00:01
All right, so we want to take liquid water at a temperature, or sorry, ice, at an initial temperature of zero degrees celsius, and convert it to steam at a temperature of 100 degrees celsius.
00:13
And so we want to know how much heat is this going to require for part a.
00:18
So we're going to first need to melt the ice, and so that requires an amount of energy of m times the latent heat of fusion for ice, and then plus we're going to need to, heat that water up to steam through 100 degrees celsius temperature increase.
00:35
And then we're going to need to vaporize that steam.
00:38
So we can see, like, there's a common mass term here.
00:41
So this will just be lf plus c delta t plus lv.
00:46
Where this c is the heat capacity of water, like liquid water.
00:52
All right.
00:57
And so that's basically all the questions asked for.
01:02
And then b, we're taking our mass to be, let's see, 0 .22 kilograms, our latent heat of fusion.
01:13
Let's see, where is this? and another thing to note, sorry, if you want to, if you're taking it from a temperature beneath these two, then all you do is you'll add the mass of this, times the specific heat of ice, times the temperature change of ice, and then we'll do the same thing for, steam.
01:40
So we'll write cs times the temperature change for steam.
01:44
So the only thing that really changes here is that you have, you know, heat capacity of water times the temperature change of water plus the heat capacity of ice times the temperature change of ice plus the heat capacity of steam times the temperature change for steam plus the latent heat of fusion or vaporization.
02:06
So anyway, so the latent heat of fusion.
02:12
Is going to be 795 ,000 calories per kilogram i believe or sorry just no just 79 .5.
02:25
The latent heat of vaporization is in kilo calories per kilogram.
02:30
This is 539 per kilogram looks like.
02:38
And then of course our heat capacity of water is 1 ,000 calories per kilogram per degrees celsius...
