---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
---------------------------------------------------------
Alright; here it goes:
You can pour from second container to first until the first is full,
then from first to second until the second is full, and repeat this
cycle two more times to get 9 liters of water in the second container
at 35,648 degrees. But, there is a better solution:
Pour from 9-liter container to 8-liter container until the 8-liter is
full. Now there is 8 liters of 70-degree water in the first container;
and 7 liters of 10-degree water in the second. Here comes the
interesting part: Keep pouring water from second to first very slowly
until there is no water to pour. When you are finished, and there is
water all over the floor, you have the 8-liter container full of water
at exactly "T2+e^(-V2/V1).(T1-T2)" degrees. Namely, 35,012 degrees in
our puzzle.
Cihan Altay
---------------------------------------------------------
Well with these rules you don't have much choice but to keep
alternately filling each container from the other. After the first
pour the 8L is at 70 degrees. After the second pour the 9L is at 70/3.
Then 175/3, 280/9, 1855/36, 1925/54. After the 6th pour, the 9 is at
1925/54 degrees = 35 35/54 degrees. Nothing is ever closer than this,
and more pours just bring you closer to the average temperature of 42.
Joseph DeVincentis
---------------------------------------------------------
Answer
When one container becomes empty no new temperatures can be generated
any more. So it remains to oscillate between (6,9) and (8,7) litres.
After two such cycles the temperatures are (58.333, 31.111) and after
three cycles (51.5278, 35.6481).Thereafter one will not be able to
get below 35.6481, so this is the best you can do.
Christian Blatter
---------------------------------------------------------
1)Pour the water from the 9lt container into the 8 litre container till the 8
litre container is full.(ie 2 litres at 10 degrees have mixed with 6 litres at
90 degrees).So the temperature in the 8 litre jar is (90*6+2*10)/8=70 degrees .
Now there is 8 litres at 70 degrees and 7 litres at 10 degrees.
2)Now pour the 7 litres into the 8 litres drop by drop(between every drop ensure
complete mix).With a little bit of calculus,we can show that the final
temperature would be 10-(10-70)*exp(-7/8)= 35.01172
Balakrishnan V
---------------------------------------------------------
This is in response to your puzzle that was posted on mathpuzzle.com: You have
two containers with 8-liter and 9-liter capacities. Some water is in each, at
given temperatures and volumes. Get some water at a temperature as close to 35
degrees as possible. No guesswork; you must be able to know what the resulting
temperature exactly is.
Well, since there are only 2 jugs and no external sources of water, you
basically just have to pour back and forth until you get as close as possible.
let's call the 8 liter jug "jug 1" and the 9 liter jug "jug 2." here are the
resulting temperatures after the first 6 pours back and forth, trading two
liters each time:
pour 0: (before pouring) 1: 90, 2: 10
pour 1 1: 70, 2: 10
pour 2 1: 70, 2: 70/3
pour 3 1: 350/6, 2: 70/3
pour 4 1: 350/6, 2: 30
pour 5 1: 205/4, 2: 30
pour 6 1: 205/4, 2: 625/18
now, the water in jug 2 has a temperature of 34.72222... which is as close to 35
as you can get. two more pours gives you over 37 degrees in jug 2.
lemme know if i did something wrong. and sorry for the non-reduced fractions.
thanks for the puzzle!
Simon Birenbaum
---------------------------------------------------------
DENIS BORRIS
"Water Pouring Puzzle:
You have two containers with 8-liter and 9-liter capacities.
Some water is in each, at given temperatures and volumes.
Get some water at a temperature as close to 35 degrees as possible.
No guesswork; you must be able to know what the resulting temperature exactly is."
Hmmm...nothing in puzzle's wording prevents getting a thermometer and a 3rd
container, pouring a bit of "10" in it, then a chaser(!) of "90" while watching the
thermometer raise to exactly "35" and bingo!
However, I assume we must go by ye olde pouring rules: no other container,
no waste and no outside assistance: using the 2 containers only...right?
Let opening temperatures be: x = 90 and y = 10; A = 9 liter container, other = B.
operation: pour from A to B filling B, then from B to A filling A:
x = (6x + 2y) / 8, then y = (7y + 2x) / 9
Recursively yours for 3 operations, ending with filled A at 1925/54 degrees ;~35.65.
Leaves the 6 liters in B at 1855/36 degrees.
Start: [9 * 10 + 6 * 90] / 15 = 42 degrees; and now:
[9 * (1925/54) + 6 * (1855/36)] / 15 = 42 degrees; guess that's proof !
Here's a cute case:
A = 6 liter (full) at 1 degree, B = 5 liter (contents = 4 liters) at 46 degrees;
operation 1: A = 7 degrees, B = 37 degrees
operation 2: A = 11 degrees, B = 31 degrees
operation 3: A = 13 2/3 degrees, B = 27 degrees
Makes for a similar puzzle, a bit like:
"Water Pouring Puzzle:
You have two containers with A-liter and B-liter capacities.
Some water is in each: A is full, B is missing 1 liter.
After p pours back and forth, you end up with water at 27 degrees in a container.
What's p, A abd B?
No guesswork; you must be able to know that the resulting temperature is 27."
---------------------------------------------------------
Doug Gwyn:
The only useful action is to pour 2 liters from one bottle to the other,
alternately. The temperature in each bottle quickly approaches 42 degrees C
without ever quite reaching it, which is the solution to the problem.
The key missing physical factoid is how to compute the temperature of the
mixture, which is the volume-weighted average of the combining component
temperatures. In principle the absolute temperature (degrees Kelvin) should
be used for this, but it turns out not to matter in this application.
---------------------------------------------------------
Hi Ed,
Trust Cihan to come up with a new twist on an old puzzle!
After six pourings, in each case pouring 2 litres from one container to the
other, never spilling a drop, and stirring well between pourings, I think you'll
have 9 litres at about 35.65 degrees (1925/54 to be exact). Always assuming the
vessels have a negligible thermal contribution to make.
Best wishes
Chris Lusby Taylor
---------------------------------------------------------
Dear Ed,
I "solved" the Water pouring puzzle on 23 July 2006
using Excel (enclosed file).
Well, maybe I didn't "solve" it because I took for
granted that you can only pour exactly 2 liters at
a time from one flask to the other.
Am I missing someting? Should there be something smarter?
My result is
35,64814815
in the 9-liter flask after 6 single moves.
Anyway, thanks for a great site!
Anders Uddgren
Uppsala
Sweden
---------------------------------------------------------
Here is an amusing sequel: same puzzle, but the goal is to reach as close as
possible to 37 degrees Celsius.
Solution: Drink either one of the bottles (although the 90 C water will burn
you, not recommended!). After a while, pee into the empty bottle. This water
will be at body temperature (37 degrees Celsius).
Original:
If you pour 2 liters from the big container to the little one, mix
well, and then pour 2 liters back, you have the same volumes, but the temps are
now 70 and 70/3 ~ 23.333. If you then repeat this two more times they are at
1855/36 ~ 51.52777 and 1925/54 ~ 35.648, assuming I haven't made a mistake.
Of course this assumes the heat in the containers is negligible, perfect mixing,
etc. In the real world, hitting 35 degrees to within 0.648 degrees would seem
pretty good. But maybe there is a better way? I don't see how without having a
3rd container around. Hmm...
-George Bell
---------------------------------------------------------
I have the solution to the Water Pouring Puzzle. Its a lot simpler than
it seems.
The 8 liter glass with 6 liters of 90 degree water, I will call glass A.
The 9 liter glass with 9 liters of 10 degree water, I will call glass B.
Step 1.
Pour water from glass B to glass A until glass A is topped off.
Using this math:
(6liters(90 degrees) + 2liters(10degrees))/8 liters = 70 degrees
In other words, topping off A will result in 8 liters of 70 degree water
in A, and 7 liters of 10 degree water in B.
Step 2.
Pour water from glass A to glass B until glass B is topped off.
(2liters(70degrees) + 7liters(10degrees))/9 liters = 23.333 degrees
A has 6 liters of 70 degree water, and B has 9 liters of 23.333 degree
water.
Step 3.
Pour water from glass B to glass A until glass A is topped off.
(6liters(70degrees)+ 2liters(23.333degrees))/8 liters = 58.333 degree water.
A has 8 liters of 58.333 degree water, and B has 7 liters of 23.333
degree water.
Step 4.
Pour water from glass A to glass B until glass B is topped off.
(2liters(58.333degrees)+7liters(28.333degrees))/9 liters = 34.999 degree
water.
A has 6 liters of 58.333 degree water, and B has 9 liters of
approximately 35 degree water.
~William Riley
---------------------------------------------------------
Hi,
First, I will assume an ideal scenario where we lose no heat to the surroundings.
There are 3 possible predictable actions for each step: empty 8L container,
empty 9L container, which both lead to dead ends, leaving only pouring from one
to the other.
This is a linear question and thus quickly solved: The following lines are the
tempratures at each step, the left being the 8L container, and the right 9L. In
() is how much is in each container. Note that the total "Liter degrees Celsius"
is constant at 630. Equilibrium is 42 degrees, meaning that the solution will be
in the big container, and we will know it as soon as we pass 35 degrees.
(6,9) 90 10
(8,7) 70 10
(6,9) 70 23.33
(8,7) 58.33 23.33
(6,9) 58.33 31.11
(8,7) 51.53 31.11
(6,9) 51.53 35.65
next line will be bigger.
The exact values:
23.33 = 70/3
58.33 = 175/3
31.11 = 280/9
51.53 = 1855/36
35.65 = 1925/54 = 35 + 35/54
The amounts you can get are:
9L = everything
7L = pour 2 into other container (losing other 2)
8L = empty other container, pour from big into small
1L = like 8L
--Yaacov Yoseph Weiss
---------------------------------------------------------
Ed,
1. Pour 2 liters from the 9-liter flask to the 8-liter flask. This leaves
8 liters of 70 deg C water in the 8-liter flask and 7 liters of 10 deg C
water in the 9-liter flask.
2. Continue pouring from the 9-liter flask to the 8-liter flask, allowing
the latter to overflow, until the 9-liter flask is empty. Volume and energy
balances on the 8-liter flask, once simplified and integrated, give the end
temperature of the 8-liter flask as 35.012 deg C. The exact value is (70
deg C - 10 deg C * (1 - exp(7/8)))/exp(7/8).
Thanks for sharing this interesting puzzle!
Jason Dieterle
---------------------------------------------------------
Hi Ed,
The solution to the water containers problem is T=1925/54, or approximately
35.648 degrees C, which is obtained in the larger container after pouring 2
liters back and forth between containers three times. This is a Markov process
with solution given by T_i = 42 - 32*(7/12)^i, where T_i is the temperature of
the water in the larger container after being poured back and forth i times.
David Terr
---------------------------------------------------------
Daniel Sigmon
8 cylinder 9 cylinder
T V T V
90 6 10 9
70 8 10 7
70 6 23.33- 9
58.33- 8 23.33- 7
58.33- 6 31.11- 9
51.5277- 8 31.11- 7
51.5277- 6 35.648148- 9
Pouring back and forth between the 2 cylinders the sixth pour yields the closest
temperature to 35 degrees C
(Assuming no further cooldown or warm up due to ambient temperatures and staying
in degrees celcius since unspecified beyond given temperature) roughly 35
and 11/17 degrees celcius
---------------------------------------------------------
I think I have a solution to the Water Pouring Puzzle. It seems impossible to
get any water to exactly 35 degrees, although this is hinted at anyway in the
request for an exact answer.
My answer is 3073/81 or 37.9382716.
The next closest result was 280/9 or 31.1111111... ad infinitum.
Feel free to E-mail me back and tell me whether you agree. It's also interesting
to note that something happened in Japan in 1931 regarding this exact number.
I'm not sure what it was as my computer couldn't decode the language, but try
googling for the decimal 37.938... There is only 1 result.
Thanks for the great site!
Ashley Dickson
---------------------------------------------------------
brook miscoski
I guess we assume the water has the same density and so on so that it's just a
math problem. After that, if we aren't allowed to do anything to the container
we just pour 2 litres into the 6 of 8, and then 2 litres into the 7 of 9, ad
nauseum until we arrive at a temperature of about 35.6478 (who knows what
decimals I've messed up in the problem) in the 9 litre container. Any further
and of course we just go towards 42. If we're allowed to be somewhat creative
and mark the containers, you can get 35.3125 degrees Celsius by just constantly
discarding the 2 markable (initially empty) litres from the 8 litre container.
But, if we can do that we can go a step further and mark both containers each
time a new level is inferred, and use a combination of mixing and discarding to
reach exactly 35 degrees Celsius:
(Mark 2 top litres in 8 litre)
1. Combine in the 8 litre for 70C (7 left @10C in 9 litre)
2. Discard top 2 litres from 8 litre, 6 left @70C
3. Combine in 8 litre for 55C (5 left @10C in 9 litre)
4. Combine in 9 litre for 30C (Mark 4 top litres in 8 litre, 4 left @55C)
5. Combine in 8 litre for 42.5C (5 left @ 30C in 9 litre)
6. Discard top 4 litres from 8 litre, 4 left @ 42.5C
7. Combine in 8 litre for 36.25C (1 left @30C in 9 litre)
8. Discard top 4 litres from 8 litre, 4 left @36.25C
9. Combine in either container for 35C.
Perhaps this can be done more quickly, but I'm done with it for now, especially
as I don't even know whether marking is legal or considered "guessing," etc. If
it had to be done simply by the series of interchanges with no discarding, I'll
be interested if there is some nice way of doing that series that doesn't take 3
or so minutes for calculations (would be much worse without a calculator, which
usually means there is a better way).
---------------------------------------------------------
I'm really curious to see the answer to this puzzle, because it seems
impossible to me. As I figure it, there are only two possible
actions we can take:
1) Pour as much water as possible from one container to the other.
2) Dump the contents of one container on the ground.
However, if we ever do 2), we'll be left with only one container of
water, and therefore we will have lost all ability to modify its
temperature. So all we can do is pour water back and forth from one
container to the other until equilibrium is reached at 42 degrees.
I guess I'm missing something?
-Greg Janée
---------------------------------------------------------