Amount of Propane Required to Melt 1L Aluminum

[Voltux]

I started off yesterday learning about specific heats, and measuring my ice cubes to figure out the correct number to cool my tea to a proper drinking temperature.

Now today I was curious how much propane would be required to melt 1L of Aluminum since I'm also interested in casting Aluminum. I was hoping that someone could check my work, and then tell me how I can get more "Real world" results, or if this would even be practical at all for my level of math/physics/chem understanding.

Also, if you have any concepts you think I should learn next I would appreciate any suggestions. I have some Physics, and Chemistry books that I am reading for furthering my education.

Aluminum
Latent Heat of Fusion: 398 J/g
Specific Heat: 0.91 J/g°C
Density: 2.70 g/cm^3
Melting Point: 660.32°C
Ambient Temperature: 20°C

Mass of 1 Liter Aluminum (Mass = Density * Volume) (1mL = 1cm^3)
2.70g/cm^3 * 1000mL = 2700g

Required Energy for +640.32°C (Mass * Specific Heat * Temperature Difference)
2700 * 0.91 * (660.32-20)
2700 * 0.91 * 640.32
2457 * 640.32 = 1,573,266.24

1,573,266.24 J Required to Heat 20°C Aluminum to 660.32°C

Required Energy to Change Solid -> Liquid (Mass * Latent Heat of Fusion)
2700 * 398 = 1,074,600 J

Total Energy to Heat, and Melt Aluminum (Raise Temp + State Change)
1,573,266.23 + 1,074,600 = 2,647,866.23 J

2,647,866.23 J
735.5184 Wh (Joules/3600)
2509.6929443 BTU (Joules/1055.05585262)

Propane
Molar Mass: 44.1g/mol
Energy Density: 2220kJ/mol
Liquid Density at 25°C: 0.493g/cm^3
Mass Per Liter: 493g/Liter

Moles Per Liter (Mass Per Liter/Molar Mass)
493/44.1 = 11.17913832mols

Energy Density per Liter (Mol * Energy Density)
11.17913832 * 2,220,000 = 24817687.0704J

Amount of Aluminum Melted per Liter of Propane (Energy per Liter Propane/Energy to Melt Aluminum)
24817687.0704/2,647,866.23 = 9.3727118044L Melted per Liter of Propane

 

[Vanadium 50]

The first thing you could do to get more "real world" results is to use significant figures. 24817687.0704? Good heavens!

 

[mfb]

Well, in real setups, you don't get 100% efficiency. You will heat a lot of air and other parts of your experiment as well. The efficiency really depends on the setup.

Working with units everywhere and with less [strike]in[/strike]significant figures would be useful.

 

[Catria]

9.372L of aluminum melted using just 1L of propane? Holy moly... Doing it for real will likely result in maybe 20-25% efficiency; you'd be lucky to be able to melt 3L with that much propane for real.

 

[pmacias]

Thank you for sharing your calculations and for your interest in learning more about physics and chemistry. Your calculations appear to be correct and it is great that you are using your knowledge of specific heats and energy to explore real-world applications.

To obtain more "real world" results, you may want to consider factors such as heat loss during the melting process, as well as the efficiency of the propane burner in transferring heat to the aluminum. Additionally, you could also investigate the effect of varying the ambient temperature and the starting temperature of the aluminum on the amount of propane required.

As for suggestions for further learning, it would be beneficial to continue exploring the properties of different materials and their behavior under different conditions. You could also look into the principles of thermodynamics, which govern the transfer of heat and energy, and how they apply to practical scenarios such as melting aluminum. Best of luck in your studies!