00:02
Hello students, so the first step is the cooling of steam from 145 degree celsius to 100 degree celsius and to calculate it we have q1 is equal to the heat released m is 18 .015 grams, c is 2 .01 joule per gram per degree celsius and temperature is minus 45 degree celsius and then what we have to do is when we calculate the q1 it comes out to be minus 1631 .220.
00:54
Then in the second step we have to cooling of water from 100 degree celsius to 0 degree celsius so q2 is mc delta t again and m is 18 .015 grams, c is 4 .18 joule per gram per degree celsius and temperature is minus 100 degree celsius when we solve it q2 comes out to be minus 7531 .47 joule.
01:37
Then in the third part is cooling water at 0 degree celsius to ice at 0 degree celsius and q3 comes out to be ml so m is 18 .015 and l is latent heat that is 33 .5 joule per gram.
02:08
So gram gets cut by gram and we get it in joule which is minus 6010 .02 joule.
02:18
Then in the fourth step we have to do cooling of ice from 0 degree celsius to minus 50 degree celsius and for this q4 is again mc delta t, m is 18 .015 grams, c is 2 .09 into delta t is minus 50 so it comes out to minus 1883 .97 joule.
02:57
Then q total when we add it comes out to be q1, q2, q3 and q4...