00:01
I believe this is all the information we're given.
00:03
We've got a bunch of things we're going to calculate for all of this.
00:07
First, mass of h2o, and that'll equal 93 .3 grams minus 13 grams.
00:21
I'm going to go ahead and i'm not very happy with sig figs on this.
00:29
So this will be 80 .3 grams, and that's assuming 13 grams should be 13 .0 grams.
00:46
You do need to be pretty careful of those.
00:48
Next, the temperature change of water, and that'll equal 60 .8 degrees c minus 23 .5 degrees c.
01:12
That'll be 37 .3 degrees c.
01:18
The q of the water will equal mc delta t.
01:28
So q will equal mass.
01:34
C for water is 4 .184 joules per gram degrees c times our delta t is 37 .3 degrees c.
01:48
80 .3 times 4 .184 times 37 .3 is 12 ,532.
01:59
I'll round later.
02:14
Okay, so but my reportable number here will be 12 ,500 joules.
02:32
What is the heat lost by food in calories? i wondered if they wanted me to do this in calories.
02:50
I'll go ahead and do it like this.
03:12
That'll be 12 ,500 joules times 4 .184 joules per calorie.
03:37
One calorie is 4 .184...