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Hypnos_16
Dec6-11, 01:41 PM
1. The problem statement, all variables and given/known data

10.11g of CaO (molar mass = 56.08 g / mol) is dropped into a styrofoam coffee cup containing 157.0 g of H2O at 25.0ºC. If the following reaction occurs, calculate the final temperature of the water assuming that the cup acts as a perfect insulator.
(Assume that the solution has the same specific heat capacity as water.)

CaO(s) + H2O(l) → Ca(OH)2(aq) ∆H = –64.8 kJ

m = 10.11 g of CaO + 157.0 g of H2O
∆T = (Tf - Ti) = (Tf - 25.0ºC
Cs = 4.184
q = -64800 J

2. Relevant equations

q = mCs∆T

3. The attempt at a solution

q = mCs∆T
-64800 = (167.11g)(4.184)(Tf - 25ºC)
-64800 = 699Tf - 17480
-64800 + 17480 = 699Tf
-47320 = 699 Tf
Tf = 67.7ºC

However i know that isn't right,
The answer i was given was to be 42.8ºC
So i'm confused as to what i did wrong...

ehild
Dec7-11, 09:30 AM
64.8 kJ heat is released by 1 mol of CaO during the reaction. In the cup, only 10.11 g CaO reacts with water. So how much energy is released?


ehild.