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How much energy is needed to vaporize 0.22 kg of water?

Answer 1

#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#

Let's assume that the water is ready to be vaporized, i.e. it's already at the boiling point, and that we just need to put in enough energy to vaporize it. We need to know the enthalpy of vaporization for water (also called the latent heat of vaporization). This can be found from standard tables as:

#Delta H_(vap,water) = 2257 kJ//kg#

Now we just need to use this like a conversion factor to find the energy we are looking for knowing the mass of the water we are working with:

#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#

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Answer 2

Cooper Adams

The energy needed to vaporize 0.22 kg of water can be calculated using the formula:

Energy = mass * latent heat of vaporization

The latent heat of vaporization of water is approximately 2260 kJ/kg.

So, Energy = 0.22 kg * 2260 kJ/kg = 497.2 kJ

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.