It will probably depend on the size of the pieces... but at what point should I stop assuming a linear relationship?
I was prompted to ask in a previous question.
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2Actually, you should be cooking real bacon instead :-P – David Z Aug 18 '11 at 06:43
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i wish i had some canadian bacon – Timtam Aug 18 '11 at 13:12
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1I welcome all forms. – user420667 Aug 20 '11 at 03:33
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My experimental evidence comes from baked potatoes.
1 potato takes 5 minutes to be at my liking
2 potatoes come out the same at 10 minutes
8 potatoes come out the same at about 35 minutes, i.e. less time per potato.If I let them go to 40 minutes (5x8) they become dry and wizened.
The total mass plays a role then.
anna v
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On the basis of simple heat balance, we should expect the microwave to have higher power consumption in any case that reduces the cook time per object, right? – Alan Rominger Aug 18 '11 at 14:49
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@Zass: Yes, and it requires a magnetron that can support the power losses. – dmckee --- ex-moderator kitten Aug 18 '11 at 15:27
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this fried my mind. I would think that more mass would mean more power required, meaning longer cooking times... unless somehow the residual heat was still cooking them. – user420667 Aug 24 '11 at 07:47
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@user420667 I think it has to do with black body radiation and the fact that the microwaves penetrated the bulk , in contrast to radiant heat ovens. It means the centres are heated. There is a smaller surface to radiate away the heat if concentrated in a large mass rather than individual with air between. – anna v Aug 24 '11 at 09:24
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To some approximation the whole cavity can be thought of as full of microwave radiation at some fixed power density. (Of course, we all know this approximation is weak 'cause some food has hot and cold spots, right?)
In that case, adding a second separate piece of food should have no effect, and you should use the same time for two pieces as for one. See the part of anna's answer about two potatoes.
However, this relies on the magnetron being able to keep the cavity flooded at a constant energy density even while the food is absorbing some of the power. It may be that the power will sag, in which case it will take a bit longer; that's the "many potatoes" part of anna's answer.
For a long time my rule of thumb was for $N$ pieces multiple the time by $1.5^{N-1}$, but I have begun to wonder if the base should be lower with my current machine.
Additional consideration: if your "extra" food is stacked or placed close together parts of it may shield other parts which almost certainly calls for more cooking time (perhaps at lower power).
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+1 for stacking. But it suggests taking into account about the depth that the microwaves penetrate (I've heard 1/8 inch for water.) and the rate of heatflow within the substrate to be cooked. – user420667 Aug 24 '11 at 08:00
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Normally the heat losses increase with increase of the surface so you may have to cook even longer if the free surface is doubled. 8 potatoes put close to each other may have smaller free surface and get cooked faster than 5x8.
Ah, if you stuff your MW oven with too much of food that needs high temperature for cooking, then your temperature may become so low that cooking will take much more time.
Vladimir Kalitvianski
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Power input is constant and does not depend on the MW oven content. – Vladimir Kalitvianski Aug 18 '11 at 16:13
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Let's assume that heat losses are relatively insignificant comparing to heating power.
1:Microwave oven's power is likely to stay constant, but it's efficiency may increase with more bacon slices.
2:Roughly speaking, it will take time $t_1$ to heat up the food to 100 celsius , typically followed by time $t_2$ for the food to cook to whatever condition you find tasty (of course in practice it will also cook some during the heating up, and the $t_2$ may be zero). Doubling the power per bacon will only cut down the $t_1$ two times, but it won't cut down $t_2$ (and at the end of heating the bacon will be less cooked).
Both of those things suggest that microwave oven will need less than twice the time for 2 bacons, even if heating time is twice as long.
However, if the bacon is considered cooked by the time it reaches 100 celsius in both cases the relationship will be closer to linear.
Dmytry
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I'm guessing it's more efficient b/c not all the power is absorbed with one piece of bacon, is that you're thinking? – user420667 Aug 24 '11 at 07:54
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Bacon in the microwave oven is a real-world problem that does not reduce easily to a physical model. It is not clear what cooking bacon in the microwave actually means. Does it have to reach a certain internal temperature? Does it need be exposed to some above threshold temperature for a certain period of time to kill bacteria? Or maybe it needs to rapidly absorb a certain amount of energy per gram to break down the fibers and soften up?
It would be easier if you asked about (and/or experimented with) boiling up water in the microwave. Water boils when it reaches 100 degrees Celsius, and that is very clear. In this way, effects of cooling, feedback of load on efficiency, etc., can be isolated and investigated in detail. You can experiment with boiling up twice as much water in a glass vs boiling up two glasses of water (test the impact of cooling). You can try to boil up 1 cup vs 1 gallon (correct for the effect of cooling and learn if the extra load affects the power output). To avoid superheated water, you need to stop every minute and stir (see Superheated water on Mythbusters). And you must use fresh tap water every time so it contains the same amount of air.
drlemon
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Oh brother, me too! That's why I wrote two paragraphs instead of putting two cups of water in the microwave. – drlemon Aug 25 '11 at 20:40