Answer:
662480 J
Explanation:
The heat required to convert 0.8Kg of ice at -35°c to steam at 100°c is the sum of the heat required to raise the temperature of the ice from -35°c to 0°C, the latent heat of fusion of ice and the heat required to raise the temperature of the ice from 0°c to 100°c.
Latent heat of fusion for ice= 3.36× 10^5 J Kg-1
Specific heat capacity of ice= 2060 J kg−1K−1
Specific heat capacity of water= 4200 JKg-1K-1
Heat required to raise the temperature of the ice to fusion point
H= 0.8 × 2060 × [0-(-35)]
H= 57680 J
Latent heat of fusion of ice;
H= mL
H= 0.8 ×3.36× 10^5
H= 2.688×10^5 J
Heat required to raise the temperature of the water to 100°C
Since mass is conserved, the mass of ice that has melted has been turned into an equivalent mass of water.
Therefore mass of water formed= 0.8Kg
H= 0.8× 4200 × 100
H= 336000J
Therefore total heat required;
H= 57680 + 2.688×10^5 + 336000
H= 662480 J
