I have 0.3 kg of corn syrup at room temperature (294 K). The specific heat of corn syrup is [2.72 kJ/(kg K)][1]. I want to heat it up in a 1,000 W microwave until it reaches 394 K. I know that  q=mcΔT  so  q = 0.3 kg * 2.72 kJ/(kg K) * 100 K  q = 81.6 kJ = 82 seconds in a 1,000 W (1 kJ/second) microwave.   From my experience, it takes longer than 82 seconds to heat up corn syrup that much. How should I account for extraneous variables?  From Chris H's Answer: 82 seconds to raise the temperature + 111 seconds to boil off the water = 3 minutes, 13 seconds in a 1,000 Watt microwave.   [1]: https://www.teachengineering.org/content/uoh_/activities/uoh_magic/uoh_magic_lesson01_activity1_worksheet_new_answers.docx 
A major factor to note is that to get the corn syrup to 394K you don't just have to heat it, you have to concentrate it by boiling off some of the water.  You need to take into account the latent heat of vaporisation of water, about 2,260 kJ/kg in the temperature range of interest.  It's easy to see that this can be a major factor. Hard ball candy is 90% sugar (Wikipedia), while corn syrup is more like 76%.  From your 300g you thus need to drive off 49g of water; that will take 111kJ in addition to raising the temperature, more than doubling the energy (and time) required 