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An 800-g block of ice at 0.00°c is resting in a large bath of water at 0.00°c insulated

An 800-g block of ice at 0.00°c is resting in a large bath of water at 0.00°c insulated from the environment. after an entropy change of this system of 600 j/k, how much ice remains unmelted? the latent heat of fusion of water is 333 j/g.
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ewalker1917
ewalker1917
4,8(30 marks)

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

           Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

efjdjskd
efjdjskd
4,8(21 marks)
Y=3•10x is the answer I hope it’s correct

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