Bomb Calorimetry: Solve for Water Temp.

[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?

 

[AbedeuS]

I think that going off the assumption that \Delta T = T_{ball} - T_{water} might be where the deviation is arising, remember, the ball is heating up a limited amount of water up until a point when T_{ball} = T_{water}, your case would be true if there was a massive amount of water relative to the ball, because the final temperature of the system would be pretty close to the waters original temperature anyway.

I would go about finding the final temperature as:
q_{steel} = mC(x-117.82)
q_{water} = mC(x-18.44)

Since the change in heat energy must be equal to zero as you stated you find:
q_{steel}= -q_{water}
which is what you stated

Therefore:
mC_{steel}(x-117.82) = -mC_{water}(x-18.44)

Rearranging:
mC_{steel}x - mC_{steel}\times 117.82 = mC_{water}\times 18.44 - mC_{water}x

mC_{steel}x+mC_{water}x = mC_{water}\times 18.44 + mC_{steel}\times 117.82

x(mC_{steel}+mC_{water}) = mC_{water}\times 18.44 + mC_{steel}\times 117.82

x = \frac{mC_{water}\times 18.44 + mC_{steel}\times 117.82}{mC_{steel}+mC_{water} }

Substituting:

x = \frac{120\times 4.184\times 18.44 + 30.14\times 0.474\times 117.82}{30.14\times 0.474+120\times 4.184} = 21.1896^{\circ}C

 

[Mitchtwitchita]

Thanks again AbedeuS!