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1kg. Water (at 30oC) is mixed with 2kg. Water (at OoC) calculates the change in entropy

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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

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