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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)?

Respuesta :

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

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