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

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To calculate the power needed for water evaporation at a rate of 0.25 g/s, the heat of vaporization (2260 kJ/kg) is used. The equation E = mL_v helps determine the energy required, where E is energy, m is mass, and L_v is the latent heat. The initial calculation yielded E = 0.565, but it was noted that the heat of vaporization should be converted to joules for SI unit consistency. After conversion, the final power can be calculated using P = E/t. Accurate unit conversion is crucial for the correct power calculation.
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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?
 
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You can use ##E=mL_v## and ##P= \frac{E}{t}##

Please show your progress :smile:
 
adjacent said:
You can use ##E=mL_v## and ##P= \frac{E}{t}##

Please show your progress :smile:

E=(.00025)(2260) -> E=.565 -> .565/1=.565. Hows that look to you?
 
Bgerst103 said:
E=(.00025)(2260) -> E=.565 -> .565/1=.565. Hows that look to you?
It looks correct. :smile:
EDIT:Wait, the latent heat of vaporisation should be in SI units.Change the kJ to joules.
 
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