Let the final temperature reached be t℃
Specific heat capacity of water = c = 1000 J kg− ℃−
Given masses = 1 kg and 2 kg
Respective temperatures = 30℃ and 0℃
By zeroth law of Thermodynamics,
Heat lost by 0℃ water = Heat gained by 0℃ water
i.e. 1 × c × (30-t) = 2 × c ×(t-0)
30 - t = 2t
30 = 3t
t = 10
Change in entropy of 30℃ water = {1×1000×(10–0)}/(10+273)
= 35.33 j/k
Change in entropy of the 0℃ water = -{(2×1000×(30–10)}/(10+273)
= 212.01 j/k
However the net change in the entropy of the total system(of 30℃ water and 0℃ water)
= 35.33 - 212.01 j/k
= 176.88 j/k