixing
∴ 3 cpw (80 – 0) + 4cpw (15 – 0) = 7 cpw (tm – 0)
[cpw = Specific heat of water at constant pressure]
or 240 cpw + 60cpw = 7cpw tm
or 240 + 60 = 7 tm
∴ tm =^3007 = 42.85°C
Initial entropy of the system,

= 3cpw loge

80 273

273

F +

H

I

K + 4cpw loge^

15 273

273

F +

H

I

K

= 0.7709cpw + 0.2139 cpw = 0.9848 cpw

Final entropy of the system

= (3 + 4) cpw loge 42.85 273

273

F +

HG

I

KJ

= 1.0205 cpw

Net change in entropy,

∆S = Final entropy – Initial entropy

= 1.0205cpw – 0.9848 cpw = 0.0357 cpw

= 0.0357 × 4.187 kJ/K [Q cpw = 4.187 kJ/kg K]

= 0.1495 kJ/K

Hence, net change in entropy = 0.1495 kJ/K. (Ans.)

Example 5.48. A mass ‘m’ of fluid at temperature T 1 is mixed with an equal mass of the

same fluid at T 2. Prove that the resultant change of entropy of the universe is 2 mc

(T T )/2

TT

12

12

+

and also prove that it is always positive.