Calculation in change in entrophy when 5
kg water at 30°C is mixed with 15 gm
water at 10°C
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Let the final temperature reached be be t ℃
Specific heat capacity of water = c = 1000 J kg − ℃ −
Respective masses = 2 kg
Respective temperatures = 0℃ and 60℃
Since at 0℃ also it's water , there's no need to consider any change regarding latent heat.
Let's assume that the system is isolated (actually it has to be since we don't know the temperature of the surroundings and so cannot detect any change regarding that).
By 0th law of Thermodynamics ,
Heat lost by 60℃ water = Heat gained by 0℃ water
i.e. 2×c×(60-t) = 2×c×(t-0)
=> 60-t = t-0
=> 2t = 60
=> t = 30
Change in entropy of 0℃ water = {2×1000×(30–0)}/(30+273)
~ 198.02 J/K
Change in entropy of the 60℃ water = -{(2×1000×(60–30)}/(30+273)
~ -198.02 J/K
However the net change in the entropy of the total system(of 0℃ water and 60℃ water) = 198.02 -198.02 = 0 J/K
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