Physics Help

Heat is added to 20kg of water that is at an initial temperature of 20?????????C. The water is now at its boiling point. The heat required to vaporize all of the water is:

a. 20 kJ
b. 2311 kJ
c. 6700 kJ
d. 46220 kJ (think that might be a typo, it's probably supposed to be 4620 kJ)

Can anyone help?

I already know how to use Q = mc#916;T to find the heat required to bring the water to its boiling point. I don't know what the heat of vaporization for water is or what to do with it once I know it. Anyone?

You just need to find the heat of vaporization of water and multiply that by 20kg. Then add that to the mc(delta)T from 20-100 degrees, unless they only want to know what it takes to turn 100 degree water into 100 degree steam.

google? A physics or chemistry textbook should have it too.

hmm, actually I found it and supposedly it's 2260, but when you multiply that by 20kg it gives you 45200kJ...?

eh whatever that's close enough to what my friend says the answer is (D)

I bet you have to convert to 20kg into moles depending on the units...I'd say just skip the problem if it's taking this long though[lol]

ha okay. thanks for your help man.

too heat 1 gram of water 1 degree you need 1calorie(4.128j).
You got 2000grams of water so you need 8.256.000j (8.256kj) to heat it 1dergee.
You need to heat it to 100 dergee to let it vaporized.
Meaning 80*8.256=660.480kj needed to go to a 100 dergee.
I could be wrong about 100 dergee as vaporizing point but heat
2kg 1dergee you need 8.256kj.
And I don't know about the . Meaning thousend or a indication of
Number below 1.

Greats
A M?????????H that just got awake.

Oops doublepost
(stupid iPhone)