00:01
I'm outside so forgive me for if there's strange noises here.
00:04
I'm babysitting a couple of dogs.
00:07
We're supposed to consider 1 .15 grams of rubbing alcohol that evaporates from a 65 gram aluminum block.
00:16
The aluminum block is initially at 25 degrees c.
00:19
What's the final temperature of the block after the evaporation of the alcohol? so step one, let's first determine q for the rubbing alcohol.
00:39
And that will equal the mass 1 .15 grams times my molar mass.
00:56
So that is c3h8o and its molar mass is 60 .09.
01:06
I'll go 0 .095 for this one grams of c3h8o.
01:16
And its heat of vaporization is 45 .4 kilojoules per mole.
01:23
And this number is the heat of vaporization.
01:32
And this will equal 0 .86986888688.
01:54
That'll be kilojoules.
01:58
That's our first step.
02:01
So the heat for the q for the will equal negative 0 .8688.
02:15
So let's do that.
02:19
My t final will equal, let's do a delta t, will equal q over mc and the c for aluminum is 0 .903 j over g degrees c...