- #1
360modina
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I want to take a liquid metal, Aluminum at 700oC and pour it into a steel mold starting at 20oC. For the purpose of the question let’s say that the heat transfer is 100% with no losses to the surroundings.
Here are some numbers:
Al
Mass = 200 g = 0.2 Kg
Ti, Al = 700oC
Cp = 900 J/Kg-K
ΔHf = 10.67 kJ/mol
Melting point of Al = 660oC
Steel
Mass = 3Kg
Ti, st = 20oC
So the liquid aluminum will cool once it hits 660, but the steel mold should never change state due to its higher melting point.
At this point I want to find the final temperatures of the Aluminum and the Steel mold.
Here's what I have so far can someone please check my theory/number placement:
(900 J/Kg-K)(0.2Kg)(700oC-660oC)+(200g)(1 mol/26.96g) + (900 J/Kg-K)(0.2Kg)(Tf, Al - 20oC)
= (486 J/Kg-K)(3Kg)(Tf, St - 20oC)
So my guess is both final temperatures will be the same, I just wanted to know if this work looks correct or if I need to change/move some numbers.
Here are some numbers:
Al
Mass = 200 g = 0.2 Kg
Ti, Al = 700oC
Cp = 900 J/Kg-K
ΔHf = 10.67 kJ/mol
Melting point of Al = 660oC
Steel
Mass = 3Kg
Ti, st = 20oC
So the liquid aluminum will cool once it hits 660, but the steel mold should never change state due to its higher melting point.
At this point I want to find the final temperatures of the Aluminum and the Steel mold.
Here's what I have so far can someone please check my theory/number placement:
(900 J/Kg-K)(0.2Kg)(700oC-660oC)+(200g)(1 mol/26.96g) + (900 J/Kg-K)(0.2Kg)(Tf, Al - 20oC)
= (486 J/Kg-K)(3Kg)(Tf, St - 20oC)
So my guess is both final temperatures will be the same, I just wanted to know if this work looks correct or if I need to change/move some numbers.