Explain how you could calculate the heat released in freezing 0.250 mol water?

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To calculate how much heat is released from 0.250 mil of water when it freezes.

The mass of 0.250 mol of water = has a mass of .250{2 for H and 16 for O} = .250*18 = 4.5 gram.

If the water freezes at zero degree Celsius , it has to release a latent heat of 334J/gram when it converts from water to ice (from liquid to solid state).

So m gram ( (1/18) mol) of water has to release a heat of mass*L heat of fusion = 334m J.

So 4.5 gram of water releases a heat energy of 4.5 *334 J = 1503 J of heat energy while freezing.

If you want the calculation of heat release for water at 20 water to change its state ice, then it has release heat energy to come dowm from 20 celsius degree to zero in liquid state and then release heat while coveting from zero degree liquid state to zero degree solid state of ice . So it has to relase 4.5gram*(20-0) Joules + 4.5 L Joules = 4.5(20+L) Joules = 4.5(20+334) = 4.5*354 Joules = 1593 Joule of heat energy.

The amount of heat removed from water for freezing it is directly proportional do the mass of ice formed. The exact amount of heat released or extracted or released by the water frozen is given by:

Heat extracted = (Mass of ice formed)x(Latent heat of freezing for ice)

1 mol of water 18.0152 grams

Therefor mass of 0.25 mol of water = 0.25x18.0152 = 4.5038 grams

And latent heat for freezing of ice is 80 calories per gram.

Therefore:

Latent heat released by 4.5038 grams (0.25 mol) water in freezing is given by:

Heat released = 4.5038x80 = 360.304 calories

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