Food as Energy

Basal metabolism

We need a constant input of food energy to keep our internal organs functioning and to allow us to move – this is called our basal metabolism.

For mammals, Mb ≈ 50 (A/m2) W;   for humans, surface area A ≈ 1.7 m2, i.e. ~ 85 W for sleeping

Fast walk or cycle (20 km/h): M ~ 700 W

Overall mean M ~ 100 – 150 W (2000 – 3000 kcal/d)

Where does this energy come from?

Most of the energy required for agriculture comes from sunlight, right? Wrong. Most of the energy comes from burning oil.

The usable energy content of beef and bread is about 2500 kcal/kg or 10 MJ/kg (+/- 30%). For chicken and fish the number is about half that [note] Beef calories and nutrients, [2019-10-04].[/note] [note] Peter Rez, Simple Physics of Energy Use, Oxford UP 2017.[/note]

Overall for a typical Western diet, the ratio of embodied energy of the food to the dietary energy in the food $r$ is about 6, i.e. for every joule (or calorie) of dietary energy, six joules (or calories) of oil have been burned. The ratio for vegetables is generally more favourable; that for meat (especially beef) is much worse. However, the picture is complicated, as can be seen from a British study by David Coley et al.[note] Coley, D., Goodliffe, E. and Macdiarmid, D. (1998) The Embodied Energy of Food: The Role of Diet, Energy Policy, 6, 455–459.[\note]:

  • Baked beans $r \approx$ 10
  • Sausages $r \approx$ 10
  • Yoghurt $r \approx$ 9
  • Beef $r \approx$ 8
  • Pork $r \approx$ 6
  • Eggs $r \approx$ 6
  • Pasta $r \approx$ 4
  • Rice $r \approx$ 3
  • Milk $r \approx$ 3
  • Potatoes $r \approx$ 2
  • Bread $r \approx$ 1

Note however, that these figures are highly variable, depending on circumstances.

What about Animals?

The metabolic rate, $\Gamma$, for animals is related to the mass of the animal by Eqn.1:

$\Gamma = 4 M^{3/4} \tag{1}$

Such a relation where a physical property of an animal scales with mass to some power that is not unity is called an allometric relation[note] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004[/note].  It is obvious that an elephant which is large and consumes vast amounts of food will have a larger metabolic rate than a mouse that is much less massive and consumes much less. Considering, however, the metabolic rate per unit mass of these animals (Eqn.2):

$\dfrac{\Gamma}{M} = 4 M^{-1/4} \tag{2}$

we find to sustain itself an elephant needs less energy per unit mass than a mouse.  This explains why even though elephants eat a larger quantity of food compared to mice, mice actually need to eat more compared to their body weight than elephants.  Our goal is to understand how heat loss contributes to smaller animals having a higher metabolic rate per unit mass than larger animals.

As mentioned some of the energy generated inside an animal goes to heat which is radiated out through the surface area of the animal.  The rate of heat loss is proportional to the surface area, while the mass of an animal is proportional to its volume. The rate at which heat is radiated per unit mass then is proportional to surface area per unit volume.

For simplicity, if we take an animal to be spherical with radius, $r$, then we can relate the heat loss per unit mass to size (Eqn.3):

$\frac{4 \pi r^{2}}{\frac{4}{3} \pi r^{3}} \propto \dfrac{1}{r} \tag{3}$

As animals increase in size the rate of heat loss per unit mass decreases and less energy is needed to sustain the animal per unit mass.  It might appear an obvious advantage then to be a large animal. Large animals, however, still consume more energy on the whole than smaller animals so that they need more food. If food is not abundant larger animals cannot sustain themselves[note] Ahlborn, Zoological Physics, p.12-16 , Springer, 2004[/note].

In conclusion, larger animals have a smaller heat loss per unit mass and require less food per unit mass even though they eat larger amounts than smaller animals.


Bicycling and Calories

How long do I have to ride my bicycle to burn off a doughnut? This example looks at the relationship between physical exercise and calorie consumption.

In magazine articles and on the web, you often find numbers relating exercise and calorie consumption. In this example, we will explore where the numbers come from.

For example, you find that a 68 kg person cycling at 15 km/h for one hour burns approximately 400 Calories[note]Many web-based resources. See for example: or[/note],[note]Knight, Jones, Field, “College Physics: a strategic approach”, 1st edition, p. 339, Pearson Addison-Wesley (2007)[/note] (corresponding to approximately 1.5 – 2.0 donuts[note]Tim Hortons Nutritional Information, [2019-10-16].[/note]). Let’s check.

Our calculation is based on the principle that the energy in food is like other forms of energy, so it can be transformed into mechanical energy. However, like other real-life engines, our body cannot transfer 100% of the chemical energy in food into mechanical energy (motion). We need two additional bits of information: the amount of energy that the body consumes during sedentary periods (typically 100 W; this is mostly needed to keep the internal organs functioning) and the efficiency to convert chemical energy into mechanical energy (approximately 25%). The two numbers given above vary somewhat with the person and the activity but can conveniently be used to get a good estimate.

So why does bicycling consume energy, even on a flat road and a day without wind? You have to overcome rolling friction and air drag, since the relative wind speed is 15 km/h.


  • level road, no wind (relative wind speed = 15 km/h)
  • 68 kg person plus 10 kg bike
  • rolling friction: rubber on concrete: μ r = 0.02 (table 5.1[fn value=2][/fn]).


    • Force due to air drag: FD = ¼ ρ A v 2, with density of air ρ = 1.28 kg/m 3

(The simplified equation above follows from the more general formula for air drag FD = ½ ρ CD A v 2 by using CD = 0.5, which is typical for everyday moving objects.)

We need to estimate the frontal area A of person (+ bike)
A = (1.5 m) · (0.6 m) = 0.9 m 2.
v = 4.2 m/s, so the distance traveled is d = 4.2 m each second.
So F = 5.1 N and the work done is W = F d = 21 J, so P = 21 W

    • Rolling friction: f r = μ r m g = (0.02) · (68 kg+10 kg) · (9.8 m/s2) = 15.3 N,

so the work done by rolling friction is W = (15.3) · (4.2 m) = 64 J and the corresponding power is P = 64 W.

  • So we need a total of 85 W of mechanical power to maintain our speed.
  • Now we want to relate this to the total metabolic power. We need to multiply our result by 4 and add 100 W to account for our efficiency and the energy consumed by our internal organs. Result = 440 W.
  • Finally, we have to perform a unit conversion to relate our result to food calories: 1 Cal = 1 kcal = 4200 J.
  • 440 W = 440 J/s correspond to (440 J/s) · (3600 s/h) = 1.58 MJ/h or 377 Calories burned in one hour.
  • You can replenish your energy by eating almost two doughnuts (62 g each) or six timbits (17 g each).

Our result is close to the published value showing that the calorie consumption in bicycling is mainly due to overcoming air drag and rolling friction.

Finally, how long will someone have to bike to burn off eating a donut?  A chocolate glazed donut from Tim Horton’s has 260 Calories and is 70 g.  If you eat one it will take you (260 Cal)(1 h/377 Cal)(60 min/h) = 41 min of biking  at 15 km/h to burn it off.



Updated (CEW) 2019-10-16