will it take off?

[daTeacha]

Egon -- I have picked a number of the ball and I know if it is light or heavy. Would you like to tell me what is on the left side and right side of the balance so I can tell you what happens? Presuming your method works, you will know what ball I picked and the nature of its discrepancy in a very few posts.

Patrick -- I fear I was a bit harsh yesterday in my comments. Sorry. I just think we need to more exactly define the parameters of a problem when we address it rather than leaving the door open for factors not mentioned originally.

 

[Egon]

daTeacha:

Couldn't answer that question but as for finding one out of twelve balls that is a different weight than the others by using a scale three times has at least three different approaches that will result in finding the off weight ball. There may be more but these have eluded me!

 

[BillyP]

daTeacha:

Couldn't answer that question but as for finding one out of twelve balls that is a different weight than the others by using a scale three times has at least three different approaches that will result in finding the off weight ball. There may be more but these have eluded me!

Egon, you gave the hint of solving the problem several posts ago, 4

 

[Egon]

{4-4-4} {5-5-2} {6-6} will all work.

This would be perfect for a flow diagram for FORTRAN programing.

 

[SPYDERLK]

Quote --Taking the coffee cup as an example, the use of the cream as a heat sink of sorts to store the heat at a lower temperature makes sense, but you failed to account for the increase in radiant surface area, which also makes sense. I pointed that out as a question for which I was seeking an answer. If you're going to interpret any question about your answers as some kind of a challenge to the accuracy of your comments, it makes it pretty hard for us ignorant types to learn anything.--

So, are we competing? Bad us. --- daTeacha, you have a good education and raise good questions. Some have an intuitive feel that instantly bypasses some of these questions. This may cause a certain aura of impatience and each may view the other as splitting hairs. A communication problem. -- It is very difficult to explain what one sees as intuitively obvious to another who hasnt to same feel for the problem definition. Adding the cream first is the way to go. I am an intuitive physicist and it is intuitively obvious to me. However, following your thot, there is a good possibility that with the introduction of a bit of intuitive tedium, a "cup" could be designed and filled to just the right level with coffee, such that adding the cream early would be the wrong choice. It would be a strange cup. It wouldnt suit lips. Youd have to drink it with a straw I think.
Larry

 

[daTeacha]

Puzzling over the billiard balls -- Are Egon's solutions all going to result in not only the identification of the odd ball, but also tell whether it is lighter or heavier than the rest? I think Egon thought he was told if the ball was light or if it was heavy. He was told it was different, but not in which way.

I ask because 6 vs 6 simply tells that one side goes down and the other goes up, which is foregone given that one of the balls is different, and useful only if you know ahead of time whether it is light or heavy.

Also, given that the problem involves the use of a basic 2 pan balance and not some exotic 3 pan design, I'm thinking the 4 - 4 - 4 solution is also based on knowing in advance if the ball is light or heavy. The same probably holds true for the 5 - 5 - 2solution.

The thing that makes this problem interesting is that you only know the ball is different. If you accept the challenge of the problem, you are to not only identify the odd ball, but also to tell if it is light or heavy.

If all three of your solutions will work when you don't know the nature of the difference, I would very much like to see them here. The only one I know does not start with 6 vs 6.

 

[patrick_g]

Da Teacha, I just checked the back of my lisc and confirmed that I was liscensed to do all of the things you suggest I do and MORE! You probably are too. I regret not being able to chat with you face to face as I am more that just sure we would have fun and part friends. Unfortunately there are certain difficulties in this sort of communications (especially with my meager skills) that interferes. Let me assure you I am not at all distressed and enjoy the exchange.

If we were always in agreement there would be little fun. I sometimes alter the circumstances dymanically and adaptively to increase the fun. After convincing someone of my position I have been known to take up their original position and make a case for it trying to refute my previous argument. We aren't being graded. This isn't even pass/fail, this is RECESS.

Now as to the coffee, I assert that your argument is flawed and that apparently you don't understand the exponential decay of the coffee temp toward the room temp. The heat loss of the coffee per unit time is way faster when the differential is the greatest and rapidly drops over time (an exponential decay, not linear.) That is why adding the cream immediately is the superior strategy. Lets forget the spray foam (got a chuckle on that, great idea... deux ex machina) and the burning napkins. I thought about a large Fresnel lense but that requires dining al fresco or a window seat as concentrating fluorescent light output is too puny.

If neither of us has a Gilbert chemistry set or Erector set or any special "Inspector Gadget" surgical implants then of the original choices the "stir in the cream immediatley strategy" is the winner by a good margin.

I gather from the latest posts that the billiard ball thingy is a rehash of the counterfit coin problem where one coin is different but we don't know if it is heavier or lighter and we are using just 12 of the balls. This is as old as the Fox, Goose, and Grain problem (or older.)

Larry, I can't think of a cup design that would favor delaying the addition of the cream. Can you give an example? Even a spherical vacuum bottle filled with a fine gauge needle would favor the early addition since it still has finite heat loss which is in the form of exponential decay. The heat loss by evaporation from a "regular" cup follows the form of exponential decay but would be elliminated by the above "special" spherical vaxcuum cup. The answer would remain the same although of less importance since the coffeee in a vacuum cup (spherical mini-dewar) would stay hot so well the loss during a quick lunch would probably not matter.

Now, about the hot water freezing faster than cold. That is an assertion that needs defense. It is like saying why is it that black cows fly faster that red cows. It is not established that either black or red cows can fly.

Lets take an example where the wind speed is essentially zero and the RH is 50% and we have two ice cube trays suspended on well insulated holders in identical domestic freezers. One ice cube tray is filled with hot water and the other with cold, from the tap. The hot is 125 F and the cold is 50 F. (typical household values in winter.)

Insert trays into the freezers and start the clock. The remote reading thermometer on the cold sample starts to fall. The temp of the hot samoplke starts to fall. At some point the hot water will have cooled to 50 F.

Why is it not reasonable to expect the time to freeze the hot water after it has fallen to 50F to be didfferent from the water that started at 50F?

I assert that with no evidence to counter the assumptioin that the time to freeze from 50 F on down to be thte same. Since it took time to get from 125 F to 50 F the time to freeze hot water is greater.

Pat

 

[SPYDERLK]

Pat, daTeacha - - Ok heres a "cup" with dubious potential. It is normal up to near the top, then has an extreme flair outward on the lip that close to doubles its diameter with little change in depth. The coffee fill would be to the top of the normal classic cup shape. The addition of cream would result in a solution that now extended well out onto the flaired lip to the extent that the surface exposed to air was increased 2 to 4 times normal. Evaporation would be favored - and heat of vaporization is high. Now intuitive tedium... The solution on the lip is somewhat isolated from the well of the cup and w/o some optimization favoring a natural convective flow, it will not cool the well well. In order to provide convective flow the lip would incorporate radial channels sloping downward back to rejoin the well. These would serve to collect the denser cooled fluid and provide a flowpath back to the well. This would in turn favor warm surface flow back out onto the lip.

As for the ice. Pls re read post 647 and add the following contributory factors. Most would fill an ice tray with equal volumes of hot vs cold water. The hot cube already starts with less water. Then there is evaporation. Then there is a higher surface area to volume ratio. The key point is there is less ice from the hot water. See what I mean?
Larry

 

[daTeacha]

Well, let's see here. I graduated from high school in '63, so I heard the billiard ball problem in '62. That makes it fairly old, but it's still an interesting problem. As you said, Pat, about like the Fox, Goose, and Grain. Or the town on the island with seven bridges.

If we want to pursue a cup design that would bring about faster radiation than the traditional cup shape as the cream is added, consider something like an old fashioned "Yard Glass" as used in British Pubs to serve up a yard of ale. These containers, if you are not familiar with them, resemble nothing quite so much as an Florence flask with a very long neck. If you minimize the size of globe at the bottom and the diameter of the neck, while maximizing the flare at the top, is it conceivable the increase in surface area resulting from increasing volume would offset the heat sink characteristics of the cream? This would in effect be the opposite of your sherical DeWar flask with the tiny opening.

Oh yeah, if you've never tried drinking ale or whatever out of a yard glass, you owe it to yourself to do it at least once, preferably in the company of good friends in a room that can put up with a spill or two.

 

[daTeacha]

On the water -- I see what you mean, but the density of water doesn't increase by a lot while it's still a liquid. The temperature of maximum density is 4C. At 100C, the density is 95.8% of that, so you're dealing with only a 5% or so decrease in the mass of water to be frozen.

The specific heat of water is 4.18 kj / kg C degree. The heat of fusion is 333 kj / kg , more or less. The heat of vaporization is about 2200 kj / kg.

I have no idea how much water is in a typical ice cube tray, but if I had to guess, I'd go with about 600 ml or .6 kg.

To cool .6kg of water from 100C to 50C would mean the freezer needs to remove about 10 000 kj of heat by my calculations. How much heat would be lost due to vaporization during the cooling process would depend on the humidity of the air in the freezer.

At this point, your exponential rate of heat loss based on temperature differences takes over, so you guys can take it from here.