- #1

Mitchtwitchita

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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?