[MitchTwitchita]

Hey guys, I'm having a real tough time with this question:

A 30.14-g stainless steel ball bearing at 117.82 degrees C is place in a constant pressure calorimeter containing 120.0 mL of water at 18.44 degrees C.  If the specific heat of the ball bearing is 0.474 J/g x C, calculate the final temperature of the water.  Assume the calorimeter to have negligible heat capacity.

qsteel +qwater = 0
qsteel = -qwater

qsteel = ms(deltaT)
=(30.14 g)(0.474 J/g x C)(18.44 degrees C - 117.82 degrees C)
=-1420 J

Therefore qwater =1420 J
1420 J = (120 g)(4.184 J/g x C)(Tfinal - 18.44 degree C)
Tfinal = 1420/502 + 18.44
=21.27

However, the answer in the book is 21.19 degrees celsius.  I don't think that I'm doing this one properly.  Can anybody set me straight?

[enahs]

You are not.
The final temperature of the ball bearing is not that of the water. It is when the two come to equilibrium.

You have to setup your equation and solve for x, but in this case x is the final temperature in both expressions.