How much heat in kJ is needed to convert 866g of water at 10.0C to steam at 126C?

30fg for help

what do you mean 126C? once water hits 100C it boils o.O

at least at normal room temp.

If you mean also heat the steam once the water has all evaporated, then you can do the following:
calculate the amount of energy needed to heat the water from 10C to 100C, then the energy needed to convert the water to steam, then the energy needed to heat the steam from 100C to 126C. Add them all up.

http://en.wikipedia.org/wiki/Heat_capacity



This post was edited by khemist on Dec 1 2013 01:13pm

>.>

Wat.

You need Cp of water and Cp of steam. They may be the same but they might not be.

h = m*Cp*(T2-T1)

Then you need heat of vaporization of water (energy needed to turn water to steam). A quick google search says that this is 2260 J/g but if you are provided with a number use it.

So just h = 866g * Cp * (100-90) is heat to get water to 100
then heat of vaporization h = 2260 * m
then h = 866 * Cp * (126-100) to get the steam to 126

This post was edited by Dontrunaway on Dec 1 2013 01:16pm

well so far my answer is 2,359.71kJ is what I got earlier

If you use these values:

Specific heat water - 4.187 kJ/kgK
Specific heat water vapor - 1.996 kJ/kgK
Heat of Vaporization - 2260 kJ/kg

You would get 0.866 * 4.187 * 90 + 2260 * 0.866 + 1.996 * 0.866 * 26

2328.44 kJ most of which is just converting water to steam at 100 Celsius

if you show me all the steps like Q= (4.18)(866g)(100C)(1kJ/1000J) =

Ill give you the fg

my value for steam was 1.84 J/GC by the way

This post was edited by airsoft986 on Dec 1 2013 01:31pm

What about values for Cp water and heat of vaporization?

4.187 kJ/kgK * 90 K * 0.866 kg = 326.33 kJ
2260 kJ/kg * 0.866 kg = 1957.16 kJ
1.84 kJ/kgK * 26 K * 0.866 kg = 41.43 kJ
Total: 2324.92 kJ

I didn't show conversions from J/g to kJ/kg because it's 1:1

This post was edited by Dontrunaway on Dec 1 2013 01:36pm

Heat of vaporization wasnt given.
According to my notes, specific heat water - 4.18 and steam is 1.84

heat of vaporization has to be given since youre changing states from boiling water -> steam -> to 126C


specific heat would be used to calculate changes in temperature when the water is still in the same state, that is, steam


This post was edited by nlin on Dec 1 2013 02:51pm

this is right, clarifying here



mc(deltaT) + [ q = m (heat of vaporization)] + mc(deltaT)

the 2nd term is the Heat required to "convert" the water into steam
the 3rd term would cover the heat changing the steam from 100C to 126C