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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

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    If this requires boiling, then it's a very bad idea in a microwave: fda.gov/radiation-emitting-products/…
    – FuzzyChef
    Jun 10, 2021 at 0:17
  • Dear all, if you can address why the calculation is not correct, please post it as an answer, not as a comment. It is OK to have incomplete answers, don't worry about that.
    – rumtscho
    Jun 10, 2021 at 8:26

2 Answers 2

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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

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Honestly, I'd go the trial-and-error route with that if you are going to do this for practical applications (e.g. guesstimating how long you need to put your syrup in the microwave)

If you want to actually build an accurate mathematical model, some things you should be taking into account:

1000W is the nominal potency of the microwave, not exactly the amount of energy it gives off from the magnetron. You also have a loss of energy within the oven itself, as not all the energy emitted will reach and be absorbed by your object.

Furthermore, the penetration of the microwaves will account for uneven cooking depending on the surface area of your object and the surface area x volume ratio, so you might have to interrupt your process to mix the liquid to an even temperature of 394K, or have a surface temperature higher than the core temperature.

As warned in the comments, please be VERY careful when heating up liquids in the microwave. If you have glass or ceramic beads that you can put in the liquid to heat up, that's good.

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    Even more than that, if your magnetron puts 1000W into the cavity, that doesn't mean it puts it into the food. Some will be absorbed by dirt, moisture that you've evaporated that then condenses on the walls, etc. Some heat will also be convected and conducted from the dish
    – Chris H
    Jun 10, 2021 at 16:30

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