Dead heat

I just found out tonight that my apartment’s central heating is provided in part by a local crematorium. Surprisingly, this little nugget of weird news, first reported in Dagens Nyheter last month, has so far flown below the radar screens of the internet’s weird news industry.

Perhaps its because it all makes eminent sense if you look at the details. Basically, it’s an environmentally friendly gesture: The smoke produced by the crematorium’s ovens needs to be cooled, from over 1,100 degrees celsius to around 150 degrees, for filters to efficiently remove mercury. Using electric fans costs a lot of money, but piping water through the smoke cools it while heating the water. So for the past 5 years the crematorium has offered to let itself be connected to the city’s central heating grid, an offer Stockholm Energi, the local energy company, refused, fearing public queasiness.

Until now. Several local bishops have said they see no ethical problems with such a setup; on the contrary, both sides see environmental benefits: Stockholm Energi needs less fuel to heat Stockholm, and the crematorium spends less energy cooling its furnaces, saving 50-75% on its energy bill in the process.

But the crematorium insists it’s about saving energy, not money. Furthermore, the technical director of the local church administration assures us that the heat generated by the ovens does not come from bodies—bodies are not good fuel, he says, because they are mainly just bags of water. The heat is generated in part from the coffins, but mainly from the gas that is burned to cremate the bodies.

He also assures us that the steam from the cremated bodies does not end up flowing through my heater here; the processes are kept separate. In that case I’m all for it. In fact, wouldn’t it be even more efficient to have Stockholm Energi burn the bodies in their own powerplants, obviating the need for crematoria altogether? And also, I would feel a bit better if the local old people’s homes get a deep discount on their next heating bill.

(Matthew, what do you think, could you pitch this as an A-head and get a free trip to Stockholm out of it?)

[Mon, Dec 09 2002 – 16:40] Charles Kenny (www) (email) Now I understand why the global environment is in crisis. It’s all these people living so long. With the average age of death rising so fast, there aren’t enough people dying to run the furnaces, so we’re having to use more coal. Its the damn Swedes who are responsible for global warming.

Why, by the way, are Swedish bodies full of Mercury? Sounds awful unhealthy.

[Tue, Dec 10 2002 – 07:23] Stefan (email) Well, googling crematorium and mercury got me this shocker:

The amount of mercury in the mouth of a person with fillings was on average 2.5 grams, enough to contaminate 5 ten acre lakes to the extent there would be dangerous levels in fish.

[Tue, Dec 10 2002 – 13:48] Felix (www) (email) I just discovered that the boiling point of Mercury is a mere 357∞C, so I imagine it’s very easy for the boiling point of the amalgams used in fillings to be well below the 1100∞C to which the coffins are heated when burned. But if I were a Swede, I wouldn’t worry too much: I don’t think the inside of my mouth ever reaches 357∞C, not even when I’m eating pizza straight out of the oven.

[Tue, Dec 10 2002 – 14:05] Charles Kenny (www) (email) Why can’t we use something in teeth that doesn’t pollute 5 ten acre lakes (and just think how many five acre lakes)? Wouldn’t silly putty do instead?

[Tue, Dec 10 2002 – 21:25] Felix (www) (email) Interesting question, Charles. Stefan, there’s your question for this morning: assume, for the sake of argument, that all lakes are perfect hemispheres. How many 5-acre lakes would it take to hold the volume of water in five 10-acre lakes?

[Wed, Dec 11 2002 – 04:00] Stefan (email) A little bit more than 14 (~14.14) 5 acre lakes, or the square root of 8 amount of times a 10 acre lake, to be exact.

[Wed, Dec 11 2002 – 05:28] kartika (email) Stefan

My maths must be a bit skewed- but did you use the Pie, r squared formula? Ok… for the surface area- but how deep are the lakes to calcuate volume?

Confused Kartika

[Wed, Dec 11 2002 – 11:35] eurof (email) no, i’m afraid that wasn’t an interesting question.

[Wed, Dec 11 2002 – 13:05] Felix (www) (email) Sorry, Stefan, you confused me with that root-8 stuff. Can you explain just how you got to your 14.14 answer?

[Wed, Dec 11 2002 – 14:01] Stefan (email) Well, given area A, r = sqrt( A / pi ) [because A = pi * r^2]

Now plug this definition of r into the formula for volume V = 4/3 ( pi * r^3) and you get

V = 4/3 (pi * ( A / pi )^(3/2))

From this formula, which relates the area of the cross-section of a sphere to its volume, it’s clear that multiplying the area by X increases the volume by X^(3/2). if you double the area, then, volume increases by 2^(3/2) or the square root of 8. Hence you need the square root of 8 5-acre lakes to equal the volume of a 10 acre lake.

But I’m not sure if Felix is just baiting me. Nevertheless, I had fun.

[Wed, Dec 11 2002 – 14:10] Stefan (email) But Felix, I feel that lakes tend to be more like cones, rather than hemispheres. What if these lakes, albeit circular, were conelike and had a depth equal to their radius? How many 5 acre lakes would we need then to hold the same volume as a 10 acre lake?

[Wed, Dec 11 2002 – 16:36] Matthew (www) (email) what if there are ducks in the lake? did you factor the water displacement?

[Thu, Dec 12 2002 – 02:50] kartika (email) Where did the 4/3 in your volume formula come from Stefan? That seems most suspicious. You are having us all on aren’t you?

A mathematically challenged K

[Thu, Dec 12 2002 – 07:09] Stefan (email) That 4/3 ratio is the result of doing some calculus but in any case is irrelevant for the purposes of our calculations. For example, for the cone with both radius r and height r, the formula for volume is V = 1/3 (pi * r^3). In both cases, volume V is always directly proportional to A^(3/2) where A is the area of a particular cross section.

Felix and I were discussing this last night. Because the area of any cross section is always two-dimensional, and the volume always three-dimensional, the relationship between the two is always of order (3/2).

This means the square root of 8 answer applies equally to cones, as well as any shape of lake you care to dream up.

[Thu, Dec 12 2002 – 11:53] Charles Kenny (email) Frankly, I’m worried about the idea that depth and area are assumed to be so closely correlated. Take Loch Ness and the Aral Sea for example. I bet that the far more secure answer to Felix’ original question is ‘probably somewhat more than 10, but with the potential for wide variation.’

[Thu, Dec 12 2002 – 18:39] Stefan Geens (email) Charles, the answer is the same regardless of the shape of the lakes, just so long as the big and small lakes are the same shape. Admittedly, this last requirement is a bit utopian, shall we say. But then the quote leaves a lot of questions unanswered. What if the lakes have a lot of fish, or no fish? Salt water would be more dense than fresh water too… And does the author mean dangerous levels for the fish, or for humans? etc.. Most unsatisfactory quote.

[Fri, Dec 13 2002 – 13:04] Charles Kenny (email) Talking of more dense… If they’re the same shape this means there is a 100 percent correlation between depth and area –i.e., you can predict depth with 100 percent accuracy if you know the area. I call that ‘closely correlated,’ but then I’m only a proto-social-scientist, so what would I know.

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