Answer:
1. 111.54 kJ
2. 944.68 kJ
3. 339 kJ
4. 94.92 kJ
5. 21.0578 kJ
Step-by-step explanation:
1. The heat required to melt a given quantity of ice is known as the latent heat of fusion,
, of ice
, of ice = 330 J/g
The heat required to melt a given mass of ice = Mass of ice, m × latent heat of fusion,
, of ice
The heat required to melt 338 grams of ice = 338 g × 330 J/g = 111,540 J = 111.54 kJ
2. The latent heat of vaporization, l, of water = 2260 J/g
The heat required to convert, m grams of water to steam at
The heat required to convert 418 grams of boiling water into steam = 418 g × 2260 J/g = 944680 J = 944.68 kJ
3. To convert 150 grams of boiling water to steam, we have;
Heat required = 150 g × 2260 J/g = 339,000 J = 339 kJ
4. To convert 42.0 grams of boiling water to steam, we have;
Heat required = 42.0 g × 2260 J/g = 94,920 J = 94.92 kJ
5. The specific heat capacity of steam is 1.996 J/(g·°C)
The heat, Q, required to raise a given mass, m, of steam by Δt °C is given as follows;
Q = m × c × Δt
The heat required to raise the temperature of 422 grams of steam from 110.0 °C to 135.0 °C is therefore;
Q = 422 g × 1.996 J/(g·°C) × (135.0 °C - 110.0 °C) = 21057.8 J
The heat required to raise the temperature of 422 grams of steam from 110.0 °C to 135.0 °C = 21057.8 J = 21.0578 kJ