[ARGOS++]

[mr cool]

Ok here we go....
Time | temp
.5.        35.1
1.         57.1
1.5.       63
2.          65.1
2.5.       89
3.         91
3.5.       95
4.         101
4.5.      104
5.         104
5.5.      104
6.         104
6.5.      104
7.         104
7.5.      106
8.         106
8.5.       107
9.          108
9.5.       110
10.        113
10.5.     115
11.       117
11.5.    120     
12.       130
12.5.    135
13.       142
13.5.     145
14.        148

Some stuff from the graph....
Y = 5.7563x + 62.563
R² = 0.8402     

[ARGOS++]

[mr cool]

Thanks anyways. I called some of my chem friends and they had no clue how to do it either so i will probably go into school early tomorrow. 

[mr cool]

Here is how we did it...

Take the heat of vaporization X mass of H20
(540 cal/g)(20.00)g = 18900 cal
Divide that by the time it took to vaporize all the water
10800cal/g 3.6 min = 3000 calories

The ammount if energy doesnt change so 18900/ mass / deltaT
(i forgot where i got 18900 from but it is how much energy was used in the final portion of the graph.)
It came out to around 7 something just for those who were wondering about this yesterday. So i guess my chem teacher > yours :p
 

[ARGOS++]

Dear reezyfbaby;

Thank you for all that information!

So it is basically the method I explained in my former reply (see: #15) to get from the plateau the heat transfer of the Bunsen burner and from the slope the amount.

Even as it is a very rudimentary and very imprecise method, it is worth to be remembered.

Good Luck!
                    ARGOS⁺⁺