How much heat energy would be released if 78.1 g of water at 0.00 °c were converted to ice at - 57.1 °c. give your answer as a positive number in kilojoules (kj) ?

To convert 78.1 g of water at 0° C to Ice at - 57.1°C; we can do it in steps;

1. Water at 0°C to ice at 0°C

The heat of fusion of ice is 334 J/g;

Heat = 78.1 * 334 = 26085.4 Joules

2. Ice at 0°C to - 57.1°C

Specific heat of ice is 2.108 J/g

Heat = 78.1 * 2.108 J/g = 164.6348 Joules

Thus the total heat energy released will be; 26085.4 + 164.6348

= 26250.0348 J or 26.250 kJ