How many grams of water at 50C must be added to 10g of ice at -10C to result in only liquid water at 0C?
The answer is 17g

Things to do:
1. Make ice to 0C
q=(10g)(2.11 J/gK)(10) = 211

2. Change ice to water.
q=(deltaH)(n)

3. Find the total q
Do this by adding the q from #1 and the q from #2

4. Change water from 50C to 0C
qtotal = (m)(4.18j/gK)(50)

So I got stuck on number 2. I don't remember how to find delta H or n.
Also if any of my steps are wrong PLEASE PLEASE correct me, I am reviewing for my final.

Thanks!

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