Calculating thermal energy released when aluminum cools an solidifies

[thermite]

Homework Statement

While forming a 1.5kg aluminum statue, a metal smith heats the aluminum to 2700 degrees C, pours it into a mould, and then cools it to a room temperature of 23.0 degrees C. Calculate the thermal energy released by the aluminum during the process.

Homework Equations

Q = mc(delta)T
Q = mLf

Lf aluminum = 6.6 * 10^5 J/kg
heat capacity of aluminum : 9.2 * 10^2

The Attempt at a Solution

So I already know that I have to find the thermal energy released when the aluminum cools from 2700 degrees to 23 degrees, and I have to add it to the thermal energy released during the change of state.

Q = mcT
= 1.5(9.2*10^2)(-2677)
= - 3 694 260 J

Q = mLf
= 1.5(6.6 * 10^5)
= 990 000 J

Q = -3 694 260 + 990 000
= -2 704 260 J

I know that the thermal energy released is 2 704 260 J, but I don't understand why you are supposed to add the thermal energy released in the change of state (990 000 J) to the thermal energy released due to the change in temperature.

My physics teacher told me that when a substance is cooling from a liquid to a solid, the value of thermal energy should be negative (so basically you subtract), but in this case (the textbook) does not agree.

Why is this?

 

[Delphi51]

You have to do the temperature change in two stages:
* heat lost by liquid cooling to melting point
* heat lost by solid cooling to 23 degrees
The heat capacities for the liquid and solid will not be equal.

 

[gneill]

Wow! That metal smith must be doing aluminum vapor deposition! 2700 C is well above the boiling point for aluminum at 2467 C. So the question is, do we take this to be a mistake on the part of the problem author, or are there actually two phase transitions to deal with?

Offhand, who remembers the heat of vaporization of aluminum? Something around 11,000 kJ/kg maybe?