Question

How much energy is needed to turn 2 kg of water, at 100°c? Specific latent heat of vaporization of water =2300 000 J\kg.

Answer 1

By definition, the amount of energy required to vaporize $m = 2kg$ of water at 100°C is:

$$Q = \lambda m$$

where $\lambda = 2.3\times 10^6J/kg$ is the specific latent heat of vaporization of water. Thus, obtain:

$$Q = 2.3\times 10^6J/kg\cdot 2kg = 4.6\times 10^6J$$

Answer. $4.6\times 10^6J$ or $4.6MJ$.