# Water melting

1. Jun 9, 2009

### bigboss

1. The problem statement, all variables and given/known data
A 24.0 kg sample of ice is at 0.00°C. How much heat is needed to melt it? (For water Lf = 334 kJ/kg and Lv = 2257 kJ/kg.)

A)5.42*10^4 kJ B)0.00 kJ C)8.02*10^3 kJ D)2.19*10^6 kJ

2. Relevant equations

3. The attempt at a solution

2. Jun 9, 2009

bigboss, as part of PF forum rules, you need to show some sort of attempt in order for us to help you. (Or else Borek will make you wear that wig!! )

3. Jun 9, 2009

### bigboss

i beleive i have figured this out, q= M*Lf

so 24*334= 8.02*10^3

4. Jun 9, 2009

I am not sure about this. Do you know what you are doing here? Or are you just multiplying numbers until you get one of the offered answers?

Like Borek asked: what do Lf and Lv stand for? (I know; I just want you to tell me)

EDIT: Been awhile since I have dealt with heat; I think you are correct.

Last edited: Jun 9, 2009
5. Jun 9, 2009

### Staff: Mentor

Note: I asked, but then I deleted my post to not interfere.

And honestly, I don't know. I can guess, but I have never seen these symbols used. Call it cultural difference