Question

3 kg of water at 80°C is mixed with 4 kg of water at 15°C in an isolated system. Calculate the change of entropy due to mixing process.

Answer 1

hows the isolated system before mixing. When barrier is removed, the water from two compartments mix each other. Let t_m is the final equilibrium temperature after mixing.

Applying first law of thermodynamics to the isolated system : 

Total energy before mixing = Total energy after mixing

∴ 3c_pw (80 – 0) + 4c_pw (15 – 0) = 7 c_pw (t_m – 0)

[c_pw = Specific heat of water at constant pressure]

240c_pw + 60c_pw = 7c_pw t_m

240 + 60 = 7 t_m

t_m = \(\cfrac{300}7\) = 42.85°C

Initial entropy of the system,

Final entropy of the system

= (3 + 4) c_pw log_e \(\left(\cfrac{42.85+273}{273}\right)\) = 1.0205 c_pw

Net change in entropy, 

∆S = Final entropy – Initial entropy 

= 1.0205c_pw – 0.9848 c_pw = 0.0357 c_pw 

= 0.0357 × 4.187 kJ/K 

[\(\because\) c_pw = 4.187 kJ/kg K] 

= 0.1495 kJ/K 

Hence, net change in entropy = 0.1495 kJ/K