Question

The heat of fusion for water is 334 J/g, and the heat of vaporization of water is 2260 J/g. How many joules are required to convert 10.0 g of ice at 0.00 °C to steam at 100. °C? The specific heat of water is 4.184 J/g °C.

Answer 1

To convert 10.0 g of ice at 0.00 °C to steam at 100. °C, the total energy required is 30224 joules, which includes the heat for melting the ice, heating the resulting water, and vaporizing the liquid into steam.

Step-by-step explanation:

Calculating Energy for Phase Change and Temperature Increase

The process of converting 10.0 g of ice at 0.00 °C to steam at 100. °C involves both phase changes and temperature changes, each of which requires a certain amount of energy. The heat of fusion for water is 334 J/g, the heat of vaporization of water is 2260 J/g, and the specific heat capacity of water is 4.184 J/g °C.

To calculate the total energy required, we need to consider the energy for melting the ice, heating the liquid water to boiling point, and vaporizing the water to steam:

1. Melting the ice to liquid water: energy = mass × heat of fusion = 10.0 g × 334 J/g = 3340 J
2. Heating the water from 0.00 °C to 100. °C: energy = mass × specific heat × temperature change = 10.0 g × 4.184 J/g°C × 100 °C = 4184 J
3. Vaporizing the water to steam: energy = mass × heat of vaporization = 10.0 g × 2260 J/g = 22600 J

Add these three amounts to find the total energy required:

Total energy = melting energy + heating energy + vaporizing energy = 3340 J + 4184 J + 22600 J = 30224 J