Question

2. Water at the Antarctica usually exits as ice, making it very difficult to use the water or conduct experiment requiring the water to be in liquid state. A coffee addicted physicist stationed at a research base requires about 0.8 kg of ice for his coffee on the hour.
a. How much heat is required to convert 0.8 kg of ice at —35 °C into steam at 100°C?

Answer 1

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

Answer 2

Answer:

2468,080j

Explanation: