00:01
This is sort of a big problem, so i'm going to find myself a very bright color and write down some given information here.
00:08
Our starting point for this problem is negative 12 degrees c, and our ending point is 149 degrees c.
00:23
So, we will have a one, two, three, four, five step process.
00:34
This is a five step process.
00:40
Steps one, three, and five will all use q equals mc delta t, and steps two and four will use q equals m times delta h.
00:54
And the factors that it talks about, the heat transfer constants, this is what i'm using.
01:02
You will need to verify these, because i have no idea what you're using.
01:07
So, you'll have to verify that these match, if not just substitute them in.
01:15
So, i'm going to do five separate calculations, then add them all together at one atm.
01:24
How much energy? okay, let's begin.
01:27
Step one will be to heat the ice from minus 12 degrees c to zero degrees c.
01:41
And as you can see, my delta t will be equal to zero degrees c minus minus 12 degrees c equals 12 degrees c.
01:54
So, my q will be my mass, which was 91 .0 grams times my ice was 2 .087.
02:08
This is my c, 2 .087 j over g degrees c times 12 .0 degrees c.
02:19
And i think i'm going to try to decide what units i want.
02:25
I'll just stay right here.
02:26
91 times 2 .087 times 12 is 2279 .00 degree or joules, which i'm going to make a note of is 2 .279 kilojoules.
02:46
Then i'll just be able to do this however i want in a little bit.
02:53
Step two.
02:58
For step two, we're going to use q or this will be the melty ice and we will use q equals 91 .0 degree or grams times 333 .6 j over grams...