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

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Homework Help Overview

The discussion revolves around calculating the energy required to vaporize 1 liter of water, starting from a temperature of 0 degrees Celsius and considering the boiling point at 100 degrees Celsius. Participants explore the concepts of specific heat and latent heat of vaporization in the context of thermodynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the energy needed to heat water to its boiling point and the additional energy required for vaporization. There are questions about the applicability of the energy equation for steam at different temperatures and the need for specific heat values for steam.

Discussion Status

The discussion is active, with participants providing insights into the calculations needed for both heating water and vaporizing it. Some participants have offered equations and numerical examples, while others are questioning the assumptions regarding temperature and phase changes.

Contextual Notes

There is mention of varying energy requirements depending on the temperature at which vaporization occurs, as well as the need for specific heat values for steam, indicating that additional information may be necessary for complete calculations.

charlie95
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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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