00:01
Okay, so we're looking to find the final temperature of the water when it absorbs four kilojoules of energy.
00:08
So we're going to use this equation, q equals mc delta t, and delta t is our final temperature minus our initial temperature.
00:21
So we're going to just first solve for what that delta t is.
00:24
We know that our q is four kilojoules, so i'm going to put that in joules so that multiply by 1 ,000.
00:31
So 4 ,000 joules.
00:34
The mass of our water is 15 .0 grams.
00:39
The specific heat of water is 4 .184 joules per gram degree celsius and then times our delta t.
00:49
So using algebra, we'll solve for delta t.
00:53
We'll take 4 ,000 divided by 15 and divided by 4 .4 .4 .5.
01:02
4 .184 and that's going to be our change in temperature is 63 .7 degrees celsius.
01:21
So that's not going to work because once it gets to 100 if we were to add 63 to this, once it gets to 100 it's going to change to a vapor.
01:30
So still using the same idea.
01:34
We're going to take our change in temperature and say, right, once it gets to 100, how much energy would that have used? so we'll take our 15.
01:45
So now we're going to look for q.
01:47
15 grams times the 4 .184 times here, it's just 20 degrees to get our change in temperature to get to 100.
02:00
And for this, then it is 15 times 4 .184 times 20.
02:07
And that would be, q is going to be 1255 joules.
02:14
Okay, so we will have used 1 ,255 of our 4 ,000 joules.
02:20
Okay, so, all right, the next thing we need to do is we need to figure out how much energy it's going to take to change our water from liquid at 100 to vapor at 100...