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Wize University Physics Textbook (Master) > Work and Energy

Chemical Energy

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Chemical Energy


Chemical energy is energy stored in the bonds of atoms and molecules. It's often commonly referred to as food energy.
  • Food that we eat is turned into chemical energy by our digestive system and can be expended as mechanical work in our muscles, or released as heat.
  • Food energy is often measured in Calories (big "C"), but can be converted into Joules:
 1 kcal=4.18 kJ \boxed{ \ 1 \ kcal = 4.18 \ kJ\ }


Wize Concept
One kcalkcal is the amount of energy required to raise the temperature of 1 kg1 \ kg of water by 1°C1\degree C.

Watch Out!
The calories you see on a food package are actually kilocalories: 10001000 calories, or 1 kcal1 \ kcal or 1 Cal (again, big "C")


Exam Tip
Human Efficiency is ∼25%\sim25\%. That is, about 25 percent of the food energy we ingest can be used for mechanical work.
Basal Metabolic Rate is the power required to run our bodies (most of this energy is consumed by the brain and released as heat): Pmetabolic=100 WP_{metabolic}=100 \ W

Example: Eating a Doughnut


You buy a doughnut, the package says 250250 kcal. Assume that you're really hungry and your body needs all the energy it can get from it (the basal metabolic rate is 100100 W). After you eat it, to what height can you climb in 11 minute? Assume a mass of 7070 kg and an efficiency of 2525 %.

First, let's convert the energy to Joules:

250 kcal=250 kcal×4.18 kJkcal250 \ kcal=250\ kcal\times\dfrac{4.18 \ kJ}{kcal}
=1045 kJ=1045 \ kJ
=1.046×106 J=1.046\times10^6 \ J

The efficiency in using this energy is only 2525% , which means that the total energy available is:

Eavailable=1.046×106×0.25=2.615×105E_{available}=1.046\times10^6\times0.25=2.615\times10^5 (J)


Now part of this energy is used up by the body. In 6060s, this metabolic energy is:

Emetabolic=Pmetabolic⋅tE_{metabolic}=P_{metabolic} \cdot t
=100⋅60=100\cdot60
=6000=6000 (J)


The rest of the energy is used to climb to a maximum height related to the potential energy gained.



mgh=(Eavailable−Emetabolic)×0.25mgh= (E_{available} - E_{metabolic})\times 0.25

h=Eavailable−Emetabolicmgh=\dfrac{ E_{available} - E_{metabolic}}{mg}

=2.615×105−600070×9.81=\dfrac{2.615\times10^5-6000}{70 \times 9.81}

=372=372 (m)

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An average candy bar contains 250 Calories, according do the nutrition facts on the wrapper. If we assume 25% of the food we eat is available to do work (ie. 25% efficiency, then rest going to heat, etc), how fast could a 100kg person theoretically run? Note: a big C calorie on food wrappers is: 1 Cal = 1 kcal = 1000 calories (small 'c'). Also, there is a LOT more involved in this, like 'how long' this could be sustained, etc, but we are simply doing a mathematical exercise 'for fun'.