100 g of ice at 0 degree C is added to an insulated beaker containing 150 g water at the 100 degree C. The final temperature of the mixture 56.8 degree C. Calculate the entropy change of the process. Given heat of fusion of ice at 0 degree C is 80 cal/g , heat capacity of water is 1 cal/g .
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Answer:
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answer : entropy change would be 29.74 cal/K [approximately]
mass of ice, m = 100g
temperature of ice = 0°C
mass of water, M = 150g
temperature of water = 100°C
the final temperature of the mixture = 56.8°C
concept : at constant temperature, ∆S = Q/T
when temperature varies, ∆S = 2.303ms log(T_f/T_i)
step 1 : when ice melts,
heat gained, Q = mL
= 100g × 80cal/g
= 8000 cal
and temperature, T = 0°C = 273K
now ∆S1 = 8000/273 = 29.3 cal/K
step 2: water formed by molten ice increases its temperature upto 56.8°C
so, ∆S2 = 2.303 × 100g × 1cal/g × log (56.8 + 273)/273
= 230.3 log(329.8/273)
= 18.9 cal/K
step 3 : water of 100°C decreases its temperature from 100°C to 56.8°C
∆S3 = -2.303 × 150g × 1cal log(373/329.8)
= -18.46 cal/K
now net entropy change = ∆S1 + ∆S2 + ∆S3
= 29.3 + 18.9 - 18.46
= 29.74 cal/K