heat problem

By taking the specific heat capacity of steam = 2080 J/kg K, we have:
When cooling down from 125°C to 100°C, heat energy taken away is:
25 × 2080 × 0.075 = 3.9 kJ
By taking the specific latent heat of condensation of steam = 2.26 × 106 J, we have:

When condensing from steam to water at 100°C, heat energy taken away is:
0.075 × 2.26 × 106 = 169.5 kJ
By taking the specific heat capacity of water = 4200 J/kg K, we have:
When cooling down from 100°C to 0°C, heat energy taken away is:
100 × 4200 × 0.075 = 31.5 kJ
By taking the specific latent heat of freezing of water = 3.34 × 105 J, we have:

When freezing from water to ice at 0°C, heat energy taken away is:
0.075 × 3.34 × 105 = 250.5 kJ
So upon the 4th cooling procedure, total heat energy taken away should be 3.9 + 169.5 + 31.5 + 25.05 = 229.95 kJ.
Hence, if 215 kJ of energy is taken away, the resultant phases should be a mixture of ice and water as this value lies between the 3rd and 4th cooling process.