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?