00:01
Students welcome here in this question we have given that 4 .25 grams of h4 n .o3 ammonium nitrate is mixed in are dissolved in 60 grams of water okay so here when it is dissolved in this water the temperature drop is equal to 22 minus 60 point five so that is equal to 5 .1 degree celsius temperature is down here okay so initially it was 22 degrees celsius later it will drops to 16 .5 sorry 16 .9 then you got 5 .1 degrees celsius delta t we got here so we have to calculate calculate the q heat release during this reaction okay so this is the first one and the second one is q in kilo jowls per mole we have to calculate okay so see what happens so q heat released during this reaction is equal to m cp delta t so cp is for what a specific heat capacity uh capacity so what for water? it is 4 .184 joules per gram.
01:38
Okay.
01:39
So here it is m is 60 grams given and it is multiplied by 4 .184.
01:46
It is multiplied by delta t 5 .1.
01:51
As here, heat is releasing.
01:55
That's why we will take here it is a negative sign.
01:58
Okay.
01:59
So here we will give.
02:00
Get the temperature minus heat released during this reaction is 1280 jowls okay so we got the heat released during this reaction process so this is first one we got here and the second one is we have to convert this one into per mole okay so we got this one in a kilo only joules so so q is equal to, so per mole, so minus 1280 per number of moles.
02:39
So number of moles is equal to weight given divided by molecular weight.
02:46
So weight given is equal to minus 1280 divided by weight given 4 .25.
02:53
This is divided by that's why it will go upside.
02:56
Then we will get here it is 80 for...