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This question sort of comes to mind when hearing how efficient an internal combustion engine is turning chemical energy in mechanical energy (something like 20-40%) with lots of excess heat. As an analog, how efficient is (or potentially) the human body at turning food into energy? Please bare with me, I realise there LOTS of different variables (how much the person weighs vs mass, metabolism, diet, etc). But I would imagine that there shouldn't be much margin of error given that most people maintain the same constant temperature (98 F +/- 1 degree).

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The MET (Metabolic Equivalent Task) readout on your gym equipment is your body doing 1Kcal/kg/h = 4184 J/kg/h and can be reasonably accurately measured by how much oxygen a test victim uses.

Sitting still is roughly 1 met and cycling at 100 Watts is around 5.5 Mets.

So taking a man of 75kg, cycling at 100Watts (100J/s) he is having to do 5.5 * 4184 * 75 / 3600s = 480Watts so an efficency of 20%

Remember though that the person is spending 80-100Watts just staying alive doing nothing - unlike your car. There is an interesting experimental fit to how much energy you need to just stay alive, calculated about 100 years ago, the Harris-Benedict equation

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    $\begingroup$ WOW...great answer... today i found out that i need 100 watts just to stay alive!! :) $\endgroup$ – Martin Sep 30 '16 at 13:39
  • $\begingroup$ Yes, but the 100 watts to just stay alive include self repair and maybe just enough energy for replication. Try that with a car. $\endgroup$ – brianlmerritt Jan 18 '17 at 15:12
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The human muscle efficiency (the mechanical work divided by the total metabolic cost) when performing intense exercise is measured to be in the typical range of 18-26%. Manufacturers of fitness equipment use such results and typically show a guesstimate of burned calories based on the actual mechanical work delivered.

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    $\begingroup$ So as an example, if you're doing crunches without weights and lower roughly 100 lbs of your bodyweight 1 meter and back up, then you should be burning approximately 0.5 kCals per crunch right? (i.e., 100lbs * 4.45lbs/newton * 1meter * 0.000239joules/kilocalories /.20% efficiency) $\endgroup$ – speedplane Nov 13 '13 at 14:42
  • $\begingroup$ I meant squats, not crunches. $\endgroup$ – speedplane Nov 13 '13 at 14:59

protected by Qmechanic♦ Jan 30 '13 at 15:49

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