## anonymous 4 years ago Water (2690g ) is heated until it just begins to boil. If the water absorbs 5.31×10^5J of heat in the process, what was the initial temperature of the water?

1. anonymous

alright lets try solving for delta t really quick $\frac{5.31x10^5}{(2690g)(4.184)}=\Delta T$ let me know what you get.

2. anonymous

Did you figure out this one?

3. anonymous

No i cant im super confused:/

4. anonymous

you can't divide 5.31x10^5 by 4.184 and then again by 2690?

5. anonymous

that will give you the amount that the water changed, but then you have to subtract that from the final temp of the water to get where you started from

6. Xishem

What equation are you starting with zbay? I think it's helpful to provide that.

7. anonymous

I was using q=C *M Delta T but the euation thing isn't loading up