IIT JEE , AIEEE Forum - Physics , Thermal Physics

mghn

water in a container having 2 mm thick walls having thermal conductivity 0.5.container is kept in a melting ice bath at 0C.total surface area in contact with water is 0.05 m^2.a wheel is clamped inside the water and is coupled toa block of mass M.as the block goes down,the wheel rotates.it is found that after sumtym a steady state is reached in which the block goes down witha constant speed of 10cm/s and the temperature of the water remains constant at 1 C .Find the mass M of the block.assume heat flows out of the water only through the walls in contact

mgn

hey,frndz!!! plz abs my Q

svj29

sm1 ansr gud qsn.... i give up...!!!

svj29

atleast 1 can tell the basic idea...!!!!!

svj29

sm1 do it na....5 salutes assured....every1 will reply d answered posts to show off... nw prove urself here...tis boring to type bt atleast one can tell the basic concept used here...

MUDIT

I REALLY DON'T KNOW THE FORMULAS IN DETAIL BUT HAVE A LOOK AT THIS
THE PARTICULAR SYSTEM HAS REACHED ITS STEADY STATE AND THEREFORE HEAT EXCHANGE PER SECOND IS CONSTANT
I FOUND A SIMILAR PROBLEM LIKE THIS IN H.C VERMA UNSOLVED EXAMPLES
ACCORDING TO THE GIVEN DATA I THINK BY USING THE STANDARD FORMULAS OF THE BOOK YOU CAN FIND THE HEAT LOST BY THE WATER AND THE CONTAINER WHEN THEY ARE AT 1 C WHEREAS THE SURROUNDING IS AT 0 C.
NOW HAVE A LOOK AT THIS DUE TO THE BLOCK GOING DOWN THE WHEEL ROTATES THAT TRANSFERS THE LOSS OF POTENTIAL ENERGY OF THE BLOCK INTO KINETIC ENERGY OF THE WHEEL AND FINALLY ALL SORTS OF FORMS OF ENERGY IN WATER. BUT IF YOU THINK FOR A MOMENT ALL THE ENERGY IS FINALLY DESSIPATED AS HEAT TO THE SURROUNDING. SO IN SHORT I CAN SAY THAT THE LOSS IN POTENTIAL ENERGY OF THE MASS IS CONVERTED INTO HEAT ENERGY
NOW IN STEADY STATE THE BLOCK MOVES DOWN WITH A CONSTANT VELOCITY OF 10cm/s AND HENCE LOSSES A HIGHT OF 10cm IN ONE SECOND AND HENCE THE LOSS IN POTENTIAL ENERGY PER SECOND = Mg(0.1) WHICH IS EQUAL TO THE HEAT ENERGY LOST BY THE WATER AND THE BEAKER IN 1 SEC
AND HENCE WE CAN FIND OUT M
CORRECT ME IF I AM WRONG

svj29

i got this qsn smwhr... the ansr is 12.5 kg...
c n try if ya get it...

MUDIT

YUP I GOT IT AS 12.5 Kg
ITS JUST THAT THE RATE OF TRANSFER OF HEAT TO THE SURROUNDING IS GIVEN BY THE FORMULA KAT/x WHERE K IS THE THERMAL CONDUCTIVITY OF THE WALLS
A THE AREA OF THE WALLS
T THE DIFFERENCE IN THE TEMPERATURE OF THE CONTAINER AND THE SURROUNDING AND x THE THICKNESS OF THE CONTAINER
THE ABOVE VALUE COMES TO 12.5J/s AND HENCE
12.5 = Mx10x0.1 = M = 12.5 [TAKING g AS 10]

mgn

thnx frndssssss..

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