00:01
In this problem, let us first calculate the value of delhd t2.
00:08
So it will be equal to delhd t1 plus del cp t2 minus t1.
00:20
So this value will be equal to delhd 25 degree centigrade plus del cp 68 degree centigrade minus 25 degree centigrade.
00:35
So it will be equal to 10 kcal per mole plus 1650 cal per mole per degree centigrade multiplication 43 degree centigrade.
00:58
So on solving it further you can write this value as 10 ,000 kcal per mole plus 70950 kcal per mole per degree centigrade multiplication degree centigrade which is equal to 80950 kcal per mole.
01:30
Now i will calculate the value of delhd tm.
01:36
Delhd tm is equal to delhd tm by tm which is equal to delhd 68 degree centigrade by 68 degree centigrade which is equal to 80950 by 68 plus 273.
02:03
So you will get this value as 237 .4 cal per mole per kelvin.
02:14
Now i will calculate the value of delhd t2 which is equal to delhd t1 plus del cp ln t2 by t1 which is equal to 273 .4 plus plus 1650 ln 37 plus 273 by 68 plus 273.
02:54
So on further simplification you will get this value as 80 .65 cal per mole per kelvin.
03:10
Now i will calculate the value of delhd t2...