00:01
Okay, so for this question, it says we have 0 .343 grams of the compound and the formula is c10h20o2.
00:16
And this compound is undergoing combustion and we want to mirror the heat of combustion of this compound.
00:25
So we use a calorimeter to mirror the energy that is absorbed or released from the combustion.
00:31
So it tells us the calorimeter change from 24 .5 to 26 .72 celsius.
00:41
And in this calorimeter, there's two things.
00:44
One is water, the mass is 1 .17 times 10 to three grams.
00:49
And another is the bump.
00:53
And for the bump, we know the heat capacity, which is 8322 per celsius.
01:00
So from here, we can calculate two things.
01:02
The first one is to calculate the heat that's absorbed by the water, simply equals to the specific heat of water times the mass of the water times the change of the temperature.
01:17
And i just plug in the numbers, 4 .184 is a specific heat of water and the mass is 1 .17 times 10.
01:28
And then we have the change of temperature from 26 .72 minus 24 .5.
01:35
And we do the calculation.
01:37
So the energy that is absorbed by the water is 10 .868 kilojoule.
01:45
And then we can calculate the heat that is absorbed by the bump, simply equals to the heat capacity of the bump times the change of temperature...