Recently I asked on skeptics if cold pressed juice is higher in nutrients than one from centrifugal juicer and one of the claims is that the centrifugal juicer heats the juice enough to destroy nutrients. I'd like to know how much a juicer can actually heat the product. I've made my own calculations and I'd know if they're correct and how can they be improved.

The assumptions are:

  • The average juicer takes $500W$ (I looked at a few brands)
  • It takes 5 seconds to grind an apple which has a mass of $150g$ and we end up using $100g$ of it ($50g$ is waste which magically is not heated)
  • 100% of the heat generated is absorbed by the juice (for simplicity)
  • $1W$ of energy generates $1J$ of heat per second (ref)
  • $1cal$ is the amount of energy to increase water temperature by $6^\circ C$ (ref)
  • $1cal$ approximately equals $4.18J$ (ref)
  • We ignore the fact that the juicer's engine gets hot just from running because it seems the engine is usually isolated from the product or the blades.

The calculation:

  • $500W * 5s = 2,500J$
  • $2,500J / 4.17 \frac{K}{K*g} = 0.6K kg $
  • If we uniformly distribute the heat over the $100g$ of product it increases the overall temperature by $6^\circ C$

My question is threefold:

  1. Is the above calculation and conclusion correct?
  2. Am I correct when I claim the assumptions are optimistic and in real life the heat increase would be smaller?
  3. How can this calculation be improved to be more representative of real life?
  • 2
    $\begingroup$ Just terminology: In 5 seconds we use 2500W, no, you're using 500W for 5 seconds. $\endgroup$
    – jinawee
    Jun 26, 2015 at 6:32
  • $\begingroup$ @jinawee I haven't done physics in the last 6 years and back in school I never really understood them well :(. It should be better now but can I somehow improve the second line of calculation to better convey the intent or is it clear enough and should stay as-is? $\endgroup$
    – Maurycy
    Jun 26, 2015 at 6:36
  • $\begingroup$ I suspect the 500W is a peak power figure, which is needed when the blender is trying to digest a particularly tough item. Apples probably aren't the toughest fruits you'd want to squeeze, so you might need only 250W on average (1250J). $\endgroup$
    – MSalters
    Jun 26, 2015 at 14:03
  • $\begingroup$ Also, the 2500/4.17 has units 2500J/4.17 J/K g = 600 K g (Kelvin grams) or more conventionally 0.6 K kg (Kelvin kilograms). $\endgroup$
    – MSalters
    Jun 26, 2015 at 14:06
  • $\begingroup$ @MSalters Thanks. I didn't think about it not using all power and thanks for the proper way to write the units. I updated the question with MathJax, I just hope I didn't make any mistakes. $\endgroup$
    – Maurycy
    Jun 26, 2015 at 17:58

1 Answer 1


You calculation seems valid as a rough approximation, giving the highest reasonable number.

The actual number will be somewhat smaller, because you assumed that all the energy used in converted into heat and all the heat goes into the juice.

Some points to see why the actual number should be smaller (but actual approxiations for these are hard to do!):

  • The blender will not be 100% efficient. Some of the 500W used will be converted into heat inside the mashine. You mentioned the blender getting hot - this heat needs to be subtracted from your total $2.5 kJ$.
  • Grinding up the apple requires some energy, although the actual ammount depends on a lot of factors like whether it was peeled or not, whether you cut it in pieces before blending, etc.
  • I expect the "waste", which is less viscuous than the juice, to take up most of the heat. Since the juice is easy to move around, it will not experience a lot of friciton.

So after some consideration, the juice will heat up by less than $6^\circ C$, but a more precise estimation is very hard to do.


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