00:01
In this problem we're going to talk about heat.
00:02
And there are basically two types of heat we're going to be interested in in our problem.
00:08
First, the heat necessary to change the temperature of an object.
00:15
I'm going to call it just q and it's equal to the mass of the object times the specific heat, c of the object, times the change in temperature.
00:27
So if the temperature is raised, the heat necessary, to do so is positive if the temperature is lowered the heat that the object loses is negative.
00:41
Also, there is the latent heat i'm going to call a q -o -l that is necessary to change the face of matter of an object.
00:55
So, for example, if we have solid ice and we want to transform it to liquid water, that process happens.
01:03
At zero kelvin, so there is no heat related to the difference in temperature, but there is a heat, there is a heat, i'm sorry, related to the fact that the water is changing its face.
01:20
So ql, the latent heat, is equal to the latent heat constant l times the mass of the object, and now we can go on to our problem.
01:31
Basically we have a mass of 12 grams of water that is initially at a temperature of minus 10 degrees celsius and we want the final state of the water to be vapor at a temperature of 100 degrees celsius so notice some things first i'm going to write here the specific heat of the water which is 4 .182 joules per kelvin gram.
02:11
And also the latent heat of fusion, that is when the water changes from being ice to being liquid water, it's equal to 334 joules per gram.
02:28
And the latent heat of vaporization, the name tells it all, it's when the water converts from being liquid to being gas.
02:36
This is equal to 2 ,260 joules per gram.
02:45
Okay, so notice that we have the following process.
02:51
We have water at minus 10 degrees celsius.
02:54
Then we have the heat...