Physics: Final Temperature after Equilibrium Problem?

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A 93 g piece of copper is heated in a furnace to a temperature T. The copper is then inserted into a 150 g copper calorimeter containing 198 g of water. The initial temperature of the water and calorimeter is 16°C, and the final temperature after equilibrium is established is 38°C. When the calorimeter and its contents are weighed, 1.2 g of water are found to have evaporated. What was the temperature T?

My previous answer, 231°C, was incorrect.
Please help! Looking for final numerical answer and a short explanation. Thanks!
 
First let's solve for T leaving out the evaporated water, which for this step ends up at 38 C like the rest of the mixture:
Basic equation for final (f) state of an N-component mixture expresses the equality of final and initial heats:
mfcfTf = Σ(i=1,N)(miciTi). Since mfcf = Σ(i=1,N)(mici),
Σ(i=1,N)(mici)Tf = Σ(i=1,N)(miciTi) ==>
If we select component 1 as the unknown, this leads to
Σ(i=2,N)(mici)Tf+m1c1Tf = Σ(i=2,N)(miciTi)+m1c1T1 ==>
m1c1(T1-Tf) = Σ(i=2,N)(mici(Tf-Ti)) ==>
T1 = Tf+Σ(i=2,N)(mici(Tf-Ti))/(m1c1) = 581.429048972088 C

Now we compute the additional heat needed to bring 1.2 g water from 38 C to 100 C and evaporate it:
ΔQ = m(cΔT+Hevap) = 0.0012*(4186*62+2.27E6) = 3035.4384 J
(where Hevap = heat of vaporization, J/kg)
Next compute the equivalent temperature change in the copper piece:
ΔT(cu) = ΔQ/(m1c1) = 3035.4384/(0.093*386) = 84.5573123850911 C
Total initial temp. = 581.429048972088 + 84.5573123850911 = 665.986361357179 C.
 
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