"Aluminum metal melts at 658.5°C. What would be the final temperature if 28.0 grams of solid aluminum at this temperature were placed into 520 grams of liquid aluminum at 1000.0°C? The heat of fusion for aluminum is 396 J/g, and the specific heat of liquid aluminum is 0.481 J/g°C"
After doing q_ΔH(fus) + q_aluminum(s) + q_aluminum(l) = E, I got about 937°C. Is this right? I saw the other question answered, but I don't think it's right to just say "The answer is at melting point". If it is, please provide a detail explanation because I do not see how it is at melting point where the liquid. I thought the melting point is the initial temperature for the solid aluminum; hence using that to find the temperature.
q_ΔH(fus) = (28g * 396J/g)
q_aluminum(s) = (28g * 0.481J/g°C * (T_f - 658.5°C)
q_aluminum(l) = (520g * 0.481J/g°C * (T_f - 1000°C)
3 answers
Well 976°C. Minor calculation errors, but still. I do not know if this is correct or not.
I think you are right but I don't think it is 976 C. Perhaps 940.5? I think the other post is in error. You have 520 x 0.481 x (658.5-1000) = 85,416 J available and that is enough to melt all of the solid Al at 658.5 since that takes only 11088 J. The excess heat then raises the temperature of the melted Al at 658.5 to some higher T which I think is about 940 C.
I got an exact answer of 940.485447 without rounding with sig figs till the end. Thank you for clarifying.