"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"
(28g * 396 J/g) + (28g * 0.481 J/g°C * (T_f - 658.5°C(?)) = -[520g * 0.481 J/g°C * (T_f - 1000.0°C(?)]
Kinda confused on this, thanks.
first, see what would happen if all the solid aluminum melted, could it absorb the heat of the liquid Al?
heat to melt Al=396*28=11Kj
heat to lose by liquid: 520*.481=250J, so, the liquid Al cannot melt all the solid, so the final temp is at the melting point .
Are you sure bobpursley? Because it say "...if 28.0 grams of solid aluminum at this temperature were placed into 520 grams of liquid aluminum..." -- Do we not imply "at this temperature" is in fact the melting point of aluminum?
You not only imply that it is at the melting point, the sentence states that. So you have solid Al floating in liquid Al at 658.5 C
Sorry, wrong choice of words. It does state that. However, is the answer really just 658.5°C? I thought the 658.5°C is the initial temperature of the solid Al; therefore, finding the final temperature in a similar fashion that I wrote on my original post.
To solve this problem, we can use the principle of heat transfer. The heat gained by the solid aluminum when it melts plus the heat gained by the solid aluminum as it heats up to the final temperature should be equal to the heat lost by the liquid aluminum as it cools down to the final temperature.
Let's break down the equation:
(28g * 396 J/g) represents the heat gained by the solid aluminum as it melts. This is calculated by multiplying the mass of the solid aluminum (28g) by the heat of fusion for aluminum (396 J/g).
(28g * 0.481 J/g°C * (T_f - 658.5°C)) represents the heat gained by the solid aluminum as it heats up to the final temperature T_f. This is calculated by multiplying the mass of the solid aluminum (28g) by the specific heat of liquid aluminum (0.481 J/g°C) and the temperature change from its initial temperature (658.5°C) to the final temperature (T_f).
-[520g * 0.481 J/g°C * (T_f - 1000.0°C)] represents the heat lost by the liquid aluminum as it cools down to the final temperature T_f. This is calculated by multiplying the mass of the liquid aluminum (520g) by the specific heat of liquid aluminum (0.481 J/g°C) and the temperature change from its initial temperature (1000.0°C) to the final temperature (T_f).
Now, let's solve the equation for T_f:
(28g * 396 J/g) + (28g * 0.481 J/g°C * (T_f - 658.5°C)) = -[520g * 0.481 J/g°C * (T_f - 1000.0°C)]
First, we can simplify the equation by distributing the negative sign:
(28g * 396 J/g) + (28g * 0.481 J/g°C * (T_f - 658.5°C)) = -(520g * 0.481 J/g°C * T_f - 520g * 0.481 J/g°C * 1000.0°C)
Next, we can simplify further by multiplying the terms:
(28 * 396) + (28 * 0.481 * T_f - 28 * 0.481 * 658.5) = -(520 * 0.481 * T_f - 520 * 0.481 * 1000.0)
Combine like terms:
11088 + 13.488 * T_f - 13.488 * 658.5 = -250.4 * T_f + 250.4 * 1000.0
Now, let's isolate T_f:
Combine the terms with T_f:
13.488 * T_f + 250.4 * T_f = 250.4 * 1000.0 - 11088 + 13.488 * 658.5
Combine the constants:
263.888 * T_f = 250.4 * 1000.0 - 11088 + 13.488 * 658.5
Simplify the equation:
263.888 * T_f = 249760 - 73024 + 8892.648
Combine the terms:
263.888 * T_f = 190628.648
Lastly, solve for T_f:
T_f = 190628.648 / 263.888
T_f ≈ 723.18°C
Therefore, the final temperature of the aluminum would be approximately 723.18°C.