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

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

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

Answer»

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

Discussion

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