[achibaby1974]

ok. so the problem is this:

A 680. g piece of copper tubing is heated to 92.8°C and placed in an insulated vessel containing 43.0 g of water at 31.5°C. Assuming no loss of water and heat capacity for the vessel of 10.0 J/K, what is the final temperature of the system (c of copper = 0.387 J/g·K)?

The formula is for it (MASS copper)(SPECIFIC HEAT copper) (TF-TI) = (MASS water)(SPECIFIC HEAT water) (TF-TI) right? so we're solving for TF. My question is what about the heat capacity of the vessel being 10? What do I do with that? Does it matter? I did it the first time by adding 10 with the water side. But I guess I'm wrong.

I did:
(Mass)(Spht) (Tf-Ti) = (Mass)(Spht)(Tf-Ti)  + (Ht cap of vessel?)
(680)(.387)(TF-92. = (43)(4.184)(TF-31.5) + 10       (<------- this is from the vessel!!!)
263.16TF - 24421.25 = 179.912TF - 5667.2 + 10
263.16TF - 24421.25 = 179.912TF - 5677.2
83.248TF = 18744.05
TF = 225.16
3 sig fig so that's 226???
it's wrong. i'm pretty sure there's something wwong with the fact that i added the heat capacity of the vessel? where does it really go?
TF = 225 

[Borek]

Hint: heat capacity of the vessel is J/K...