[gvsu88]

A 29.85 g aluminum block is warmed to 65.91 °C and plugged into an insulated beaker containing 97.61 g water initially at 16.73 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

I'm pretty sure that what I need to do is, q = -q, as in q Al cooling = - water heating, but usually when i have these problems I'm at least given a specific heat, and my professor will usually give us that constant if we need to use it, so is there something that I'm missing here? I'm not really sure how to start this without more information given. Thanks

[lhunt]

I'm not trying to be snarky here. You could look up the SH for Al. Water is a consistent SH of 4.184 J/g⁰C.

I'm very new to Chemistry. I was trying to figure out how to do this problem using what I know. So change in T =Heat/(mass)(SH), if you mix up the formula to solve for T?

[gvsu88]

Like I said, I know how to do it using specific heat. I just think that my prof would have given that value in the problem if I needed it, because he usually does. So I wanted to know if there was another way to solve this.

[gvsu88]

never mind. I figured it out. Apparently I did have to look those constants up. I just didn't expect that because he's never made us do that before.