Chemistry Help - A 30.78 g stainless steel ball bearing at 117.82°C is placed in a constant...?
A 30.78 g stainless steel ball bearing at 117.82°C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44°C. If the specific heat of the ball bearing is 0.474 J/g·°C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.
The heat lost by the stainless steel ball is equal to the heat gained by the water. Therefore, we can use the following relationship where ss represents stainless steel:
mss sss Tss = -mH2O sH2O TH2O
i know what equation I'm opposed to be using, but how do I solve without the change in temperatures?
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The change in temperature is (Tf-Ti), final - beginning temp for water and (Ti-Tf) , initial - final temp for the ball bearing. This gives us:
30.78-g x (117.8 oC - Tf) x 0.474-J/g*oC = 120-g x (Tf - 18.44 oC) x 4.184-J/g*C)
First multiply the specific heats by the masses then distribute through the parentheses.:
14.58972-J/oC x ((117.8 oC - Tf) = 502.08-J/oC x (Tf - 18.44 oC)
1718.669-J - 14.590Tf = 502.08Tf - 9258.355-J ---> combine like terms
516.67Tf = 10977.0-J
Tf = 21.2 oC
Note that the steel goes from 117.82 oC to 21.2 oC - a 96.62 oC decrease while the water goes from 18.44 oC to 21.2 - a 2.76 o C increase. If we divide steel's temp change by water temp change: 96.62 / 2.76 = 35, we see that the steel's temp change was 35 times greater than water's! Why so great? Well let's look at the combined ratio of specific heats and masses:
(4.184 / .474) x (120 / 30.78) = 34.4 . We see that the change in temp is explained by the difference in specific heats and masses!
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