My professor gave me this problem and the answer but I haven\' been able to solv

Question

My professor gave me this problem and the answer but I haven' been able to solve this problem...

A 10.0 kg block of ice is at a temperature of -10.0oC. The block absorbs 4.11x106 J of heat. As a result the block of ice undergoes (a) an increase in temperature, (b) followed by a change in phase and (c) then another increase in temperature. what is the final temperature of the liquid water? Answer=13oC

...I'm assuming Lf water=33.5x104 and I'm using Q=mct +mL, but haven figure it out what I'm doing wrong... Thanks for your help!

Explanation / Answer

Step 1- Ice must drop in temperature to 00C, which takes up some of the energy.
Q=miciT=(10kg)(2090J/kg*0C)(100C)=2.09*105J (energy taken up to lower temp of ice to 00C)

Step 2-All the ice must melt, which takes up more energy, but no temperature change occurs (latent heat)

Q=miLf= (10kg)(3.33*105J)= 3.33*106J (energy required to melt ice into water)

Step 3- Excess energy is used to heat up water.

Excess energy= total-used= 4.11*106J- (2.09*105J+ 3.33*106J)= 3.539*106J

Q=mwcwT= (10kg water)(4186J/kg*0C)(T)= 13.6280C

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