# Amount of Propane Required to Melt 1L Aluminum

 P: 13 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
 Emeritus Sci Advisor PF Gold P: 16,450 The first thing you could do to get more "real world" results is to use significant figures. 24817687.0704? Good heavens!
 Mentor P: 12,023 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 insignificant figures would be useful.
 P: 56 Amount of Propane Required to Melt 1L Aluminum 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.

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