How much energy do we need to vapourize 1 liter of water?

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The energy required to vaporize 1 liter of water starting from 0 degrees Celsius to steam at 100 degrees Celsius is calculated using the formula Q = mcΔT + mlv. Here, m is the mass of the water (1 kg), c is the specific heat capacity (4.2 kJ/kg·K), ΔT is the temperature change (100 K), and lv is the latent heat of vaporization (2260 kJ/kg). The total energy needed amounts to 2680 kJ. For steam at temperatures above 100 degrees Celsius, additional energy must be calculated using the specific heat of steam and a steam table for accurate results.

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How much energy do we need to vapourize 1 liter of water?

How much energy do we need to vapourize 1 liter of water?

I know that to heat up 1liter of water by 1 degree, you need 1kcal(or 4,2kj/kgK).
So let us say that th water is 0 degree at start temperature, and that the boiling temperature is 100 degree.
So is the answear then: 4,2*100=420KJ ??
 
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That gets the water to 100C but not vaporized. For that you need to add in waters heat of vaporization. (you can do a web search to find that number.)
 


If you are starting as liquid water at 0 degrees C, then the total energy needed would be the energy to increase the temperature from 0 deg C to 100 deg C + the energy required to change the phase of the water from liquid to gas (latent heat)

Q= mcΔT + mlv where lv is the latent heat of vaporization of water.
 


Thanks...
Q= mcΔT + mlv is just for steam at 100 degrees ? So if I want the steam to increase in temperature I need to add more energy right ??
But how is the equation if want the steam to be 200 degrees?
 
Last edited:


rock.freak667 said:
If you are starting as liquid water at 0 degrees C, then the total energy needed would be the energy to increase the temperature from 0 deg C to 100 deg C + the energy required to change the phase of the water from liquid to gas (latent heat)

Q= mcΔT + mlv where lv is the latent heat of vaporization of water.

To get 1kg of water at 0 degrees to steam at 100degrees you need:
(1Kg*4,2KJ/KgK*100K)+(1Kg*2260KJ/Kg)=2680KJ, right?
 


That's only if the water is vaporized at 100C. Of course, it can also be vaporized at all other temperature, up to the critical temperature, with different amounts of heat required.
 


To find out how much heat is required to raise the temperature of 100C steam to 200C, you'll have to find out the specific heat of steam.

A more practical approach is to use a steam table.
 

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