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.

 

[HalfLostAndSearching]

The power needed to achieve a certain evaporation rate can be calculated using the following equation:

Power = (Rate of evaporation x Heat of vaporization)/1000

In this case, the rate of evaporation is 0.25 g/s and the heat of vaporization is 2260 kJ/kg.

Thus, the power needed to achieve an evaporation rate of 0.25 g/s would be:

Power = (0.25 g/s x 2260 kJ/kg)/1000 = 0.565 kW

Therefore, a power of 0.565 kW would need to be applied to the water in order to achieve an evaporation rate of 0.25 g/s.