00:01
Hi, here this given problem is based upon the principle of caloremetry and there is an aluminum calorimeter whose mass is given as m .a .l that is 100 gram mass of the water in that calorie meter this is given as mw is 250.
00:49
Initial temperature of the calorimeter ti this is at 10 .0 degrees celsius.
00:58
Now the mass of the copper block that is m, c .u, is equal to 50 .0 gram and its temperature, this is initially at 80 .0.
01:31
Degrees celsius.
01:34
Then there is one more block, unknown block.
01:38
Mass of that unknown block, its material is unknown.
01:47
So its mass is supposed to be m -u -n for unknown and that is given as 70 .0 gram its temperature.
02:06
This is t -u -n for unknown again that is 100 degrees celsius and both of these blocks are dropped into the calorie meter so the final temperature of the mixture t f that is given as 20 .0 degrees celsius now a specific heat capacity first of all that of aluminum c c .a .l.
02:57
It is found to be equal to 0 .9 jules per gram per degree celsius, specific heat capacity of water that is known to us as cw is equal to 4 .18 jules per gram per degree celsius highest specific heat capacity.
03:20
And then for copper that is cw.
03:29
That is c.
03:29
Cu is equal to 0 .385 joules per gram per degree celsius.
03:39
So using the principle of caluritory heat lost by the blocks, two blocks, that will be equal to the heat gained by calorimeter along with the water in it...