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.

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.