Problem — Fermi estimate: the calorie of an apple

Fermi estimate: the calorie of an apple. To what height would you have to lift your own mass to spend the energy of an apple ($\sim 80$ kcal $\approx 335\;\mathrm{kJ}$), assuming $m = 70\;\mathrm{kg}$? Compare with a well-known building.

Solution

Data: apple energy $E \approx 335\;\mathrm{kJ} = 335\,000\;\mathrm{J}$, mass $m = 70\;\mathrm{kg}$.

Setup. The energy spent lifting oneself by a height $h$ is $E = mgh$. Solving for $h$:

$h = \frac{E}{mg} = \frac{335\,000}{70\cdot 9{,}81} = \frac{335\,000}{686{,}7} \approx 488\;\mathrm{m}$

$\ev{h \approx 488\;\mathrm{m}}$

Comparison. $488\;\mathrm{m}$ corresponds to a very tall skyscraper: the Empire State Building ($\sim 380\;\mathrm{m}$ to the roof, over 100 storeys) or a little more. To burn off a single apple you would have to climb roughly 130 storeys.

Physical reading. The estimate reveals a surprising fact: the chemical energy of food is extremely dense compared with gravitational mechanical energy. A small apple contains as much energy as is needed to hoist 70 kg up to half a kilometre in height. That is why it is far easier to take in calories than to burn them off through physical activity.

Collegamenti

Argomenti: Work and energy Concetti: Gravitational potential energy Competenze: Fermi estimates