00:01
So this is problem 54, chapter 15, and we're finding the increase of entropy in some liquid nitrogen.
00:08
And so heat transfer by the nitrogen when nitrogen starts boiling is qb equals mlv.
00:16
Qb is the heat transfer by the nitrogen.
00:18
M is the mass of the nitrogen, and lb is the latent heat of vaporization.
00:22
And so we can go ahead and find qb very easily.
00:24
So qb is going to be equal to 1 kilogram multiplied by 2 .0 .0 .2.
00:30
0 .01 times 10 to the 5th joules per kelvin.
00:34
And this is going to be equal to 2 .01 times 10 to the 5th joules.
00:40
And so now we can go ahead and find the delta s boil by substituting a couple of values into here.
00:47
So delta s boil is going to be equal to qb, which is 2 .01 times 10 to the 5th joules, all over negative 195 .8 degrees celsius.
01:01
And we actually need to add 273 to the bottom one to get it in kelvin to actually find it.
01:08
And this is going to be equal to 2 .60 times 10 to the third joules per kelvin.
01:16
And so now we need to actually find the delta t.
01:20
And so since it warmed up by 20 degrees, we need to find our delta t.
01:25
So delta t or 2 20 degrees celsius, excuse me...