Latent heat and specific heat in insulated container

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Homework Statement

A 1.0kg of ice at 0◦C, 3.0kg of water at 0◦C, and 5.0kg of iron at temperature T are placed in a sealed and insulated container. cFe =400J/ kg◦C ,cWater=4200J/ kg◦C , cIce= 2000 J/ kg◦C

latent heat for ice is 3.3×105 J .
The equilibrium temperature is 20◦C.

Homework Equations

Q=mcΔT
Q=ml

The Attempt at a Solution

0=Qice→water+Qice+Qwater+Qiron
=ml(ice)+mcΔT(ice)+mcΔT(water)+mcΔT(iron)
=(1.0)(330000)+(1.0)(2000)(20-0)+(3.0)(4200)(20-0)+(5.0)(400)(20-Ti)
Ti=331°C

BUT the answer says it should be 353°

 

[Orodruin]

Once you have melted the ice, it is water.