Calculating energy needed to melt aluminum cans?

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    Aluminum Energy
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To calculate the energy needed to melt 1500 aluminum cans, each weighing 13.9 g, the specific heat, latent heat, and melting point of aluminum are essential. The relevant equations are Q=mcΔT for heating and Q=mL for melting. Initial calculations yielded incorrect results, indicating a need to verify the mass conversion from grams to kilograms and ensure proper unit consistency. The total energy required combines both the heating and melting phases, but the calculations must be carefully checked for accuracy. Correctly applying these formulas will yield the total energy in joules needed for the process.
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Homework Statement


You have collected exactly 1500 aluminum cans for recycling, each with a mass of 13.9 g. How much energy is needed to melt them if their initial temperature is 25.5◦C? Assume
the specific heat, the latent heat and the melting point of aluminum are 899 J/kg ·◦C, 3.97 ×10^5 J/kg and 660.4◦C, respectively.
Answer in units of J.

Homework Equations


I'm pretty sure that the equations used are Q=mcΔT, and after getting that, using the Q=mL formula..

The Attempt at a Solution


So, for the Q=mcΔT part, I put; Q= (13.9)(899 J/kg◦C)(634.9)+(13.9)(397000 J/kg) which is 1.34*10^7. I put it in my homework online and it said it was incorrect. I later did Q=(13.9)(899)(634.9) which was 7933773.89 and then put it into Q=ml.. 7933773.89=13.9(397000) and it calculated to be 7933773.89=5518300. Do I subtract, divide? I have a feeling I'm doing it wrong. I don't know what to do after this step, my teacher was explaining it at the end of class really fast and I didn't get to finish writing down the notes.
 
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How many cans are you trying to melt?
 
Exactly 1500 Aluminum cans.
 
And how many cans did you use in your calculation?
 
Well, 1500.
 
Check again. Also check your units.
 
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