A 1 * 10^2W electric heater is used to heat 4 * 10^-1 kg of a liquid, Z, placed in a well insulated calorimeter of mass 2 * 10^-1 kg and a specific heat capacity of 4 * 10^2 J Kg^-1 C^-1. The temperature of Z is plotted against time of heating as shown in the graph below.
The graph shows that the difference in temperature is 66 degrees C. And the whole thing takes 12 minutes to reach this point.
I determined that the specific heat capacity of z is 2527 J Kg^-1C^-1
Im having real trouble figuring out what the boiling point of Z is? I've tried a few different things, but i can't find any notes on it, and i honestly don't really know or understand what im doing.
The next question is "Determine the latent heat of vaporisation of Z?" and i would have thought you'd need to work this out before you find the boiling point? Im not sure. Any help (again) would be greatly appreciated.
As far as I know, the specific heat tells you nothing about the boiling point of a liquid. I presume you're plotting temperature against time. Does the graph show the temperature rising linearly and then leveling off like the lower part of this; User posted image: <a target='_blank' href='http://www.sci-journal.org/reports/vol3no2/saltgrph.gif'>User posted image</a>
Or does it rise with a decreasing exponential like this?: User posted image: <a target='_blank' href='http://www.revisemri.com/tools/timeconst/images/1_exp_x400.gif'>User posted image</a>
If its the former, you can just read off the temperature of the level section because at the boiling point the heat you put in to the system doesn't go into increasing the temperature but actually boiling the liquid, which brings me nicely onto your next question
To find the latent heat of vapourisation you'll need to use this formula: Q = mL, where L is the latent heat you want. The Q will be worked out by the amount of time the graph is flat for times the power of the heater.
Finally, if the graph looks like my second option, we'll go back to scratching our heads!
Or does it rise with a decreasing exponential like this?: User posted image: <a target='_blank' href='http://www.revisemri.com/tools/timeconst/images/1_exp_x400.gif'>User posted image</a>
If its the former, you can just read off the temperature of the level section because at the boiling point the heat you put in to the system doesn't go into increasing the temperature but actually boiling the liquid, which brings me nicely onto your next question
To find the latent heat of vapourisation you'll need to use this formula: Q = mL, where L is the latent heat you want. The Q will be worked out by the amount of time the graph is flat for times the power of the heater.Finally, if the graph looks like my second option, we'll go back to scratching our heads!
Thankyou! That made everything so much clearer. It was indeed the former graph. Your answer really helped my understanding of how those silly graphs work. I was doing all these crazy nonsensical calculations without even really looking much at the graph. 
Again, thanks!
Again, thanks!