How many joules of energy are required to vaporize 512.0 g of water?

The heat required to evaporate 512 g of water initially at 0°C is 1,371,340.8 J

Explanation:

The given parameters are;

The mass of water given in the question = 512.0 g

The specific heat capacity of water = 4.184 J/(g·°C)

The heat of vaporization of water = 2,260 J/g

The heat, Q, required to raise the temperature of 512.0 g from 0°C to 100°C is given as follows;

Q = 512 × 100 × 4.184 = 214,220.8 J

The heat required to vaporize 512.0 g of water at 100°C = Mass × Heat of Vaporization

The heat required to vaporize 512.0 g of water at 100°C = 512 × 2260 = 1,157,120 Joules = 1,157.12 kJ

The heat required to evaporate 512 g of water initially at 0°C is therefore;

214,220.8 J + 1,157,120 J = 1,371,340.8 J.

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

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