In reading over the previous posts I realized that I am switching between the metric and Imperial systems of measure at will. This is what I get for being a Canadian who has lived in the United States for sixteen years. When doing any kind of “scientific” calculation it is of course far easier to do in the metric system, where converting between litres and cubic centimeters is simply a matter of moving a decimal place. I have no intuition for how many fluid ounces are in a cubic foot; I always have to look it up. But when it comes to carpentry and oven temperatures, I’ve learned how in the Imperial system of inches and degrees Fahrenheit. I’ll probably continue to switch back and forth indiscriminately, so, sorry about that.
On the subject of Fahrenheit, a quick reminder. As I’ve described this project to people, they often ask if I am going to try to melt iron. No, I say, the temperatures are far too high; aluminum pours at 1400°F and iron melts at 2800°F. So far, three people have said “oh, so that’s twice as hot”, without stopping to think about their high school physics. Remember, you cannot divide one temperature by another.
Why not? First off, we don’t measure temperatures on an absolute scale. There are negative temperatures. If 1400°F is half as hot as 2800°F then clearly it is negative 40 times as hot as -35°F, and 14000 times as hot as 0.1°F! Neither of those make any sense.
This reason alone is sufficient to reject the idea that 2800°F is twice as hot as 1400°F. Now, we could convert to absolute scale. 1400°F is 1033 Kelvin, 2800°F is 1811 Kelvin, so the melting iron is about 80% hotter than the pouring aluminum, right?
But that’s not quite the right way to look at it either. We’re not starting with the metals at absolute zero to begin with. Room temperature is about 70°F, so the aluminum must have 1330 degrees of heat energy added to it, and the iron must have about 2730 degrees of heat energy, and that is just about twice as much, right?
But no, that’s not quite right either. The specific heat capacity — the amount of heat energy you have to add to a metal in order to raise its temperature by a given amount — is different for every metal, and iron’s specific heat capacity is about half that of aluminum; the same amount of energy increases the temperature of iron for two degrees for every one degree that it would increase the temperature of aluminum. So even though the temperature change of the iron is twice as much, it takes half as much energy, so it’s a wash, right?
Well, no, that’s not right either; somehow the furnace has to get up to the needed temperature and the furnace has to withstand that amount of heat. How efficiently the furnace transmits that heat into the melt is maybe an interesting theoretical question, but the fact is that the vast majority of the heat energy in the furnace is heating up stuff other than the melt.
And finally, we’re still not taking into account the latent heat of fusion! Normally when you put heat into an object, the amount of heat energy that goes in turns into an increase in temperature, in a linear fashion. That is, if putting in one unit of energy raises the temperature by one degrees, then putting in two units will raise it by two degrees. This ceases to be the case when the substance is melting (or freezing) or boiling (or condensing). When the object reaches the melting point it needs extra energy, called the latent heat of fusion (*), to overcome the stick-together-ness of the solid form; this energy breaks down the crystal structure of the solid, rather than increasing the temperature. And, like the specific heat capacity, the latent heat of fusion of a substance is a characteristic of the molecular structure of that substance. Aluminum has a much higher latent heat of fusion than iron: 398 kJ per kg, compared with 272. So, even though iron has to get a lot hotter to melt, it takes a lot less energy to get it from solid to liquid.
The long and the short of it is: don’t think of temperatures as things that you can multiply and divide, and even addition and subtraction is a bit dodgy when going over the melting point boundary.
(*) This has nothing whatsoever to do with nuclear fusion; the “fusion” in question is the fusion that a liquid undergoes when it freezes. The latent heat that you must remove from liquid water to “fuse” it into ice is the exact same amount of energy as the amount you must add to melt solid water, so the latent heat of fusion and the latent heat of melting are the same amount.