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Homework Statement
A 100 g mass of supercooled water at -2ºC in thermal contact with surroundings also at -2ºC
freezes spontaneously.
Calculate the entropy change of the universe assuming that the surroundings act as a large
temperature reservoir. [The specific heat capacity of ice is 2090 J kg-1 K-1, the specific heat
capacity of water is 4183 J Kg-1 K-1, and the latent heat of fusion of water is 334.7 kJ kg-1.]
Is this process reversible?
Homework Equations
Change in entropy of the universe = change in entropy of the system + change in entropy of the surroundings
dS = dQ/T (for a reversible process)
The Attempt at a Solution
latent heat of fusion = l = 334700J/Kg
T2 = freezing temperature of water = 273K
T1 = -2 degrees Celcius = 271K
mass = m = 0.1Kg
Change in entropy of the system = (-)m*l/T2 = (0.1*334700)/273 = -122.60J/K
Change in entropy of the surroundings = (+)m*l/T1 = (0.1*334700)/271 = 123.51J/K
Change in entropy of the universe = 123.51 -122.60 = 0.9J/KThe process is irreversibleI have a feeling this is wrong because i haven't used the heat capacities of ice and water (not sure how to) any help would be appreciated, thanks.