I need help with my chemistry homework Here are the problems that I am struggling with. I need this to be answered by 11:59 tonight. 1. How many joules are required to melt 338. grams of ice?

2. How many joules are required to convert 418 grams of boiling water into steam?

3. How many joules are required to convert 150. grams of boiling water into steam?

4. How many joules are required to convert 42.0 grams of boiling water into steam?

5. How many joules are required to heat 422 grams of steam from 110.0 0C to 135.0 0C?

1. 111.54 kJ

2. 944.68 kJ

3. 339 kJ

4. 94.92 kJ

5. 21.0578 kJ

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

d: the voltage of the secondary coil is less than that of the primary coil.

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