00:01
The perimeter contains 2 .50 kilograms of water, which equals 2 .50 times 10 to the third grams of water.
00:23
We are told the initial temperature of our water is, this is all for water, is 37 .0 degrees c.
00:38
We are told that we have a 225 gram sample of ice.
00:44
We are told that our initial temperature of ice is negative 20 degrees celsius.
01:05
We added to the water and all the ice melts.
01:10
We are asked to find our final temperature after melting.
01:19
No heat is lost.
01:21
So we're told that our heat of fusion for ice is 334 joules per gram.
01:36
The heat capacity of ice is 2 .108 j.
01:45
Over g degrees c and our heat capacity of water is 4 .184 j over g degrees c.
02:06
So what we need to do first is find heat absorbed to melt the ice and that'll be two steps.
02:33
One, the first step is moving from minus 20 degrees c to zero degrees c.
02:43
And for this we'll use q equals mc delta t.
02:49
And our second step will be to melt, which we will find by q equals our heat of fusion times mass.
03:10
And that's going to be our first step, and that will give us how much heat we have present.
03:16
Okay, so for q for the first step first, q equals the mass of the ice times the specific heat of ice, times by delta t which will be 20 degrees whoops i better go 20 .0 degrees c because we're going from 0 to 20 and that will be 225 times 2 .108 times 20 that's 9486 then for my second step q will equal 225 i don't know why i put degrees c there that should be g times 334 j over g.
04:26
225 times 334.
04:29
It will be 75150.
04:38
So my q total to melt ice is equal to 9486 joules plus 75 ,150 joules.
05:06
9846, 9846, right? 9486...