If the heat of the reaction is -1.35kJ, what is the heat of the water?

Question

If 40.2g sample of iron was heated to 99.3 oC and then placed in a simple calorimeter containing 120g of water at 21.8oC. If the heat of the reaction is -1.35kJ, what is the heat of the water?

Answer 1

You need the following:

Specific heat of water is 4.184J/g*C
Formula: -q= SH *grams*change in T
Change kJ to joules

You ae solving for change in T; once you get that, add to the initial T of water. You don't need the info about iron since you are given -q

Hope that helps!

Answer 2

If you ignore the calorimeter, the heat of the water is 1.35kJ (the heat of the water + the heat of the iron = 0). However, if you take the calorimeter into account, then the solution becomes a bit more difficult.

qFe=mFe•cFe•Î”T
qFe=mFe•cFe•(Tf-TiFe)
qFe/(mFe•cFe)+TiFe=Tf
-1350J/(40.2g•0.449J/g•C°)+99.3°C=24.6°C

Since the final temperature of the Fe is the same as the final temperature of the H2O because the system is in thermal equalibrium:

qH2O=mH2O•cH2O•Î”T
qH2O=120g•4.184J/g•C°•(24.6°C-21.8°C)
qH2O=1400J=1.4kJ

The answer almost the same, but with correct sig figs, the answer is 1.4kJ.