Calculating Power Needed for Water Evaporation | Heat of Vaporization 2260 kJ/kg

[Bgerst103]

Homework Statement

Water's heat of vaporization is 2260 kJ/kg. How much power would you need to apply to water so that the evaporation rate was .25 g/s?

Homework Equations

Not sure

The Attempt at a Solution

What equation can be used to find the given rate?

 

[adjacent]

You can use $E=mL_v$ and $P= \frac{E}{t}$

Please show your progress

 

[Bgerst103]

You can use $E=mL_v$ and $P= \frac{E}{t}$

Please show your progress

E=(.00025)(2260) -> E=.565 -> .565/1=.565. Hows that look to you?

 

[adjacent]

E=(.00025)(2260) -> E=.565 -> .565/1=.565. Hows that look to you?

It looks correct.
EDIT:Wait, the latent heat of vaporisation should be in SI units.Change the kJ to joules.