00:01
We are told that we have substance x, and substance x has the following properties.
00:13
It's specific heat or specific heats for the solid, liquid, and gas phases are, let's see here.
00:37
C of s is 3 .0 joules per g degrees c.
00:44
The c of the liquid is 2 .5 j over g degrees c.
00:54
And the c of the gas is 1 .0 j over g degrees c.
01:02
The melting points and boiling points are melting point.
01:11
It looks like that is negative 15 degrees c.
01:16
The boiling point looks like 75 degrees c.
01:22
The heat of fusion and the heats of vaporization are heat of fusion looks like 5 .0 kilojoules per mole.
01:42
Heat of vaporization is 20 kilojoules per mole.
01:50
What else do i have here? got that? got that? got that.
01:55
Okay, given 250 point grams of x with a molar mass of 75 .0 grams per mole, given 250 grams, find heat energy that must be removed to convert a gas at 11717.
02:57
Degrees c to solid at minus 401 .9 degrees c.
03:08
Okay, easy peasy.
03:11
This takes a little time but it's not a hard thing to do.
03:16
This will not be to scale.
03:19
We were told that our melting point was minus 15 degrees c and our boiling point is 75 degrees c.
03:36
We are starting and starting at 117 degrees c and we're ending at minus 41 .9 degrees c.
04:04
Okay.
04:05
This will be a five -step process.
04:12
And i'm going to call these steps one, two, three, four, and five.
04:26
Steps one, three, and five will be using q equals mc delta t and for steps two and four we'll be using q equals n delta h.
04:44
Let's begin.
04:46
I'll label each step.
04:49
Step one, we are cooling the gas from 117 degrees c to the boiling point of 75 degrees c, which gives me a delta t of 117 degrees c, minus 75 degrees c equals.
05:14
Let me put this in my calculator.
05:17
117 minus 75 is 42 degrees c.
05:27
So remember we're using q equals n -c.
05:29
We'll get q equals my mass.
05:36
My problem went out of the door here, which is 250 grams times my specific heat of my gas, which is 1 .0 j over g degrees c times 42 degrees c.
05:59
I'm probably going to turn this to kilojoules because i prefer kilojoules if i've got a big number.
06:11
250 times 42 divided by 1 ,000.
06:16
This will be 10 .50 kilo jules.
06:23
I'll round later.
06:24
My second step, i'm going to be condensing gas to liquid.
06:43
And to do this problem, we're going to use q equals n -deta -h, which will be 250 grams, times, i believe it was 75 grams per mole, times my heat of vaporization.
07:06
Vaporization, 20 kilojoules per mole, will equal 250, divided by 75 times.
07:18
Times 20 is 200 kilojoules.
07:27
This can be reported to there, this can be reported to there.
07:32
Step 3, we will cool from 75 degrees c to negative 15 degrees c...