Calculating energy needed to melt aluminum cans?

  • Thread starter Thread starter MarianaMoon
  • Start date Start date
  • Tags Tags
    Aluminum Energy
AI Thread Summary
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.
MarianaMoon
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top