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You have 25 liters of water in bucket A and 25 in bucket B. The difference in temperature difference is 40K between them. Now if you take one liter from A to B and then from B to A (after mixing properly). What would be the temperature difference between A and B. If you do this activity 5 times what would be the final temperature difference. What would be temperature difference in 10 th time?

Physics
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AFTER FIRST TIME THE DIFFERENCE WILL BE 36.92307692 K
ACTUALLY I THINK I FOUND THE SERIES: t_0=40 t_1=40*(24/26)=40*(12/13) t_2=t_1*(12/13)=40*(12/13)^2 . . . t_n=40*(12/13)^n
So, after five times, t_5=26.807 K AND AFTER 10 times, t_10=17.965 K

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Other answers:

I think both @demitris and @sauravshakya are almost excatly there. However neither of you guys has clearly A) stated B) Reasoned in explaining you respective \[ \Large\color{Magenta}{\text{Recursion Equation }} \\ \\ \\ Concentration(n+1) = \\\ =\Large{F}{(Concnetrat._{\,\,hotb}(n)\,\, , Concentrat. _{\,\,coldb}(n) )} \] Of course we mean concentration of heat. now \[ Heat.Concentrat. \,\,\sim\, Temperature \]
We all understand that this is a kind of geometric decrease , yet you have endeavored to SOLVE it exactly.... Did you not ?
The night was silent in fron of him ... \[ \Huge\color{green} {\underbrace{\ddot{}}} \]
WASNT THE ANSWER GIVEN BY ME CORRECT
not sure if that was correct ... I didn't have answer and don't remember my answer either. I had got the recursion relation. pretty close \[ t_n = {25 \over 26}t_{n-1}\]
while demetris had got \[ t_n = {24 \over 25}t_{n-1}\] which is pretty close. probably weather you take from hot cold or cold to hot might make difference. Not sure though. If you are interested you can check.
I solved it assuming the water is transferred from cold to hot and back from hot to cold.
try assuming water is transferred from cold to hot and from hot to cold. also use 1 liter.
U MEAN HOT TO COLD NOW?
yep ... if results are symmetric then you must be correct.
YEP GOT THE SAME SERIES.
OK HERE IS MY METHOD. LET THE DIFFERENCE IN TEMPERATURE OF 25 liters of water in bucket A and 25 in bucket B be x KELVIN (ALSO LET A IS HOTTER THAN B).......ALSO LET TEMPERATURE OF BUCKET B BE y KELVIN THEN the temperature of bucket A is (y+x) kelvin NOW, FROM B TO A: WHEN 1 liter of water is transferred from B to A then, THE temperature of bucket B does not change but the temperature of bucket A changes. The temperature of bucket A will be (1*y+25*(y+x))/26 = y+25/26 x Now again 1 litre of water is transferred from bucket B to A.... THIS TIME the temperature of bucket A will remain constant but the temperature of bucket B will change: TEMPERATURE OF BUCKET B = {1*(y+25/26 x) + 24*y}/25 =y+(1/26)x NOW, DIFFERENCE IN TEMPERATURE=y+25/26 x -y-1/26x = 12/13 x
WHICH CLEARLY SHOWS THAT temperature difference is in geometric series with common ratio 12/13
GOT SAME RESULT WHEN ASSUMED WATER IS TRANSFERRED FROM A to B.
worked out again and ... looks like both of you are right!! Help each other with medals.

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