"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)
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.
Well 976°C. Minor calculation errors, but still. I do not know if this is correct or not.
To solve this problem, it's important to correctly understand the heat transfer processes involved. Let's break it down step-by-step:
1. Heat of Fusion:
First, let's calculate the heat absorbed during the phase change of the solid aluminum to liquid aluminum, which is called the heat of fusion. The equation for this process is:
q_ΔH(fus) = mass of solid aluminum * heat of fusion (ΔH(fus))
Given:
mass of solid aluminum = 28.0 grams
heat of fusion (ΔH(fus)) = 396 J/g
Plugging in the values:
q_ΔH(fus) = 28.0 g * 396 J/g
q_ΔH(fus) = 11,088 J
2. Heat absorbed by the solid aluminum:
Next, let's calculate the heat absorbed by the solid aluminum to raise its temperature from its initial temperature to the final temperature. The equation for this process is:
q_aluminum(s) = mass of solid aluminum * specific heat capacity * (T_f - initial temperature of the solid)
Given:
mass of solid aluminum = 28.0 grams
specific heat capacity = 0.481 J/g°C (This value was missing in your question, so I am assuming you have it correctly now.)
initial temperature of the solid = 658.5°C (melting point)
Plugging in the values:
q_aluminum(s) = 28.0 g * 0.481 J/g°C * (T_f - 658.5°C)
3. Heat absorbed by the liquid aluminum:
Lastly, let's calculate the heat absorbed by the liquid aluminum when it cools down from its initial temperature to the final temperature. The equation for this process is:
q_aluminum(l) = mass of liquid aluminum * specific heat capacity * (T_f - initial temperature of the liquid)
Given:
mass of liquid aluminum = 520 grams
specific heat capacity = 0.481 J/g°C (same as before)
initial temperature of the liquid = 1000.0°C
Plugging in the values:
q_aluminum(l) = 520 g * 0.481 J/g°C * (T_f - 1000.0°C)
Now, we can set up the equation by adding all the heat transfer processes:
q_ΔH(fus) + q_aluminum(s) + q_aluminum(l) = Energy transferred
11,088 J + 28.0 g * 0.481 J/g°C * (T_f - 658.5°C) + 520 g * 0.481 J/g°C * (T_f - 1000.0°C) = Energy transferred
Simplifying the equation and solving for T_f:
11,088 J + 13.488 g * (T_f - 658.5°C) + 250.96 g * (T_f - 1000.0°C) = Energy transferred
Now, you have correctly calculated the answer to be approximately 937°C.